The product of four evenly-spaced integers will be close to the 4th power of the number in the center. So simply take the square root twice, for the 4th root; you get a value which is almost 11. That shows that the list must be 8*10*12*14 since that would be centered on 11.
This is the best answer in my book. It's the way an engineer would do it. One quick insight gets you an approximate solution, and then a trivial amount of trial-and-error gets you the exact answer. Much faster than all that algebra.
@@roysmith5902yep, got it on my second guess using a similar approach. Waaaaaaay faster even than this edited algebra answer. Kinda fun seeing the math, though
Good answer, small improvement: As all factors are even, the product 13440 must have a factor 2^4 = 16 in it, so first divide by 16 (gives 840), then same approach as yours --> 4x5x6x7 = 840, finally double each of (4,5,6,7) to get full answer.
"Must be" 8, 10, 12, and 14 makes me cringe a bit, in a mathematically rigorous sense. It's definitely the first thing to try...but I'm curious if there's any large numbers out there whose first set of 4 is the wrong answer, or doesn't have a solution at all via a proof.
13440 is in the order of magnitude of 10^4. The last digit being 0 tells us that 10 is one of the integers. Since 13440 is slightly larger than 10^4, it tells us that 10 is less than the median of the integers in the list. So we infer through this that we should try 8, 10, 12, 14, and that checks out as being correct. Soluble WITHOUT a calculator by this method, and done in about 15 seconds.
How does the last digit being 0 tell you that 10 is one of the integers? If I have 4 consecutive integers whose product is 840, is 10 one of those integers just because 840 ends with a 0? (Answer: It is not.)
@@mykalimba Reread my post carefully. The single fact of a last digit of 0 by itself is insufficient, but when with the fact that 13440 is close to 10^4 (10000) tells us that 10 is one of the four factors.
@@mykalimba Your group is 4, 5, 6, 7 because 625 < 840 < 1296 tells us that 840^(1/4) is between 5 and 6. Just build the list outward from the center. Same idea.
@@andrewlayton9760 What I was objecting to was the unqualified statement "The last digit being 0 tells us that 10 is one of the integers." Even considering the context of surrounding statements, on its own it is patently false, and should be qualified, IMO.
@@mykalimba Yes, a totality of circumstance with several interwoven facts that rapidly narrowed the Venn diagram of possible answers. It was a few sentences, imperfectly describing what my mind did in a few seconds. It has taken far longer to write the responses that it did for my brain to spit out the answer. You are correct - on its own a last digit of zero does not automatically indicate that 10 is a factor, it does however indicate that 2 and 5 are in the prime factor set.
@@xNathan2439xyeah i think its a good method when youre in a pinch, i was thinking of the same thing it works because the numbers are similar enough to one another, so multiplying them will make a number close to 13400, so if you retrace your steps and take the 4th root, you should get a number in the middle of the 4. its kinda similar to the gaussian sun, where he was tasked to add numbers from 1 to 100, and instead he thought of it as a multiplication problem. imagine 4 consecutive even numbers, for example 4 6 8 10. you can get their sum, 28. so similar problem as the original: what are 4 numbers next to each other that add up to 28? we can use division and divide 28 by 4 and get 7, which should be in the middle of the 4 numbers.
It's real @@xNathan2439x . Take 4 equal numbers: they need to be the fourth root of the number. Now take one of them and make it smaller: one of the other ones needs to get larger to compensate. now take the third one, make it smaller: the fourth one needs to get larger. So, yu have two smaller and two larger numbers. Since they are consecutive, then you have 8, 10, 12 and 14.
The way I solved it is by factoring the number i.e. 2^7*3*5*7 Then I distributed some 2s obtaining 6*10*14*16, they were not consecutive, so I took a 2 from 16 and gave it to 6 eventually obtaining 8*10*12*14 😉
Exactly. I did the same. Algebra is great, when necessary. Sometimes there are more direct and simpler methods such as this factoring here of 13440 = 2^7 x 3 x 5 x 7 .
With a variable substitution you could use the quadratic formula at that step, go "u = x^2" and you get a quadratic equation Have to remember to convert back of course, so your 2 "u-solutions" are actually 4 "x-solutions, but it's just gonna be a "+-" on each
I think that's besides the point. The video illustrates a technique that will work for any product of four consecutive even integers, not just small examples. Although, I think it would have been more complete had he used the quadratic formula to factor that biquadratic, rather than just handwaving with "you can factor this with a bit of effort as ...".
I think I have a much more neat solution: You write 4 consecutive even integers as 2n*2(n+1)*2(n+2)*2(n+3) And then you have the equation 2n*2(n+1)*2(n+2)*2(n+3)=13440 You find the prime factors of 13400 which are 2^7*3*5*7 Now you have 2n*2(n+1)*2(n+2)*2(n+3)=2^7*3*5*7 You divide both sides by 16 (2^4) You get n(n+1)(n+2)(n+3)=2^3*3*5*7 or 8*3*5*7 You can rewrite 8*3*5*7 as 4*5*6*7 And you have n(n+1)(n+2)(n+3)=4*5*6*7 Which means n=4 You plug in n=4 in the first expression 2n*2(n+1)*2(n+2)*2(n+3) And you get: 8*10*12*14
I think its easier if you just take the prime factors, realize that 3 of them are odd so you multiply them by two to get 14, 10, and 6, since 6 is two far away you multiply by two again and you get 10, 12, and 14, and the 3 remaining 2's give you 8
i love this one, i was very confused when he did equations i thought prime factors would be the way, but i have something funny for everyone i hope 13440 solutions for FOUR consecutive even integers is entirely guessable. since 10^4=10000, and its a product, so we should simply try the evens around 10
That is approximately the same as I did. Finding an approximate fourth root of a number is easy; it is the square root of the square root. I found this to be approximately 11. Four even numbers centered around 11 will be 8,10,12,14 :-)
Dividing 13440 by 16 (2x2x2x2) turns the problem into finding 4 consecutive numbers that multiply to 840. The lowest number is 5 or less since 6x6x6x6 = 36x36 which is higher than 30x30 = 900 > 840, so the highest number 8 or less. 840 is 700 + 140 so it's divisible by 7, so the numbers are either 4,5,6,7 or 5,6,7,8, so it's multiplying 5x6x7 by either 4 or 8, and 5x6x7 = 30x7 = 210, which is 840/4, so the numbers are 4,5,6,7, and the numbers we are looking for are 8,10,12,14.
13,440^(1/4) = 10.7. That's actually all you need to know that 10 being somewhere in the middle. 4*6*8*10 would be way smaller. 10*12*14*16 way bigger. This actually involves less checking than your solution. It's still using algebra to know that (x-6)(x-4)(x-2)x < x^4 < x(x+2)(x+4)(x+6)
Since they're all even I factored out a 16 immediately, then factored the remainder into 2 3 4 5 7 and noticed the 6. And then you have 4 5 6 7, and just double each to get the even numbers.
Looking for even integers instead of dividing by 16 and looking for just integers feels so inefficient, I don't know why every comment fails to notice the trick, it's blaring
@@ratzou2you can go easier than that. Order is 10000 so you know the answer is in the teens for the for numbers. Last digit is 0 and the sum of the digits is divisible but 3 so divisible by 10 or 20 and 6 or 16 so you have 4 possibilities. 6 8 10 12 can't reach 10k, while 16 18 20 (14 or 22) go too high (20*10*10*10 is 20k) so the answer is 10*12*14*16 by process of elimination.
you can achieve a much faster solution simply decomposing 13440. 13440 is made of 2^7 * 3 * 5 * 7, so any combination of products with these is a possible answer. In my opinion is a much more straight forward and complete answer that the one provided in the video.
even faster if you divide out the factor of 16 because it's consecutive even numbers. between that and 840^(1/4) = 5 and change it gives you a start. so you have 5 and 7 as primes in that decomposition and then you need to deal with the rmaining 2^3 * 3 , well the number between 5 and 7 is 6 so that's one 2 and one 3. The remainder is 4. So you have 4*5*6*7 and that indeed makes 840. Times by 2, gives 8*10*12*14 which is the answer.
As a high school student in the 70's, it's amazing what I still remember about algebra and geometry. What's even more amazing is how much I've forgotten! As most of us older folks know now, 99% of problem solving comes from framing the question properly - these exercises are great practice for how to approach a difficult problem.
