I did another method: I decomposed the numbers in terms of prime factors: 165 = 5x3x11 120 = 5x3x2x2x2 64 = 2x2x2x2x2x2 88 = 2x2x2x11 note that 11 (and no other numbers) is common in 165 and 88, so let's take a = 11 remaining from 165, we find 5x3 (15) which can also be found in 120. so let's take b = 15 after that, we have 2x2x2 (8) cropping up 4 times, which implies that they will occupy c and d adding them together, 11 + 15 + 8 + 8, we get 42
I did that, but I started by noting that ad x bc = ab x cd, and the factor of 11 meant that 165 and 88 had to be part of different pairings. The factor of 5 helped in a similar manner.
First acknowledge that each 4 numbers are represent twice in the given multiplications. I did it about the same way, only more developed as the first step, in 4 columns I found all the pairs making each number while multiplied. Let say for the sake of beginning somewhere, we start comparison with 64, we have 1-64, 2-32, 4-16 and 8-8. It can't be the 1-64 pair because 64 isn't found anywhere else. It ain't 2-32 either because 32 isn't foud anywhere else, same with 16. So, the 8-8 pair must be two of the numbers we don't have to find. :-) Same job with 1-88, 4-22 and 2-44 from 88 leaving the 8-11 pair.... Last pairs being 11-15 and 8-15 by the same kind of process. The four relatives being 8, 8, 11 and 15, adding to 42. I like this method more because it is easy to check back and because I am good at finding dividers.
Interesting fact i just realized is that 437 can be expressed as the difference of 2 perfect squares: 437=441-4 437=21²-2² 437=(21-2)(21+2) 437=19×23 :D
This is true for all combinations of number which are evenly spaced. Because being equidistant from the median, they can be expressed as product of (median - d) and (median + d)
@@iMíccoli evenly / equally spaced from the median. 437 = 19 x 23 = (21 - 2) x (21 + 2) Here 21 is the median and the numbers 19 and 23 being equidistant from their median (21) are also evenly spaced. I used 'evenly' because the combinations will always have a difference which would always be an even number otherwise we wont be able to get to the median which is also a natural number. Here 19 and 23 are evenly spaced and their difference '4'. Another example, 40, 10 are two evenly spaced numbers whose product can also be expressed as a difference of 2 perfect squares. Median of 40 and 10 = 25. 25 being equidistant from 40 and 10 by 15 on either side. Thus 40 x 10 = 25² - 15² = 625 - 225 = 400
I thought the problem is nice but your solution is even nicer. Thanks PS: I solved it by finding factors in each number and matching those which share some factors. 8 x 11, 11 x 15, 15 x 8 and 8 x 8 The answer is 42 but method is foolproof.
I followed the same method. 2 numbers are clearly multiples of 11 (88 and 165). And the remaining two are multiple of 8 (64 and 120). So I found 8 twice, 11 and 15. Their sum = 42
I live in the UK and run puzzle evenings in a bar. Basically it’s a quiz where you need very little general knowledge. Most of the questions are mathematical but involve a lot of logic, usually integers or what I call integer fractions. I do have to throw in a few word based problems, just to keep certain customers happy. Your videos give me great inspiration. This is a great but I will need to adapt it. Thanks
Love your elegant solution. As noted in other comments, one can also solve for a,b,c,d first: 64 = 2^6 88 = 2^3 x 11 120 = 2^3 x 15 165 = 11 x 15 W.l.o.g let ab=64 ({ab, bc, cd, da} and (a+b+c+d) unchanged by a cyclic permutation of (a,b,c,d)) => a=2^(6-k) and b=2^k some k∈{1,...,5} Note cannot have k=0 or 6 since that would mean one of the other pairwise products would have share a factor of 2^6 which they do not. Since b is a power of 2, bc must involve a power of 2. So there are 2 cases: 1) bc = 88 = 2^3 x 11 = 2^k x c => c = 2^(3-k) x 11 => cd must have a factor of 11 => cd = 165 = 11 x 15 = 2^(3-k) x 11 x d => k=3, a=b=8, c=11 and d=15 => da = 15 x 8 = 120 => (a,b,c,d) = (8,8,11,15) => a+b+c+d = 42 2) bc = 120 = 2^3 x 15 = 2^k x c => c = 2^(3-k) x 15 => cd must have a factor of 15 => cd = 165 = 11 x 15 = 2^(3-k) x 15 x d => k=3, a=b=8. c=15 and d=11 => da = 11 x 8 = 88 => (a,b,c,d) = (8,8,15,11) => a+b+c+d = 42
I actually found the solution to split the four numbers (64,88,120,165) into their primfactors easier here, bacause the only four possible factors (as combination of the forementioned primefactors) became immediately obvious.