We know 10 must be one of the factors since the end result is divisible by 10 and the only other way to make it as such would be to have 5 and 2. Divide the number by 10, then see how many ups and downs from 10 we need to have to get to there
You also know that it is roughly around 10 because it is on the order of 10000. Divisible by 6 as well so 10 12 14 16 is the best option in the first place to check without doing math. (Can't be 6 8 10 12 because it wouldn't quite reach 10000 at that point.)
A REALLY GOOD strategy is that x(x+1)(x+2)(x+3) + 1 is ALWAYS a square. So dividing 13440 by 16 will give 840 and 841 = 29² And the term that is squared is (x²+3x+1)
how the fuck does one think of this, genius stuff i'm great at math and can think a lot about it but this is something beyond anything my mind could conjure
@@seankeegan8285 op knew that the product of 4 consecutive numbers plus 1 is a square and used it in this case. So, he or she just used knowledge acquired. So this is a case where some normal act is also genius. I guess.
A bit fast to see immediately, but brilliant indeed. It's nice to have a method to not only solve this problem, but all similar ones as well. First dividing all the even numbers by two and 13440 by 16 (840) is already much simpler. Without pen and paper I can't work out the 121*111 thing. Didn't see at first why it has to be a square, but I do see that 840 is 29^2-1 = (29-1)(29+1) = 28*30 or 4*7*5*6. Or check √29 is between 5 and 6, so that are the middle two numbers. When I work out the function I do get it. Rearrange to x(x+3) * (x+1)(x+2) and get (x^2+3x)(x^2+3x+2) or (u-1)(u+1) or u^2-1 with u = x^2+3x+1 With the above for 840 I get u=29 or x^2+3x-28=0, so (x+7)(x-4) and x=-7 or the correct answer x=4. the consecutive even numbers then are 8*10*12*14.
@@seankeegan8285 I didn't do math competitions, but I did do programming competitions in high school and college. In our training, we would focus on classifying problems, and then learning what algorithms applied to what classes to answer different questions. The questions would then usually require twisting one of the algorithms slightly to get some variation of the normal answer. Anyway, the important part is that a lot of this was specialized training specifically for these competitions. Like, yes theoretically it's useful for normal programming work. But practically, I've only used that knowledge a few times in my career. So don't feel bad not thinking of something if you haven't even trained for it! We wouldn't expect to be able to run a marathon without training either.
11^4 = 14,641 which is easy to remember from Pascal's triangle. Also from the expansion of (x+1)^4. 10^4 = 10,000. Knowing these two values makes the solution easy to guess.
Factoring 13431 without a calculator is harder than the original problem. I took the 4th root for 13440 to found it equals ~10.7. So All numbers have to be close to 10. Trial and error gives 8,10,12,14.
Factoring 13431 is actually not so hard as it seems, as you can see it is a palindromic number and those are very often divisible by 11, thus it would be the first factor I'd try, and WOW it gives me 1221 which is visible as again divisible by 11 from space ;)
@@z000ey Note that 1001 is divisible by 7, 11, and 13. So you can check all three factors at once using an alternating sum of groups of three digits. 13,431 mod 1001 becomes 418, which is divisible by 11. The original number, 13,440 mode 1001 is 427 which is divisible by 7, greatly aiding those using the factoring approach.
@@wroscel truly I'd never think about going it that way, using the characteristic of 1001 and looking for a mod with it... I only saw the palndromatic numbers and that I associate with 11 and it helped me (in this particular case). Twas more trial and error without an error getting it right on the first :D
I did this question by writing 2n(2n+2)(2n+4)(2n+6)=13400 Which implies n(n+1)(n+2)(n+3)=840 which means n+3 choose 4 is 840/4!=35 and then using Pascal's triangle to solve that
Step 1: Divide by 2 four times (you could divide by 16, but dividing by 2 four times is doable via mental arithmetic): 6720 3360 1680 840 Now, Find 4 consecutive integers that multiply to get 840. Then multiply all 4 of them by 2 (reversing step 1 and giving us 4 consecutive even integers). 840 is small enough that we can use brute force: Step 1: 4! = 24 Step 2: 24*(5/1) = 120 (5!/1!) Step 3: 120*(6/2) = 360 (6!/2!) Step 4: 360*(7/3) = 840 (7!/3!) 7!/3! is 4*5*6*7 Multiply each multiplicand by 2 (thus multiplying our total by 2 four times, reversing step 1): 8*10*12*14.
Quick 'n dirty way. x^4 =13440. Then raise 13440 to the power of 1/4 = 10.787... So the consecutive numbers have to be around 10. So just try a couple of arrangements.
A fast approach, (x-3)(x-1)(x+1)(x+3), it is going to be very close to x^4, jo just take the forth root of 13440, i.e,(13440)^(1/4) somewhere around 10.7, so the x is close to 11, try three cases, x=11, x=9 if product overshoots, and x = 13 if product undershoots.
Factor the number: 13440=2*2*2*2*2*2*2*3*5*7=3*5*7*2^7 The consecutive even numbers have to be divisible by either 3, 5 or 7 we can multiply all by 2 to produce 6, 10 and 14. We have 4 twos left and from the sequence 6-10-14 both the 8 and the 12 are missing, so we use 3 twos to produce 8 and we multiply the 6 by the remaining 2 resulting in 8-10-12-14.
My approach was to dismatle it to prime factors. Some 2 will repeat in every number. Exactly 4 of them, but it is not so bad, divde by 2^4. 840 as product of 4 integers is much more friendly.
Significantly easier to let a=x^2 and solve the quadratic a^2 -10a - 13431 =0 Also very simple to list the prime factors and see that there are 4 '2's and the others produce 4, 5, 6 and 7. Leading to 8, 10, 12 and 14
divide 13440 by 16 = 840. Now the problem is the product of 4 consecutive integers to make 840. You can estimate the 4th root by estimating the square-root twice, it would be around 5 or 6. Can just try 4*5*6*7 which gives 840. So ans = 8 * 10 * 12 * 14
I just prime factorized 13440 = 2^7*3*5*7 pulled out 2*4 (there are 4 even numbers) and basically saw that the remaining factors can be grouped to 4*5*6*7 -> 8*10*12*14. But the solution in the video is also a nice trick to reduce a quartic equation to a quadratic one :)
With regards to factoring (1:50), 13431 is a 5-digit palindrome in which the third digit is the sum of the first two. This will always be a multiple of 111.
I did it by dividing by 13440 by 2^4 = 840. You're now instead looking for four consecutive INTEGERS. The prime factors of 840 is 2^3 * 3 * 5 * 7. We know that two of the integers are consecutive ODD numbers (the other two consecutive even numbers), meaning it's either 3, 5 or 5, 7; the remaining odd prime must be multiplied by 2 to make it even, and 7*2 would be too far apart from the rest, so it is 5 and 7. The 3 is multiplied by 2 = 6 and we are left with 2^2 = 4. So the consecutive integers are 4, 5, 6, 7. Multiply these by 2 to get the answer to the original question, i.e. 8, 10, 12, 14
I used a 'ball park' method of quick elimination. I found the square root of 13440, which rounds up to 116. I then intuited that the smallest and largest of the four integers I was looking for, multiplied together, would almost equal the two middle terms multiplied together. the square root of 116 approximates to 11, which gives me a value between those two middle terms, which probably therefore equal 10 and 12. Given that assumption, then the other terms must be 8 and 14. I plugged them in and was proven correct. It's not pure maths, but these types of guess and try methods are often far quicker. It's the solution of a woodworker who thinks geometrically in relative proportions but is no longer sure of all his school algebra!
If we reformulate the problem as find the 4 consecutive numbers (not even) that have product 840 (since 840*2^4=13440 and 2 divides the 4 even consecutive number once each), then we can factor it as 840=2*2*2*3*5*7 but since one of the numbers has to be divisible by 4 (because there are 4 consecutive numbers) then we can rewrite it as 840=2*3*4*5*7. Notice that in the sequence there is a missing 6 which can be filled if we multiply 2*3 resulting in 840=4*5*6*7. Hence 13440=8*10*12*14.
Simple ballpark it to estimate first. 10x10x10 = 1000. 14 is closest even integer to multiply by 1000 to get 13440. Spread the first 3 numbers to fit the brief. 8x10x12. Check the result. Easy.
I needed a hint through the first step. The difference of squares helped me out. But instead of factoring the resulting binomial (since I wanted to do this by hand), I did x^4 - 10x^2 = 13431 x^2 (x^2 - 10) = 13431 Sacrilege, but hear me out. My thought process was, maybe I can factor out 13431. I knew it would be divisible by 11 so I divided it by 11, resulting in 1221. I divvied it further by 11 giving me 111 and leaving me with (x^2) (x^2 - 10) = (11)(11)(111) I saw that 11 is multiplied twice and x is multiplied twice and 11 squared minus 10 is 111 which connects the two together... ...my sacrilegious move got me the right answer regardless.