I have been watching you for about half a year. And I am proud to say you taught me well. Before I began watching you, I was a high school math failure. Now, I finally managed to solve one of your questions on my own before watching your video!
Like many others here, I solved this puzzle by looking at the prime factorization of the dual products. The principal idea was to understand that when for example ab = 64, then a and b must both be 8, and cd must equal 165=11*15. I want to share my views on the inter-dependency of the number of solutions and the order of things given in the question. When being asked only for the sum a+b+c+d, then clearly the order of (a,b,c,d) does not matter, whereas the order of assignments of the 4 numbers {64,88,120,165} to the products {ab,bc,cd,da} could theroretically result in more than one sum a+b+c+d for the answer, so we must carefully consider it. It turns out, though, that this is not the case here, and we're left with only one sum (42) for the answer. This is probably due to the cyclic symmetry in the selection of the 4 two-products. On the other hand, if we ask for all the tuples (a,b,c,d) that fulfill the requirements of the puzzle, we arrive at 8 different solutions in total! Interestingly, this means that only one third of all 24 possible permutations of (8,8,11,15) qualify as an answer.
By inspection, you can note that 165 has to be the product of two odd numbers. From our multiplication table, even*even = even. Even*odd = even, odd*odd = odd. Lasty, 64 is a power of 2. Thus any two numbers that multiple to it must be a power of two. So we know two of the numbers are powers of two. This makes it easy to eyeball the solution based on the prime factorizations of the other numbers.
This was a cool number puzzle to solve! This is how I solved this. I started writing the given numbers 64, 88, 120 and 165 as product of their prime factors: 64 = 2*2*2*2*2*2 88 = 2*2*2*11 120 = 2*2*2*3*5 165 = 3*5*11 We have four positive integers a, b, c and d so that ab, ad, bc and cd gives us those four integers above. Each integer a, b, c and d appear twice in the pairwise products. We can instantly see that the prime 11 appears twice (as factor of 88 and 165). Since 3*5 and 2*2*2 don't have common factors, we know one of the numbers a, b, c and d must be 11. Let's set: 165 = 3*5*11 = ab => a = 3*5 = 15 and b = 11. After this it is clear that: 88 = 2*2*2*11 = cb => c = 2*2*2 = 8. Finally we solve for d: 120 = 2*2*2*3*5 = ad => d = 2*2*2 = 8. We also see that these values check up for the final number 64: 64 = (2*2*2)*(2*2*2) = 8*8 = cd. Solution: a+b+c+d = 15+11+8+8 = 42. Clearly my way of solving this is different from the video, but my brain functions this way, so this is how I solved this.
This is a technique I've seen used a lot on this channel. Convert the problem into the form: (...)(...)=integer, then test each possible pair of multiplicands for validity.
Simplify by picking it apart: 2 products have 11 (prime) as a factor, so 11 is one of the letters, while 8 and 15 are the two numbers that the first pairs with. 15 doesn't divide 64, so it must be one of the pair making 120 and it must be another of the letters. 120/15 is 8, so 8 and 8 are the remaining 2 letters. 11 + 15 + 8 + 8 = 42. Many of the comments are similar to this. Prime's solution seems like it would work better for less obvious variations of this problem.
Ur solution is really excellent. But I solved in following way: multiply all so we get. (abcd)²=64*88*120*165; abcd=15*11*8*8; So following statement can be concluded: Possible pair as (ab,cd) or (bc,ad)=(64,165)or(88,120); we can also observe that ab/bc=(cd/ad)-¹; Case 1 Let ab=64 & cd=165; bc=88 & ad=120; 64/88=(165/120)-¹=8/11=a/c ; let a=8k,c=11k; so b=64/8k=8/k similarly d=15/k as b=8/k and d=15/k So maximum value of k will be HCF(8,15) that is equal to 1. So only value of k is 1 a=8,b=8,c=11,d=15 Case 2 Let ab=64 & cd=165; bc=120 & ad=88; Similarly a=8k,b=8/k,c=15k,d=11/k again only possible value of k=1 a=8,b=8,c=15,d=11. Case3 let bc=64 & ad=165; cd=120 & ab=88; Substitute in case 2 a=b b=c c=d d=a Case 2 become case 3. Similarly all other cases can be concluded.