Lots of good estimation (fourth root) or factoring solutions, but I didn't notice anyone point out that algebraically, you can make it far easier at the second stage by iterating the original trick and using y=x^2 -5 ; then the two x^2 terms you got are (y-4)(y+4) and you have y^2 - 16 = 13,440. Thus y is 116, x^2 is 121 and x is 11 all without actually solving a general quadratic.
Since 13440 is divisible by ten. 13440 ÷ 10 = 1344 Since the numbers are even and consecutive, we can divide the result by 12 1344÷12=112 And again 112÷14=8 Therefore, the consecutive numbers are equal to: 8, 10, 12, 14
It is really quick to guess and check, but you can also use rounding to find the answer quick. In my head, I thought 13440≈14000=10*10*10*14, and 12*8≈100=10*10. Swapping 8 for 16 would double the product, and swapping 14 for 6 would cut it down by more than half, so the only answer that puts us in the ballpark must be 8,10,12,14.
at 1:45 instead of finding two numbers that add to -10 and multiply to -13431, you could use the quadratic formula to get x^2=121 and -111. can't take the square root of a negative number so x=sqrt(121)=11, which you then plug into the initial equation to get the answers
Since all are even, we can divide by 16 to simplify further… We can hit and trial as 840 easy to create factorials to get to the answer… Algebra is just brute force when we cannot solve mentally… Hint - we have 13440=16*840. that clearly means we have a 5 or 10 in this group of 4 numbers… It took me less 20 secs to get to the answer…
I used prime factorization. So I simply divided by 2 until it was no longer possible so I ended with 2^7 * 105. 105 has only 3 prime factors anymore 3,5 and 7. So as we know, it is searched for 4 consecutive even numbers this means, we search for 4 consecutive numbers multiplied by 2. 2^4 is in 2^7 so it remains with (2^4) * 2*2*2*3*5*7. Sharp looking gets you 2*2, 5, 2*3, 7 -> 4,5,6,7. So the searched consecutive even numbers are the doubles: 8,10,12,14.
I did a prime factoisation of 13440. You Get 2^7*3*5*7. I now want 4 terms that are 2 apart, and are even. I can multiply all of the odd numbers by 2 to make them even, being sure to reduce the exponent of 2 so it still equals 13440, getting 6*10*14*16. My numbers are all even, but not consecutive even numbers. I can only break apart the 16, as dividing the other numbers by 2 will make them odd. Dividing 16 by 2 makes 8, so now we have 2(6*8*10*14). The only solution that works is multiplying the 2 by the 6 to get 8*10*12*14, as if you multiply the two by the other numbers, it makes them far to large.
I did this slightly differently - I'm not sure if it worked out better or worse. x(x+2)(x+4)(x+6)=13440, but instead of fully expanding, I only expanded to (x^2+6x)(x^2 + 6x + 8)=13440. Substituting z=x^2+6x that's z^2 + 8z - 13440 = 0. z=-120 or 112, giving either x^2 + 6x - 112 = 0 (x = 8 or -14) or x^2 + 6x + 120 = 0 (x has complex roots). The only positive integer root is x=8, so 8 * 10 * 12 * 14 = 13,440, matching your result.
I just factorized the number that can be expresed into 2^7 * 3 * 5 * 7, as it as to be 4 even numbers I moved a 2 into each of the odd ones turning it into 2 ^4 * 6 * 10 * 14, and then moving another 2 to the 6 so its 12 while 2^3 becomes 8.
I basically just brute forced it. First, I did the four lowest even numbers, 2, 4, 6, and 8. Then I jumped ahead to 20, 22, 24, and 26 to find a maximum that still held the same number of digits, and realized I severely overshot it, and jumped down to 10, 12, 14, and 16, seeing that was also too far. So I went back to the beginning and jumped one number ahead to 4, 6, 8, and 10, and kept going along there until I got the number with 8, 10, 12, and 14.
I did it differently, because we have a product of consecutive even integers giving 13440, then we have a product of consecutive integers giving 13440/2^4 = 840 = 2^3 × 3 × 5 × 7. Now since in any set of 4 consecutive integers we always have one even integer not multiple of 4 and one multiple of 4, we want to keep multiples of 8 outside our solution, and same for multiples of 9, now 10×11×12×13 > 10000 > 840 so we're below 8, and 4×5×6×7 is 840 so our solution is 8, 10, 12, 14
You can also note that x(x + 2)(x + 4)(x + 6) + 16 is (x^2 + 6 x + 4)^2 and shooting for the square root of 13456 = 116 is fairly easy. That tells you x^2 + 6 x = 112 whence x = 8.
There is a better way, 10 is definately one of the numbers because 20 cant be one of the numbers (20*18*16*14 is larger). Now, 14 is definately one of the terms, because that number is divisible by 14. So we have 10, 12, 14. Now we either have 8 or 16. For this, divide the number by 10, and then by 12 and then by 14, to get 8.
Solved in 15 seconds. 4th root of 13440 for the general ballpark, then I got it on the first guess. Sometimes the best solution isn't the most rigorous one.
Good example of one you can do faster in your head with common sense than on paper. 10^4 is 10000. 11^4 is 121^2 which is greater than 120^2 which is 14400. So the 4th root of the 13440 is between 10 and 11. So the numbers ought to be 8, 10, 12, 14. Multiplying them out to verify was the longest step.
13440 is close to 10⁴, and you need a multiple of 10, so that's where I begin. Divide by 10 to get 1344. Next I try 12: 1344/12 = 112. Then 8*14 = 80 + 32 = 112. So the four even integers are 8, 10, 12, and 14. I'm feeling a bit conflicted about algebraic methods. On the one hand, I've been telling students that math beyond algebra -- as well as math in physics, economics, engineering, etc. -- is mostly a few definitions, a few theorems, and lots of algebra. On the other hand, algebraic methods might over-complicate things. (Also, it sometimes doesn't occur to students to substitute a candidate answer into the equation they are solving.) 1:50 Once a comment mentioned "difference of squares", I noticed that 112 = 121 - 9. I might have used that at one point if I decided to use an algebraic method, and your method is quite nice. I guess one should avoid the quadratic formula because that only postpones the difficulty in factoring, unless one uses a calculator. The pattern 13431 suggests -- something, although it takes a while to recognize that it suggests a factor of 111.
I kinda cheated and rationalized the solution almost immediately when I realized 10 x 10 x 10 x 10 = 10000 which is reasonably close to 13440 all things considered, so I figured the digits were all around 10, maybe very slightly higher because 13440 > 10000. I tested the correct 4 on my very first try.
This is the exact same method I used, and it took me about 10-20 seconds to realize that it had to be somewhere around 10, and my first guess was 8, 10, 12, 14. I am sure using algebraic methods works for more complicated problems like this, but estimating is a great way to solve problems.
If the fourth root of 13440 is x, the 4 integers can not be higher than x, since their product will exceed 13440, nor can they be lower than x, since their product will be under 13440. We only need to know the fourth root approximately. Note that (10)⁴ = (10)²(10)² = (100)(100) = 10000, which is less than 13400. We compute (11)⁴ = (11)²(11)² = (121)(121) = 14641, which is higher than 13440. So, the fourth root must lie between 10 and 11. The even factors must include the closest even number less than the fourth root as well as the closest even number over the fourth root. These numbers are 10 and 12, which have a product of 120. We have narrowed down our possibilities for the 4 consecutive even numbers to {6, 8, 10, 12}, {8, 10, 12, 14} and {10, 12, 14, 16}. We try (6)(8)(10)(12) = (48)(120) = 5760, too small. Then, (8)(10)(12)(14) = (8)(14)(10)(12) = (112)(120) = 13440, the desired result. So, the four consecutive even numbers are 8, 10, 12, 14. Note that if calculators are not allowed, we have no trouble doing all the multiplications by hand.
13440 is divisible by 7. The factor must be divisible by 2, so one factor must be divisible by 14. 14*16*18*20 is too big. It must be smaller than that Let's try 8*10*12*14 -- Hey! that worked. If that wasn't it I only had to guess 10*12*14*16 or 12*14*16*18.
The median of x - 3, x - 1, x + 1, x + 3 is x. So the 4th root of (x - 3)(x - 1)(x + 1)(x + 3) will be close to x. The 4th root of 13440 is 10.767... , so x = 11 and the numbers are 8,10,12,14.
It took about 15 seconds. You take the 4th root of 13,440, its ~10.77 and change. The answer must be the two even integers below that and the two even integers above. i.e. 8/10/12/14.
Or the 4th root of 13,440 is 10.767… so the required numerals are going to be about 10, biased to slightly above, so 8, 10, 12, 14 which is technically speaking a guess but also correct in one. 😀
I figured one had to be 10. If one of the integers is 10, the other three must multiply to 1344. It's divisible by 3, so 12 has to be in there, so I guessed 12*14*16. That got me 26880, exactly double the original. Drop the 16 down to 8, you get 8*10*12*14.
because 2 and 5 are not consecutive(neither 8 and 4 are consecutive with 5, so you need to count 10 as the integer automatically as the number has zero as its first digit)
Factor the number and see it contains a seven. From the order of magnitude we see that the numbers are near 10. The lowest even number multiple of 7 using the factors is 14. Test 4 cases containing 14 and find the solution.
Square root of 13,440 is about 115 so two of the numbers should multiply to something close to that. From there I got 8, 10, 12, and 14 in about 3 seconds. Not exactly rigorous but I got it.
Once you get to the x^4 equation you can do a variable change where u = x^2, and solve the quadratic equation that will give you 121 as a root. Then you change variable back and you get ±11, rather than trial and error to find that 11 is a root.
My solution was to prime factor 13440 whichis 2^7*3*5*7 then take the form of the solution which would be n, n+2,n+4,n+6 and found of the solutions there should be two congruent to 0 mod 4 and two congruent to 2 mod 4 so i had that six of the seven 2's are used cause whatever is congruent to 0 has at least two 2's and and everything that is congruent to 2 has at most one 2 (because anymore will make it congruent to 0) but with the remaining primes other than 2 can only be congruent to 1 or 3 so if they arent paired up with a two they dont fit the scheme so they must have one 2 exactly logically placing the seventh two in the group of 2's congtuent so zero so we have 2*2*2, 2*x, 2*2*y, 2*z for each number we know 2*2*y is 4 more or 4 less than 2*2*2 there is no with the numbers we have to get 4 less so we can safely say y=3 and so we can just mash numbers in the last two places so that one of them is in between 2*2*2 and 2*2*3. after mashing numbers you find 8,10,12,14
I had an easier approach: take fourth root of 13440, gives 10 point something. You know that 10 must be one of the integers, because otherwise it can never end in a 0. So 10 is the second number. So the others are 8 12 14.
It might be more interesting to express the RHS constant as a variable and use the quartic formula to find a general solution? These videos are fun but they're often constant-oriened and 'trick'-oriented.
If you're scared of big numbers, you can express this as "four consecutive integers have a product of 840" by dividing each integer by 2 and the right side by 16
I stumbled across an easier way. Call the four positive integers a,b,c,d. Turns out that (b*c)-(a*d) is ALWAYS = 8. Couldn't believe it but it's easy to prove. Plug that in and do two quadratics and you're done.
I ballparked it as 10^4 = 10000, so they'd be *around* 10. 134 is close to 132, or 11^2, so that's *around* 11. So while it can't be 11*11*11*11, what 4 +ve ints would be close to that? Need the 0 at the end, so 10's definitely a factor. (else 2 and 5 as last digits, but those are odd). First guess was 8*10*12*14, which worked out. How exciting.
You have two solutions and you actually got it when you had -11 as a solution. If you subtract 3 you get -14 for X which is even. -14*-12*-10*-8 =13440
Unfortunately I guessed the answer first go- no learning achieved. I did note the need to end in 40 which for some reason seemed to guided me to 10 and 14, and I filled in the gaps as seemed plausible. I might try with again with a different sequence using excel to generate the sequence (cryptically) and display the answer required.
i solved this via prime factorization. 4 consective even numbers are all divisable by 2, 2/4 are divisible by 4 and 1/4 must be divisable by 8. ergo, 13440/128=105 whose remaining prime factors are the other prime factors of the 4 numbers. 3*5*7=105. this is 3 numbers so one of our numbers has to be a power of 2. as only one of them must be divisiable by 8 and there are no 2s left, one of the numbers is 8. the first even number divisable by 7 is 14 so 14 is one of them as well. there are 2 even numbers between 8 and 14 which are 10 and 12 and both contain one of the remaing prime factors and one is divisable by 4. note this is not the best way to solve it at all and i did use a calculator to do 13440/128 and 105/5. and i suck at spelling divisable.
They way I did it was a bit different. If they're all consecutive even integers then you can just take the 4th root of 13440. This is 11. Then you know 11 is the mid-point of the 4 integers.
13440 < 16000 = 16*10*10*10 < 16*14*12*10 therefore the integers are less than 16. similarly, 13440 > 8000 = 8*10*10*10 > 6*8*10*12 therefore integers are more than 6. therefore, integers are 8,10,12,14
I noticed it ends it 0, so it must be divisible by 10 (can't be 20, that's way too high). So the new problem is finding the 3 numbers whose product is 1344. I noticed it's divisible by 4, so I did that, and it was still divisible by 4, so I did it again and got 84. So I ended up with 12 x 7 x 4 x 4. Then I noticed that since I have a 10 and a 12, I can rearrange 7 x 4 x 4 to 14 (7 x 2) and 8 (4 x 2). Not as reliable a method, but was fast and lazy :D
I think you should have solved the x^4 - 10*x^2 - 13431 = 0 using the quadratic equation solution instead. That way is significantly easier than saying let's figure out two numbers that add to -10 and multiply to 13431
10 power 4 is 10000 and 11 power 4 is 11 square into 11 square is 121 into 121 is 14xxx something so numbers are around 11 Plus numbers include 10 coz of last 0 and include a multiple of 4 coz of last two digits being a multiple of 4 Hence 8 10 12 14
i solved it this way. i noticed that it ended in zero meaning that the solutions MUST contain a ten somwhere so i started with ten starting and worked my way back to find the solution of 8,10,12,14
i've got a tough one. 2934 consecutive nonnegative even integers have a product of 0; what are they? i'm sure nobody can solve this one /j i'll pay -$100 to whoever gets it
The product of four evenly-spaced integers will be close to the 4th power of the number in the center. So simply take the square root twice, for the 4th root; you get a value which is almost 11. That shows that the list must be 8*10*12*14 since that would be centered on 11.
This is the best answer in my book. It's the way an engineer would do it. One quick insight gets you an approximate solution, and then a trivial amount of trial-and-error gets you the exact answer. Much faster than all that algebra.
@@roysmith5902yep, got it on my second guess using a similar approach. Waaaaaaay faster even than this edited algebra answer. Kinda fun seeing the math, though
Good answer, small improvement: As all factors are even, the product 13440 must have a factor 2^4 = 16 in it, so first divide by 16 (gives 840), then same approach as yours --> 4x5x6x7 = 840, finally double each of (4,5,6,7) to get full answer.
thats amazing
"Must be" 8, 10, 12, and 14 makes me cringe a bit, in a mathematically rigorous sense. It's definitely the first thing to try...but I'm curious if there's any large numbers out there whose first set of 4 is the wrong answer, or doesn't have a solution at all via a proof.
13440 is in the order of magnitude of 10^4. The last digit being 0 tells us that 10 is one of the integers. Since 13440 is slightly larger than 10^4, it tells us that 10 is less than the median of the integers in the list. So we infer through this that we should try 8, 10, 12, 14, and that checks out as being correct. Soluble WITHOUT a calculator by this method, and done in about 15 seconds.
How does the last digit being 0 tell you that 10 is one of the integers? If I have 4 consecutive integers whose product is 840, is 10 one of those integers just because 840 ends with a 0? (Answer: It is not.)
@@mykalimba Reread my post carefully. The single fact of a last digit of 0 by itself is insufficient, but when with the fact that 13440 is close to 10^4 (10000) tells us that 10 is one of the four factors.
@@mykalimba Your group is 4, 5, 6, 7 because 625 < 840 < 1296 tells us that 840^(1/4) is between 5 and 6. Just build the list outward from the center. Same idea.
@@andrewlayton9760 What I was objecting to was the unqualified statement "The last digit being 0 tells us that 10 is one of the integers." Even considering the context of surrounding statements, on its own it is patently false, and should be qualified, IMO.