List each of the four numbers as products of two factors.. Remove those products where the factor appears in no other list e.g. 165 = 33x5 cannot be one of the pairs, as 33 appears in no toher pair. This leaves 64 = 8x8, 88 = 11x8, 120 = 15x8, 165 =15x11. So a,b,c,d are some permutation of [8,8,11,15]. The sum is 42.
Its simple but i agree sir should make a vid explaining better than this Lim h>0 (x+h) ^n -x^n /h Then using the expansion X^n -x^n + ncn-1 x^n-1 ×h ... /h X^n cancels out and n c n-1 = n (N) x^n-1 h+ .... /h Then divide by h nx^n-1 +n c n-2 x^n-2 h... Now put h=0 and all of the terms with h will turn into 0 Thus we're left with nx^n-1 thanks
@@figgi_myestrio5092 i watched his series about 5 times it it very intuitive and helpful to be clear i understand how the power rule comes but I specifically didn't understand it's proof by the method of the definition
A solution that wouldn't work with different numbers but still works here 64 is either 8² or 64x1 so we 100% get 2 of the 4 numbers, then by dividing them with the other numbers we get the remaining 2.
Prime factorization is another way to solve this. 64=2^6, so two of the unknown numbers are only powers of two and the sum of those powers is six. 165=3*5*11, so two of the numbers have no power of two. 88=2^3*11 Since we know that all four of the numbers are either powers of two or have no power of two, We know that the two numbers whose product is 88 are 2^3=8 and 11. This, then, allows us to determine the other two numbers from the remaining factors of 64 and 165: 8 and 15. So the four unknown numbers are 8, 8, 11, 15, whose sum is 42.
I like this kind of question. It requires our logical thinking and mathematical sense. Unlike those questions, we need to apply sophisticated math theorems or knowledge of special functions. For example, I was stuck with some questions that required the Lambert function W(x) at first, which is not included in the high school syllabus in my home country, and I never heard of it before.
I just realised it was a FOIL rectangle and hence the sides are integers (a + c) (b + d). I did NOT consider the cases of 1 and 437 because it seemed too far fetched to be true. After splitting 437 = 19 x 23 I realised whatever the case it must be one or the other (because you can flip a rectangle) so 19 + 23 = 42
By prime factorization we have 2⁶, 2³×11, 2³×3×5 and 5×3×11. So i realized that since 3×5×11 is odd then we must have 2 odd numbers because of the conditions in the problem but 2⁶ is even therefore we must also have 2 even numbers in the list. Now let a and b be even, c and d odd .The trick part in my method is to realize that a=b=2³ but why? Okay so since 2⁶ is a power of 2 then of course a and b are powers of 2 but a,b2² because we've already showed a,b are even and c,d are odd so take for example 2³×11 if a or b is 2² then in 2³×11 we would still have a factor of two remaining which means that c or d is 22, contradiction because 22 is even. Since we get 2²
I took a different approach looking at the factors of these numbers, which proved to be a faster way to solve this specific problem. It is easy to see that 88 is 2³×11. Since the only other number containing the factor 11 is 165 and 165 is odd, the common factor must be 11 (and not 22, 44 or 88). After that, it only takes seconds to find the factors. The first two factors come from 88 and are 8 and 11. The third factor is 165 : 11 = 15. The fourth factor is 64 : 8 = 8. Surprisingly, we wouldn't even need to know the fourth number; we could calculate it with the only combination of factors that have only been used once*, i.e. 8×15 = 120. Therefore, the sum is 8+8+11+15 = 42. But then again, we should have known this from the start since 42 is the answer to life, the universe and everything! 