@@mykalimba Yes, a totality of circumstance with several interwoven facts that rapidly narrowed the Venn diagram of possible answers. It was a few sentences, imperfectly describing what my mind did in a few seconds. It has taken far longer to write the responses that it did for my brain to spit out the answer. You are correct - on its own a last digit of zero does not automatically indicate that 10 is a factor, it does however indicate that 2 and 5 are in the prime factor set.
take the fourth root of 13440, get 10.7. Now this number is between the seecond and third integer. Therefore, it's 8*10*12*14
Is this real or a joke? And why if it's real.
@@xNathan2439xyeah i think its a good method when youre in a pinch, i was thinking of the same thing
it works because the numbers are similar enough to one another, so multiplying them will make a number close to 13400, so if you retrace your steps and take the 4th root, you should get a number in the middle of the 4.
its kinda similar to the gaussian sun, where he was tasked to add numbers from 1 to 100, and instead he thought of it as a multiplication problem.
imagine 4 consecutive even numbers, for example 4 6 8 10. you can get their sum, 28. so similar problem as the original: what are 4 numbers next to each other that add up to 28? we can use division and divide 28 by 4 and get 7, which should be in the middle of the 4 numbers.
It's real @@xNathan2439x . Take 4 equal numbers: they need to be the fourth root of the number. Now take one of them and make it smaller: one of the other ones needs to get larger to compensate. now take the third one, make it smaller: the fourth one needs to get larger. So, yu have two smaller and two larger numbers. Since they are consecutive, then you have 8, 10, 12 and 14.
You’re not supposed to use a calculator
Brilliant
The way I solved it is by factoring the number i.e. 2^7*3*5*7
Then I distributed some 2s obtaining
6*10*14*16, they were not consecutive, so I took a 2 from 16 and gave it to 6 eventually obtaining
8*10*12*14 😉
thats how i did it aswell, its so much easier and quicker this way
I did it like that too. It usually makes these kind of exercises easy to answer.
Same there !
I sorta did that, but I divided both sides by 16 first and got 4,5,6,7 and then doubled them.
Exactly. I did the same. Algebra is great, when necessary. Sometimes there are more direct and simpler methods such as this factoring here of 13440 = 2^7 x 3 x 5 x 7 .
finding -121 and +111 is wayy harder than just brute forcing the original problem 😭
With a variable substitution you could use the quadratic formula at that step, go "u = x^2" and you get a quadratic equation
Have to remember to convert back of course, so your 2 "u-solutions" are actually 4 "x-solutions, but it's just gonna be a "+-" on each
I think that's besides the point. The video illustrates a technique that will work for any product of four consecutive even integers, not just small examples. Although, I think it would have been more complete had he used the quadratic formula to factor that biquadratic, rather than just handwaving with "you can factor this with a bit of effort as ...".
I think I have a much more neat solution:
You write 4 consecutive even integers as 2n*2(n+1)*2(n+2)*2(n+3)
And then you have the equation 2n*2(n+1)*2(n+2)*2(n+3)=13440
You find the prime factors of 13400 which are 2^7*3*5*7
Now you have 2n*2(n+1)*2(n+2)*2(n+3)=2^7*3*5*7
You divide both sides by 16 (2^4)
You get n(n+1)(n+2)(n+3)=2^3*3*5*7 or 8*3*5*7
You can rewrite 8*3*5*7 as 4*5*6*7
And you have n(n+1)(n+2)(n+3)=4*5*6*7
Which means n=4
You plug in n=4 in the first expression 2n*2(n+1)*2(n+2)*2(n+3)
And you get:
8*10*12*14
I think its easier if you just take the prime factors, realize that 3 of them are odd so you multiply them by two to get 14, 10, and 6, since 6 is two far away you multiply by two again and you get 10, 12, and 14, and the 3 remaining 2's give you 8
i love this one, i was very confused when he did equations i thought prime factors would be the way, but i have something funny for everyone i hope
13440 solutions for FOUR consecutive even integers is entirely guessable. since 10^4=10000, and its a product, so we should simply try the evens around 10
That is approximately the same as I did. Finding an approximate fourth root of a number is easy; it is the square root of the square root. I found this to be approximately 11. Four even numbers centered around 11 will be 8,10,12,14 :-)
Dividing 13440 by 16 (2x2x2x2) turns the problem into finding 4 consecutive numbers that multiply to 840. The lowest number is 5 or less since 6x6x6x6 = 36x36 which is higher than 30x30 = 900 > 840, so the highest number 8 or less. 840 is 700 + 140 so it's divisible by 7, so the numbers are either 4,5,6,7 or 5,6,7,8, so it's multiplying 5x6x7 by either 4 or 8, and 5x6x7 = 30x7 = 210, which is 840/4, so the numbers are 4,5,6,7, and the numbers we are looking for are 8,10,12,14.
Very enjoyable. I thought right away of doing x-4, x-2, x, and x+2, but yours is way better. Love difference of squares!
13,440^(1/4) = 10.7. That's actually all you need to know that 10 being somewhere in the middle. 4*6*8*10 would be way smaller. 10*12*14*16 way bigger. This actually involves less checking than your solution. It's still using algebra to know that (x-6)(x-4)(x-2)x < x^4 < x(x+2)(x+4)(x+6)
Since they're all even I factored out a 16 immediately, then factored the remainder into 2 3 4 5 7 and noticed the 6. And then you have 4 5 6 7, and just double each to get the even numbers.
Looking for even integers instead of dividing by 16 and looking for just integers feels so inefficient, I don't know why every comment fails to notice the trick, it's blaring
@@ratzou2you can go easier than that. Order is 10000 so you know the answer is in the teens for the for numbers. Last digit is 0 and the sum of the digits is divisible but 3 so divisible by 10 or 20 and 6 or 16 so you have 4 possibilities. 6 8 10 12 can't reach 10k, while 16 18 20 (14 or 22) go too high (20*10*10*10 is 20k) so the answer is 10*12*14*16 by process of elimination.
This is the way
you can achieve a much faster solution simply decomposing 13440. 13440 is made of 2^7 * 3 * 5 * 7, so any combination of products with these is a possible answer.
In my opinion is a much more straight forward and complete answer that the one provided in the video.
even faster if you divide out the factor of 16 because it's consecutive even numbers.
between that and 840^(1/4) = 5 and change it gives you a start.
so you have 5 and 7 as primes in that decomposition and then you need to deal with the rmaining 2^3 * 3 , well the number between 5 and 7 is 6 so that's one 2 and one 3. The remainder is 4. So you have 4*5*6*7 and that indeed makes 840. Times by 2, gives 8*10*12*14 which is the answer.
At that point you may as well just use intuition that the answers must be near 10 (10^4 = 10000) and guess. Should get it on the second try.
As a high school student in the 70's, it's amazing what I still remember about algebra and geometry. What's even more amazing is how much I've forgotten! As most of us older folks know now, 99% of problem solving comes from framing the question properly - these exercises are great practice for how to approach a difficult problem.
We know 10 must be one of the factors since the end result is divisible by 10 and the only other way to make it as such would be to have 5 and 2. Divide the number by 10, then see how many ups and downs from 10 we need to have to get to there
You also know that it is roughly around 10 because it is on the order of 10000. Divisible by 6 as well so 10 12 14 16 is the best option in the first place to check without doing math. (Can't be 6 8 10 12 because it wouldn't quite reach 10000 at that point.)
A quick eyeball says this number is a little more than 10^4. My first guess would be 8x10x12x14. And it checks.
A REALLY GOOD strategy is that x(x+1)(x+2)(x+3) + 1 is ALWAYS a square. So dividing 13440 by 16 will give 840 and 841 = 29²
And the term that is squared is (x²+3x+1)
That's genius.
how the fuck does one think of this, genius stuff
i'm great at math and can think a lot about it but this is something beyond anything my mind could conjure
@@seankeegan8285 op knew that the product of 4 consecutive numbers plus 1 is a square and used it in this case. So, he or she just used knowledge acquired. So this is a case where some normal act is also genius. I guess.
A bit fast to see immediately, but brilliant indeed. It's nice to have a method to not only solve this problem, but all similar ones as well.
First dividing all the even numbers by two and 13440 by 16 (840) is already much simpler. Without pen and paper I can't work out the 121*111 thing.
Didn't see at first why it has to be a square, but I do see that 840 is 29^2-1 = (29-1)(29+1) = 28*30 or 4*7*5*6. Or check √29 is between 5 and 6, so that are the middle two numbers.
When I work out the function I do get it. Rearrange to
x(x+3) * (x+1)(x+2) and get
(x^2+3x)(x^2+3x+2) or
(u-1)(u+1) or u^2-1 with u = x^2+3x+1
With the above for 840 I get u=29 or x^2+3x-28=0, so (x+7)(x-4) and x=-7 or the correct answer x=4. the consecutive even numbers then are 8*10*12*14.
@@seankeegan8285 I didn't do math competitions, but I did do programming competitions in high school and college. In our training, we would focus on classifying problems, and then learning what algorithms applied to what classes to answer different questions. The questions would then usually require twisting one of the algorithms slightly to get some variation of the normal answer.
Anyway, the important part is that a lot of this was specialized training specifically for these competitions. Like, yes theoretically it's useful for normal programming work. But practically, I've only used that knowledge a few times in my career. So don't feel bad not thinking of something if you haven't even trained for it! We wouldn't expect to be able to run a marathon without training either.
11^4 = 14,641 which is easy to remember from Pascal's triangle. Also from the expansion of (x+1)^4.
10^4 = 10,000.
Knowing these two values makes the solution easy to guess.
Lovin’ this! You are on a mission, thanks. Great communications, Andy.
Factoring 13431 without a calculator is harder than the original problem. I took the 4th root for 13440 to found it equals ~10.7. So All numbers have to be close to 10. Trial and error gives 8,10,12,14.
love this approach
Factoring 13431 is actually not so hard as it seems, as you can see it is a palindromic number and those are very often divisible by 11, thus it would be the first factor I'd try, and WOW it gives me 1221 which is visible as again divisible by 11 from space ;)
@@z000ey Note that 1001 is divisible by 7, 11, and 13. So you can check all three factors at once using an alternating sum of groups of three digits. 13,431 mod 1001 becomes 418, which is divisible by 11. The original number, 13,440 mode 1001 is 427 which is divisible by 7, greatly aiding those using the factoring approach.
@@wroscel truly I'd never think about going it that way, using the characteristic of 1001 and looking for a mod with it... I only saw the palndromatic numbers and that I associate with 11 and it helped me (in this particular case). Twas more trial and error without an error getting it right on the first :D
I did this question by writing 2n(2n+2)(2n+4)(2n+6)=13400
Which implies n(n+1)(n+2)(n+3)=840 which means n+3 choose 4 is 840/4!=35 and then using Pascal's triangle to solve that
Very nice!
Step 1: Divide by 2 four times (you could divide by 16, but dividing by 2 four times is doable via mental arithmetic):
6720
3360
1680
840
Now, Find 4 consecutive integers that multiply to get 840. Then multiply all 4 of them by 2 (reversing step 1 and giving us 4 consecutive even integers).
840 is small enough that we can use brute force:
Step 1: 4! = 24
Step 2: 24*(5/1) = 120 (5!/1!)
Step 3: 120*(6/2) = 360 (6!/2!)
Step 4: 360*(7/3) = 840 (7!/3!)
7!/3! is 4*5*6*7
Multiply each multiplicand by 2 (thus multiplying our total by 2 four times, reversing step 1):
8*10*12*14.
Quick 'n dirty way. x^4 =13440. Then raise 13440 to the power of 1/4 = 10.787... So the consecutive numbers have to be around 10. So just try a couple of arrangements.
A fast approach, (x-3)(x-1)(x+1)(x+3), it is going to be very close to x^4, jo just take the forth root of 13440, i.e,(13440)^(1/4) somewhere around 10.7, so the x is close to 11, try three cases, x=11, x=9 if product overshoots, and x = 13 if product undershoots.
wow, this is neat!
Factor the number:
13440=2*2*2*2*2*2*2*3*5*7=3*5*7*2^7
The consecutive even numbers have to be divisible by either 3, 5 or 7 we can multiply all by 2 to produce 6, 10 and 14. We have 4 twos left and from the sequence 6-10-14 both the 8 and the 12 are missing, so we use 3 twos to produce 8 and we multiply the 6 by the remaining 2 resulting in 8-10-12-14.
My approach was to dismatle it to prime factors. Some 2 will repeat in every number. Exactly 4 of them, but it is not so bad, divde by 2^4. 840 as product of 4 integers is much more friendly.
Significantly easier to let a=x^2 and solve the quadratic a^2 -10a - 13431 =0
Also very simple to list the prime factors and see that there are 4 '2's and the others produce 4, 5, 6 and 7. Leading to 8, 10, 12 and 14
I would never have thought of this in a million years.
divide 13440 by 16 = 840. Now the problem is the product of 4 consecutive integers to make 840. You can estimate the 4th root by estimating the square-root twice, it would be around 5 or 6. Can just try 4*5*6*7 which gives 840. So ans = 8 * 10 * 12 * 14
I just prime factorized 13440 = 2^7*3*5*7 pulled out 2*4 (there are 4 even numbers) and basically saw that the remaining factors can be grouped to 4*5*6*7 -> 8*10*12*14. But the solution in the video is also a nice trick to reduce a quartic equation to a quadratic one :)
With regards to factoring (1:50), 13431 is a 5-digit palindrome in which the third digit is the sum of the first two. This will always be a multiple of 111.
I did it by dividing by 13440 by 2^4 = 840. You're now instead looking for four consecutive INTEGERS.
The prime factors of 840 is 2^3 * 3 * 5 * 7.
We know that two of the integers are consecutive ODD numbers (the other two consecutive even numbers), meaning it's either 3, 5 or 5, 7; the remaining odd prime must be multiplied by 2 to make it even, and 7*2 would be too far apart from the rest, so it is 5 and 7. The 3 is multiplied by 2 = 6 and we are left with 2^2 = 4. So the consecutive integers are 4, 5, 6, 7.
Multiply these by 2 to get the answer to the original question, i.e. 8, 10, 12, 14
I used a 'ball park' method of quick elimination. I found the square root of 13440, which rounds up to 116. I then intuited that the smallest and largest of the four integers I was looking for, multiplied together, would almost equal the two middle terms multiplied together. the square root of 116 approximates to 11, which gives me a value between those two middle terms, which probably therefore equal 10 and 12. Given that assumption, then the other terms must be 8 and 14. I plugged them in and was proven correct. It's not pure maths, but these types of guess and try methods are often far quicker. It's the solution of a woodworker who thinks geometrically in relative proportions but is no longer sure of all his school algebra!
If we reformulate the problem as find the 4 consecutive numbers (not even) that have product 840 (since 840*2^4=13440 and 2 divides the 4 even consecutive number once each), then we can factor it as 840=2*2*2*3*5*7 but since one of the numbers has to be divisible by 4 (because there are 4 consecutive numbers) then we can rewrite it as 840=2*3*4*5*7. Notice that in the sequence there is a missing 6 which can be filled if we multiply 2*3 resulting in 840=4*5*6*7. Hence 13440=8*10*12*14.
Simple ballpark it to estimate first. 10x10x10 = 1000. 14 is closest even integer to multiply by 1000 to get 13440. Spread the first 3 numbers to fit the brief. 8x10x12. Check the result. Easy.
Using its divisibility test: 40 divides 13440 (= 13400 + 40) because 40 divides 40 and 13400 is an even number of hundreds.
So 13440 = 40 * 336 = 40 * 6 * 56 and we're done:
13440 = (5 * 6 * 7 * 8) * 2 * 2 * 2
= 8 * 10 * 12 * 14.
Problems like this: just factor using divisibility rules.
Alternately: every product of 4 consecutive even integers is divisible by 64, so observe that
13440 = 64 * 21
= 64 * 3 * 7 * 10
= 16 * 6 * 10 * 14
= 8 * 12 * 10 * 14
and we're done after reordering.
I needed a hint through the first step. The difference of squares helped me out.
But instead of factoring the resulting binomial (since I wanted to do this by hand), I did
x^4 - 10x^2 = 13431
x^2 (x^2 - 10) = 13431
Sacrilege, but hear me out.
My thought process was, maybe I can factor out 13431. I knew it would be divisible by 11 so I divided it by 11, resulting in 1221.
I divvied it further by 11 giving me 111 and leaving me with
(x^2) (x^2 - 10) = (11)(11)(111)
I saw that 11 is multiplied twice and x is multiplied twice and 11 squared minus 10 is 111 which connects the two together...
...my sacrilegious move got me the right answer regardless.
Exactly how I did it too :) (but I did see using +-1 and +-3 and the difference of squares).
Lots of good estimation (fourth root) or factoring solutions, but I didn't notice anyone point out that algebraically, you can make it far easier at the second stage by iterating the original trick and using y=x^2 -5 ; then the two x^2 terms you got are (y-4)(y+4) and you have y^2 - 16 = 13,440. Thus y is 116, x^2 is 121 and x is 11 all without actually solving a general quadratic.
Since 13440 is divisible by ten.