😛 Your approach is more rigorous, though, and less reliant to find these easy to isolate factors such as the 11 in this example. But then, I'm wondering whether there would be only one solution if we had more combinations, probably not. I suspect, but haven't proofed, that (a+b)(c+b)=x would have more than one exact possible value for a+b and c+d respectively if x were not a product of two primes. So maybe whenever there is a specific solution, my approach does work in all cases? Only a weak conjecture that would have to be shown to be true. By the way, why do you us Z⁺ for the positive integers as opposed to N? Those two are identical, from what I know, so it doesn't matter but using N would be the more natural (no pun intended!) choice. Not a criticism, just curious. ☺ Awesome channel, by the way. If I still were tutoring math students, I guess I could use some of your problems as useful teaching tools. Love you channel! PS: I also love how you seem to radiate a natural positiveness with your smile and soothing voice. I usually don't notice these things too much but they stand out with you! * well, since 8 accounts for two of the factors, it needs to show up four times
I just looked for common factors in all. 88 was most telling, as it's 8*11. Oh looky here, 165 is 11*15.. So we got 8, 11, and 15. 120 is 15*8 and 64 is 8*8, so everything just fell into place. Factors are 8, 8, 11, 15, adding them up is 42. QED. 😂
I'm so glad that I figured that out in like, 5 seconds If 64 is, let's say, ab then it's probably 8*8 Also, we have 88, which proves our conjecture, as 8*11 equals 88 so one of the numbers is gonna be 8 And we have 120, which is 8*x where x is obviously 15. So we get 8+8+11+15.
Let me tell you my way. we have ab, bc, cd and da and it does not matter which number is a or b or c or d. That means If I know one of these number is equal to something, I can think like it is a. So lets start. 64 = 2⁶ = 2³.2³ 88 = 8.11 = 2³. 11 120 = 8.15 = 2³ . 3.5 165 = 11.15 = 11 . 3.5 So I think that you saw there are some relationships between others. for example lets take 11 as our number because its prime. We can see 11 at 88 and 165. And there is no other number divides both of them together except 11. So I can say : a = 11 and if a=11, than I can say than one of the numbers (in this case I'll take ab) must be 88 and so I can say b = 8 and also I have to say other number (which is da) must be 165 and that means d = 15 So I got there 3 things : a = 11 b = 8 c = d = 15 so thats gonna be really easy to find c because we know bc and cd must 120 and 64 and we know that both have 2³ as I showed before. So that means c = 8 and all are done. a = 11 b = 8 c = 8 d = 15 and lets take the sum : 11 + 8 + 8 + 15 = 42 Have a nice day 😊❤
Hm prime factor decomposition + trial & error combination. Knowing that (a+c)(b+d) has to be sorted out, lead me to 8, 11 & 8, 15 pair. And suddenly all adds up correctly. Since having the sum, does not exactly mean, you have a solution - could have been impossible as well, f.e. take 65 and 87 instead. Would sum up fine, but not leading to a*b=65 sortof.
Nice... I did prime factoring 64, 88, 120, 165. 64 = 2^6 88 = 11*(2^3) 120 = (2^3)*3*5 165 = 3*5*11 Now it is easy to see the numbers of a, b, c and d.
I see a lot of people using prime factorization and I'm starting to think I'm the only one who actually used your method. But I didn't solve for a, b, c, and d.
I worked it out a different way, b/d = 8/11 so b and d must be multiples of 8 & 11 respectively, then I tested all possible combinations and found only 1 in which all numbers were integers, in which a + b + c + d = 42, when b = 8, d = 11, a = 8 and c = 15
It's actually easier to figure out what a, b, c and d are and directly compute. Two of the products have 11 as a factor (88 and 165), one of which is odd and the other only has 2 as the other prime factor, so you know that w.l.g. a = 11 and that means w.l.g. b = 8. Dividing 165 by 11 gives you another one, d = 15 and you conclude that c = 8.
ab = 64 bc = 88 cd = 120 da = 165 ab + bc + cd + da = b(a+c) + d(a+c) = 152 + 285 = 437 (b+d)(a+c) = 437 = 19 * 23 or 1 * 437 --> a + c cannot be 1 and b + d cannot be 1 because this would mean one of a, b, c, d would have to be 0 b + d + a + c = 19 + 23 = 42 --- deep thought
Very nice "Good morning!" mental math kata. Three and five occur twice each as factors, together; then eleven on its own and the other two numbers must be powers of 2 with a product of 64. That's 15 + 11 + 8 + 8 = 42.