13440 ÷ 10 = 1344
Since the numbers are even and consecutive, we can divide the result by 12
1344÷12=112
And again
112÷14=8
Therefore, the consecutive numbers are equal to:
8, 10, 12, 14
It is really quick to guess and check, but you can also use rounding to find the answer quick.
In my head, I thought 13440≈14000=10*10*10*14, and 12*8≈100=10*10.
Swapping 8 for 16 would double the product, and swapping 14 for 6 would cut it down by more than half, so the only answer that puts us in the ballpark must be 8,10,12,14.
at 1:45 instead of finding two numbers that add to -10 and multiply to -13431, you could use the quadratic formula to get x^2=121 and -111. can't take the square root of a negative number so x=sqrt(121)=11, which you then plug into the initial equation to get the answers
Since all are even, we can divide by 16 to simplify further… We can hit and trial as 840 easy to create factorials to get to the answer… Algebra is just brute force when we cannot solve mentally… Hint - we have 13440=16*840. that clearly means we have a 5 or 10 in this group of 4 numbers… It took me less 20 secs to get to the answer…
I used prime factorization. So I simply divided by 2 until it was no longer possible so I ended with 2^7 * 105. 105 has only 3 prime factors anymore 3,5 and 7. So as we know, it is searched for 4 consecutive even numbers this means, we search for 4 consecutive numbers multiplied by 2. 2^4 is in 2^7 so it remains with (2^4) * 2*2*2*3*5*7. Sharp looking gets you 2*2, 5, 2*3, 7 -> 4,5,6,7. So the searched consecutive even numbers are the doubles: 8,10,12,14.
I did a prime factoisation of 13440. You Get 2^7*3*5*7. I now want 4 terms that are 2 apart, and are even. I can multiply all of the odd numbers by 2 to make them even, being sure to reduce the exponent of 2 so it still equals 13440, getting 6*10*14*16. My numbers are all even, but not consecutive even numbers. I can only break apart the 16, as dividing the other numbers by 2 will make them odd. Dividing 16 by 2 makes 8, so now we have 2(6*8*10*14). The only solution that works is multiplying the 2 by the 6 to get 8*10*12*14, as if you multiply the two by the other numbers, it makes them far to large.
I did this slightly differently - I'm not sure if it worked out better or worse.
x(x+2)(x+4)(x+6)=13440, but instead of fully expanding, I only expanded to (x^2+6x)(x^2 + 6x + 8)=13440. Substituting z=x^2+6x that's z^2 + 8z - 13440 = 0. z=-120 or 112, giving either x^2 + 6x - 112 = 0 (x = 8 or -14) or x^2 + 6x + 120 = 0 (x has complex roots).
The only positive integer root is x=8, so 8 * 10 * 12 * 14 = 13,440, matching your result.
I just factorized the number that can be expresed into 2^7 * 3 * 5 * 7, as it as to be 4 even numbers I moved a 2 into each of the odd ones turning it into 2 ^4 * 6 * 10 * 14, and then moving another 2 to the 6 so its 12 while 2^3 becomes 8.
I basically just brute forced it. First, I did the four lowest even numbers, 2, 4, 6, and 8. Then I jumped ahead to 20, 22, 24, and 26 to find a maximum that still held the same number of digits, and realized I severely overshot it, and jumped down to 10, 12, 14, and 16, seeing that was also too far. So I went back to the beginning and jumped one number ahead to 4, 6, 8, and 10, and kept going along there until I got the number with 8, 10, 12, and 14.
I just got lucky and guessed 8, 10, 12, 14. For the tests that ask these questions, guessing and coming up with the answer can sometimes be faster.
I did it differently, because we have a product of consecutive even integers giving 13440, then we have a product of consecutive integers giving 13440/2^4 = 840 = 2^3 × 3 × 5 × 7. Now since in any set of 4 consecutive integers we always have one even integer not multiple of 4 and one multiple of 4, we want to keep multiples of 8 outside our solution, and same for multiples of 9, now 10×11×12×13 > 10000 > 840 so we're below 8, and 4×5×6×7 is 840 so our solution is 8, 10, 12, 14
An easier way to do it is to let y=x^2 in x^4-10x^2-13431 = 0, giving a quadratic in y. Then apply the quadratic formula rather than factoring.
You can also note that x(x + 2)(x + 4)(x + 6) + 16 is (x^2 + 6 x + 4)^2 and shooting for the square root of 13456 = 116 is fairly easy. That tells you x^2 + 6 x = 112 whence x = 8.
It is 2a*2b*2c*2d so 16*a*b*c*d take two times the squareroot so you get 2*(>5) so b=5 so it is 8 10 12 14.
Much quicker to multiply consecutive even integers by trial-and-error than realize that 111 and 121 satisfy the factorization!
There is a better way, 10 is definately one of the numbers because 20 cant be one of the numbers (20*18*16*14 is larger). Now, 14 is definately one of the terms, because that number is divisible by 14. So we have 10, 12, 14. Now we either have 8 or 16. For this, divide the number by 10, and then by 12 and then by 14, to get 8.
Solved in 15 seconds. 4th root of 13440 for the general ballpark, then I got it on the first guess. Sometimes the best solution isn't the most rigorous one.
Good example of one you can do faster in your head with common sense than on paper. 10^4 is 10000. 11^4 is 121^2 which is greater than 120^2 which is 14400. So the 4th root of the 13440 is between 10 and 11. So the numbers ought to be 8, 10, 12, 14. Multiplying them out to verify was the longest step.
13440 is close to 10⁴, and you need a multiple of 10, so that's where I begin. Divide by 10 to get 1344. Next I try 12: 1344/12 = 112. Then 8*14 = 80 + 32 = 112. So the four even integers are 8, 10, 12, and 14.
I'm feeling a bit conflicted about algebraic methods. On the one hand, I've been telling students that math beyond algebra -- as well as math in physics, economics, engineering, etc. -- is mostly a few definitions, a few theorems, and lots of algebra. On the other hand, algebraic methods might over-complicate things. (Also, it sometimes doesn't occur to students to substitute a candidate answer into the equation they are solving.)
1:50 Once a comment mentioned "difference of squares", I noticed that 112 = 121 - 9. I might have used that at one point if I decided to use an algebraic method, and your method is quite nice. I guess one should avoid the quadratic formula because that only postpones the difficulty in factoring, unless one uses a calculator. The pattern 13431 suggests -- something, although it takes a while to recognize that it suggests a factor of 111.
I kinda cheated and rationalized the solution almost immediately when I realized 10 x 10 x 10 x 10 = 10000 which is reasonably close to 13440 all things considered, so I figured the digits were all around 10, maybe very slightly higher because 13440 > 10000. I tested the correct 4 on my very first try.
This is the exact same method I used, and it took me about 10-20 seconds to realize that it had to be somewhere around 10, and my first guess was 8, 10, 12, 14. I am sure using algebraic methods works for more complicated problems like this, but estimating is a great way to solve problems.
If the fourth root of 13440 is x, the 4 integers can not be higher than x, since their product will exceed 13440, nor can they be lower than x, since their product will be under 13440. We only need to know the fourth root approximately. Note that (10)⁴ = (10)²(10)² = (100)(100) = 10000, which is less than 13400. We compute (11)⁴ = (11)²(11)² = (121)(121) = 14641, which is higher than 13440. So, the fourth root must lie between 10 and 11. The even factors must include the closest even number less than the fourth root as well as the closest even number over the fourth root. These numbers are 10 and 12, which have a product of 120. We have narrowed down our possibilities for the 4 consecutive even numbers to {6, 8, 10, 12}, {8, 10, 12, 14} and {10, 12, 14, 16}. We try (6)(8)(10)(12) = (48)(120) = 5760, too small. Then, (8)(10)(12)(14) = (8)(14)(10)(12) = (112)(120) = 13440, the desired result. So, the four consecutive even numbers are 8, 10, 12, 14. Note that if calculators are not allowed, we have no trouble doing all the multiplications by hand.
Could you use pythagorean's formula at 2:05 to solve for x^2 instead of guessing and checking?
13440 is divisible by 7. The factor must be divisible by 2, so one factor must be divisible by 14.
14*16*18*20 is too big. It must be smaller than that
Let's try 8*10*12*14 -- Hey! that worked.
If that wasn't it I only had to guess 10*12*14*16 or 12*14*16*18.
Even with the x=-11, we get -8, -10, -12, & -14. The product will give 13440 as answer and satisfy the solution with all negative integers
The median of x - 3, x - 1, x + 1, x + 3 is x. So the 4th root of (x - 3)(x - 1)(x + 1)(x + 3) will be close to x. The 4th root of 13440 is 10.767... , so x = 11 and the numbers are 8,10,12,14.