Im making this comment before I watch the video I think that the "in some order" is just some red herring when order doesn't matter at all, it's just multiplying and adding
Hey, I was wondering where I could find the shirt you are wearing? A previous link didn't work so I was wondering if there is an official store? Thank you, Sir. Love your videos.
So if you tried to pick a random order of ab, bc, cd, and da setting then equal to any of the 4 numbers to make a system of equations and then solve for a, b, c, and d, shouldn't you get the same sum no matter what order combination you went in? Or would only 1 of them actually work where a, b, c, and d are positive integers while the other combinations have non positive integer values or no solution?
I had the solution quicker than you stated the problem! Way to complicated way to do it. two of the productes are multiples of 11 (8x11 and 15x11) one being 8x8 and the remaining one 15x8. That's all folks!
I did another method:
I decomposed the numbers in terms of prime factors:
165 = 5x3x11
120 = 5x3x2x2x2
64 = 2x2x2x2x2x2
88 = 2x2x2x11
note that 11 (and no other numbers) is common in 165 and 88, so let's take a = 11
remaining from 165, we find 5x3 (15) which can also be found in 120. so let's take b = 15
after that, we have 2x2x2 (8) cropping up 4 times, which implies that they will occupy c and d
adding them together, 11 + 15 + 8 + 8, we get 42
That was my thoughts on how to do it as well.
That's what I did too. The main thing was observing that 11 was a prime factor of two of the numbers, so I worked backwards from there.
that was my first thought too
I did that, but I started by noting that ad x bc = ab x cd, and the factor of 11 meant that 165 and 88 had to be part of different pairings. The factor of 5 helped in a similar manner.
First acknowledge that each 4 numbers are represent twice in the given multiplications. I did it about the same way, only more developed as the first step, in 4 columns I found all the pairs making each number while multiplied. Let say for the sake of beginning somewhere, we start comparison with 64, we have 1-64, 2-32, 4-16 and 8-8. It can't be the 1-64 pair because 64 isn't found anywhere else. It ain't 2-32 either because 32 isn't foud anywhere else, same with 16. So, the 8-8 pair must be two of the numbers we don't have to find. :-) Same job with 1-88, 4-22 and 2-44 from 88 leaving the 8-11 pair.... Last pairs being 11-15 and 8-15 by the same kind of process. The four relatives being 8, 8, 11 and 15, adding to 42.
I like this method more because it is easy to check back and because I am good at finding dividers.
Interesting fact i just realized is that 437 can be expressed as the difference of 2 perfect squares:
437=441-4
437=21²-2²
437=(21-2)(21+2)
437=19×23 :D
This is true for all combinations of number which are evenly spaced. Because being equidistant from the median, they can be expressed as product of (median - d) and (median + d)
@@adityamohan7366 oh that's interesting but I didn't understand the evenly spaced part. Evenly spaced from what?
@@iMíccoli evenly / equally spaced from the median.
437 = 19 x 23 = (21 - 2) x (21 + 2)
Here 21 is the median and the numbers 19 and 23 being equidistant from their median (21) are also evenly spaced.
I used 'evenly' because the combinations will always have a difference which would always be an even number otherwise we wont be able to get to the median which is also a natural number. Here 19 and 23 are evenly spaced and their difference '4'.
Another example,
40, 10 are two evenly spaced numbers whose product can also be expressed as a difference of 2 perfect squares.
Median of 40 and 10 = 25.
25 being equidistant from 40 and 10 by 15 on either side.
Thus 40 x 10 = 25² - 15² = 625 - 225 = 400
@@adityamohan7366:0 really cool. Thanks for the explanation I understand now :D.
I thought the problem is nice but your solution is even nicer. Thanks
PS: I solved it by finding factors in each number and matching those which share some factors. 8 x 11, 11 x 15, 15 x 8 and 8 x 8
The answer is 42 but method is foolproof.
I followed the same method. 2 numbers are clearly multiples of 11 (88 and 165). And the remaining two are multiple of 8 (64 and 120). So I found 8 twice, 11 and 15. Their sum = 42
I did it similarly!
I live in the UK and run puzzle evenings in a bar. Basically it’s a quiz where you need very little general knowledge. Most of the questions are mathematical but involve a lot of logic, usually integers or what I call integer fractions. I do have to throw in a few word based problems, just to keep certain customers happy. Your videos give me great inspiration. This is a great but I will need to adapt it. Thanks
Love your elegant solution. As noted in other comments, one can also solve for a,b,c,d first:
64 = 2^6
88 = 2^3 x 11
120 = 2^3 x 15
165 = 11 x 15
W.l.o.g let ab=64 ({ab, bc, cd, da} and (a+b+c+d) unchanged by a cyclic permutation of (a,b,c,d))
=> a=2^(6-k) and b=2^k some k∈{1,...,5}
Note cannot have k=0 or 6 since that would mean one of the other pairwise products would have share a factor of 2^6 which they do not.
Since b is a power of 2, bc must involve a power of 2. So there are 2 cases:
1) bc = 88 = 2^3 x 11 = 2^k x c
=> c = 2^(3-k) x 11
=> cd must have a factor of 11
=> cd = 165 = 11 x 15 = 2^(3-k) x 11 x d
=> k=3, a=b=8, c=11 and d=15
=> da = 15 x 8 = 120
=> (a,b,c,d) = (8,8,11,15)
=> a+b+c+d = 42
2) bc = 120 = 2^3 x 15 = 2^k x c
=> c = 2^(3-k) x 15
=> cd must have a factor of 15
=> cd = 165 = 11 x 15 = 2^(3-k) x 15 x d
=> k=3, a=b=8. c=15 and d=11
=> da = 11 x 8 = 88
=> (a,b,c,d) = (8,8,15,11)
=> a+b+c+d = 42
I actually found the solution to split the four numbers (64,88,120,165) into their primfactors easier here, bacause the only four possible factors (as combination of the forementioned primefactors) became immediately obvious.
I have been watching you for about half a year. And I am proud to say you taught me well. Before I began watching you, I was a high school math failure. Now, I finally managed to solve one of your questions on my own before watching your video!
Like many others here, I solved this puzzle by looking at the prime factorization of the dual products. The principal idea was to understand that when for example ab = 64, then a and b must both be 8, and cd must equal 165=11*15.
I want to share my views on the inter-dependency of the number of solutions and the order of things given in the question. When being asked only for the sum a+b+c+d, then clearly the order of (a,b,c,d) does not matter, whereas the order of assignments of the 4 numbers {64,88,120,165} to the products {ab,bc,cd,da} could theroretically result in more than one sum a+b+c+d for the answer, so we must carefully consider it. It turns out, though, that this is not the case here, and we're left with only one sum (42) for the answer. This is probably due to the cyclic symmetry in the selection of the 4 two-products.
On the other hand, if we ask for all the tuples (a,b,c,d) that fulfill the requirements of the puzzle, we arrive at 8 different solutions in total! Interestingly, this means that only one third of all 24 possible permutations of (8,8,11,15) qualify as an answer.
By inspection, you can note that 165 has to be the product of two odd numbers. From our multiplication table, even*even = even. Even*odd = even, odd*odd = odd. Lasty, 64 is a power of 2. Thus any two numbers that multiple to it must be a power of two. So we know two of the numbers are powers of two. This makes it easy to eyeball the solution based on the prime factorizations of the other numbers.
88 and 165 are each multiples of 11, so d=11. 120 and 165 are each multiples of 15, so c=15. The greatest common factor of 88 and 64 is 8, so a=b=11.
This was a cool number puzzle to solve! This is how I solved this. I started writing the given numbers 64, 88, 120 and 165 as product of their prime factors:
64 = 2*2*2*2*2*2
88 = 2*2*2*11
120 = 2*2*2*3*5
165 = 3*5*11
We have four positive integers a, b, c and d so that ab, ad, bc and cd gives us those four integers above. Each integer a, b, c and d appear twice in the pairwise products. We can instantly see that the prime 11 appears twice (as factor of 88 and 165). Since 3*5 and 2*2*2 don't have common factors, we know one of the numbers a, b, c and d must be 11. Let's set:
165 = 3*5*11 = ab => a = 3*5 = 15 and b = 11.
After this it is clear that:
88 = 2*2*2*11 = cb => c = 2*2*2 = 8.
Finally we solve for d:
120 = 2*2*2*3*5 = ad => d = 2*2*2 = 8.