It took about 15 seconds. You take the 4th root of 13,440, its ~10.77 and change. The answer must be the two even integers below that and the two even integers above. i.e. 8/10/12/14.
Or the 4th root of 13,440 is 10.767… so the required numerals are going to be about 10, biased to slightly above, so 8, 10, 12, 14 which is technically speaking a guess but also correct in one. 😀
I figured one had to be 10. If one of the integers is 10, the other three must multiply to 1344. It's divisible by 3, so 12 has to be in there, so I guessed 12*14*16. That got me 26880, exactly double the original. Drop the 16 down to 8, you get 8*10*12*14.
smart!
Why did you thought one of the factors had to be 10?
because 2 and 5 are not consecutive(neither 8 and 4 are consecutive with 5, so you need to count 10 as the integer automatically as the number has zero as its first digit)
oh it says positive even integers, no other outcomes than 10 anyway
@@Enesprays what about 20, 30, etc?
Factor the number and see it contains a seven.
From the order of magnitude we see that the numbers are near 10.
The lowest even number multiple of 7 using the factors is 14.
Test 4 cases containing 14 and find the solution.
Square root of 13,440 is about 115 so two of the numbers should multiply to something close to that. From there I got 8, 10, 12, and 14 in about 3 seconds. Not exactly rigorous but I got it.
Once you get to the x^4 equation you can do a variable change where u = x^2, and solve the quadratic equation that will give you 121 as a root. Then you change variable back and you get ±11, rather than trial and error to find that 11 is a root.
That's a good idea. I just got lazy and entered the quartic equation in Wolfram.
My solution was to prime factor 13440 whichis 2^7*3*5*7 then take the form of the solution which would be n, n+2,n+4,n+6 and found of the solutions there should be two congruent to 0 mod 4 and two congruent to 2 mod 4 so i had that six of the seven 2's are used cause whatever is congruent to 0 has at least two 2's and and everything that is congruent to 2 has at most one 2 (because anymore will make it congruent to 0) but with the remaining primes other than 2 can only be congruent to 1 or 3 so if they arent paired up with a two they dont fit the scheme so they must have one 2 exactly logically placing the seventh two in the group of 2's congtuent so zero so we have 2*2*2, 2*x, 2*2*y, 2*z for each number we know 2*2*y is 4 more or 4 less than 2*2*2 there is no with the numbers we have to get 4 less so we can safely say y=3 and so we can just mash numbers in the last two places so that one of them is in between 2*2*2 and 2*2*3. after mashing numbers you find 8,10,12,14
Since it's 0 at units place we need 10 or 20 compulsorily and
just try 10 12 14 16 > 13440
Next try 8×10×12×14 = 13440 😅😅
Tip: If you multiply 10*12*14 first, you can save that result for later just in case you over-estimate the answer.
For all those suggesting the quadratic formula, doesn't anyone complete the square anymore? x^4-10x^2+25=13456
I had an easier approach: take fourth root of 13440, gives 10 point something. You know that 10 must be one of the integers, because otherwise it can never end in a 0. So 10 is the second number. So the others are 8 12 14.
It's got 5 digits, is reasonably close to 10,000 (10^4) and is a multiple of 10, so 8*10*12*14 just seems like a really obvious guess!
bro said "work harder, not smarter"
In all seriousness this was very cool, I thought of this but didn't get very far.
It might be more interesting to express the RHS constant as a variable and use the quartic formula to find a general solution? These videos are fun but they're often constant-oriened and 'trick'-oriented.
If you're scared of big numbers, you can express this as "four consecutive integers have a product of 840" by dividing each integer by 2 and the right side by 16
I love how if you did trial and error it would have been like 3 more tries until you got it but great channel
4th root of 13440 gives you the average value of the 4 numbers, aprox 10, i tried with close numbers, 8x10x12x14 = 13440
How exciting
I stumbled across an easier way. Call the four positive integers a,b,c,d. Turns out that (b*c)-(a*d) is ALWAYS = 8. Couldn't believe it but it's easy to prove. Plug that in and do two quadratics and you're done.
I ballparked it as 10^4 = 10000, so they'd be *around* 10.
134 is close to 132, or 11^2, so that's *around* 11. So while it can't be 11*11*11*11, what 4 +ve ints would be close to that? Need the 0 at the end, so 10's definitely a factor. (else 2 and 5 as last digits, but those are odd).
First guess was 8*10*12*14, which worked out. How exciting.
You have two solutions and you actually got it when you had -11 as a solution. If you subtract 3 you get -14 for X which is even. -14*-12*-10*-8 =13440
Unfortunately I guessed the answer first go- no learning achieved. I did note the need to end in 40 which for some reason seemed to guided me to 10 and 14, and I filled in the gaps as seemed plausible. I might try with again with a different sequence using excel to generate the sequence (cryptically) and display the answer required.
i solved this via prime factorization. 4 consective even numbers are all divisable by 2, 2/4 are divisible by 4 and 1/4 must be divisable by 8. ergo, 13440/128=105 whose remaining prime factors are the other prime factors of the 4 numbers. 3*5*7=105. this is 3 numbers so one of our numbers has to be a power of 2. as only one of them must be divisiable by 8 and there are no 2s left, one of the numbers is 8. the first even number divisable by 7 is 14 so 14 is one of them as well. there are 2 even numbers between 8 and 14 which are 10 and 12 and both contain one of the remaing prime factors and one is divisable by 4.
note this is not the best way to solve it at all and i did use a calculator to do 13440/128 and 105/5. and i suck at spelling divisable.
I kinda understand the process, but I would never have thought of doing it.
Then you didn't understand. Which is weird. Because it is pretty straightforward.
@@samueldeandrade8535 maybe to you
Instead of factorising.... No matter how many powers of equn how long is the equation, diffratiate it.. ull get the answer easyly
They way I did it was a bit different. If they're all consecutive even integers then you can just take the 4th root of 13440. This is 11. Then you know 11 is the mid-point of the 4 integers.
13440 < 16000 = 16*10*10*10 < 16*14*12*10 therefore the integers are less than 16. similarly, 13440 > 8000 = 8*10*10*10 > 6*8*10*12 therefore integers are more than 6. therefore, integers are 8,10,12,14
I noticed it ends it 0, so it must be divisible by 10 (can't be 20, that's way too high). So the new problem is finding the 3 numbers whose product is 1344. I noticed it's divisible by 4, so I did that, and it was still divisible by 4, so I did it again and got 84. So I ended up with 12 x 7 x 4 x 4. Then I noticed that since I have a 10 and a 12, I can rearrange 7 x 4 x 4 to 14 (7 x 2) and 8 (4 x 2). Not as reliable a method, but was fast and lazy :D
13440 is a multiple of 10, the only even solution. divide 1344 by 12, you get 112, and the rest is simple.
I think you should have solved the x^4 - 10*x^2 - 13431 = 0 using the quadratic equation solution instead. That way is significantly easier than saying let's figure out two numbers that add to -10 and multiply to 13431
10 power 4 is 10000 and 11 power 4 is 11 square into 11 square is 121 into 121 is 14xxx something so numbers are around 11
Plus numbers include 10 coz of last 0 and include a multiple of 4 coz of last two digits being a multiple of 4
Hence 8 10 12 14
The "include a multiple of 4" is funny. Because you simply can't avoid a multiple of 4. Hahahahaha.
i solved it this way. i noticed that it ended in zero meaning that the solutions MUST contain a ten somwhere so i started with ten starting and worked my way back to find the solution of 8,10,12,14
10^4 = 10000, so slightly larger on average. And divisible by 10. So 8, 10, 12, 14.
13440. sqrt, sqrt, = 10.76 (assume this is the midpoint. 8 x 10 x 12 x 14? yes. done.
4th root of 13440 is 10.767, so 8,10,12,14 is almost certainly the answer
My friend challenge me on this question, he said that the professor can’t solve this problem, it’s:
Integral x/x^5+1 dx
i've got a tough one. 2934 consecutive nonnegative even integers have a product of 0; what are they? i'm sure nobody can solve this one /j
i'll pay -$100 to whoever gets it
Doesn’t exist - I’ll send you my PayPal
This particular example isn't too hard to get by guessing, 10*10*10*10 is 10,000 so for 13,000 the 4 numbers are going to have to straddle 10.
Couldn't we let Y=X^2 and solve via quadratic?
Who's else here because of his voice?