We also see that these values check up for the final number 64:
64 = (2*2*2)*(2*2*2) = 8*8 = cd.
Solution: a+b+c+d = 15+11+8+8 = 42.
Clearly my way of solving this is different from the video, but my brain functions this way, so this is how I solved this.
This is a technique I've seen used a lot on this channel. Convert the problem into the form: (...)(...)=integer, then test each possible pair of multiplicands for validity.
437 = 441-4, a difference of two squares. 21^2-2^2 = (21+2)(21-2) = 23*19.
Simplify by picking it apart: 2 products have 11 (prime) as a factor, so 11 is one of the letters, while 8 and 15 are the two numbers that the first pairs with. 15 doesn't divide 64, so it must be one of the pair making 120 and it must be another of the letters. 120/15 is 8, so 8 and 8 are the remaining 2 letters. 11 + 15 + 8 + 8 = 42. Many of the comments are similar to this. Prime's solution seems like it would work better for less obvious variations of this problem.
Ur solution is really excellent.
But I solved in following way:
multiply all so we get.
(abcd)²=64*88*120*165;
abcd=15*11*8*8;
So following statement can be concluded:
Possible pair as (ab,cd) or (bc,ad)=(64,165)or(88,120);
we can also observe that
ab/bc=(cd/ad)-¹;
Case 1
Let ab=64 & cd=165;
bc=88 & ad=120;
64/88=(165/120)-¹=8/11=a/c ;
let a=8k,c=11k;
so b=64/8k=8/k
similarly d=15/k
as b=8/k and d=15/k
So maximum value of k will be HCF(8,15) that is equal to 1.
So only value of k is 1
a=8,b=8,c=11,d=15
Case 2
Let ab=64 & cd=165;
bc=120 & ad=88;
Similarly a=8k,b=8/k,c=15k,d=11/k
again only possible value of k=1
a=8,b=8,c=15,d=11.
Case3
let bc=64 & ad=165;
cd=120 & ab=88;
Substitute in case 2
a=b
b=c
c=d
d=a
Case 2 become case 3.
Similarly all other cases can be concluded.
I was looking for another good explanation. Thank you
List each of the four numbers as products of two factors..
Remove those products where the factor appears in no other list e.g. 165 = 33x5 cannot be one of the pairs, as 33 appears in no toher pair.
This leaves 64 = 8x8, 88 = 11x8, 120 = 15x8, 165 =15x11.
So a,b,c,d are some permutation of [8,8,11,15].
The sum is 42.
Sir you should do a video on proving power rule by the definition. Its really complicated in our textbooks something to do with the binomal theorem
Its simple but i agree sir should make a vid explaining better than this
Lim h>0 (x+h) ^n -x^n /h
Then using the expansion
X^n -x^n + ncn-1 x^n-1 ×h ... /h
X^n cancels out and n c n-1 = n
(N) x^n-1 h+ .... /h
Then divide by h
nx^n-1 +n c n-2 x^n-2 h...
Now put h=0 and all of the terms with h will turn into 0
Thus we're left with nx^n-1 thanks
Watch 3b1b's essence of calculus he derived it in it i think, it's pretty intuitive too
@@figgi_myestrio5092 i watched his series about 5 times it it very intuitive and helpful to be clear i understand how the power rule comes but I specifically didn't understand it's proof by the method of the definition
A solution that wouldn't work with different numbers but still works here
64 is either 8² or 64x1 so we 100% get 2 of the 4 numbers, then by dividing them with the other numbers we get the remaining 2.
Prime factorization is another way to solve this.
64=2^6, so two of the unknown numbers are only powers of two and the sum of those powers is six.
165=3*5*11, so two of the numbers have no power of two.
88=2^3*11
Since we know that all four of the numbers are either powers of two or have no power of two, We know that the two numbers whose product is 88 are 2^3=8 and 11. This, then, allows us to determine the other two numbers from the remaining factors of 64 and 165: 8 and 15.
So the four unknown numbers are 8, 8, 11, 15, whose sum is 42.
I actually started this by cutting the four numbers into prime factors. From there, the answer is obvious as well.
Thanks for sharing. Enjoyed the problem, and I actually did it the same way you did.
I like this kind of question. It requires our logical thinking and mathematical sense. Unlike those questions, we need to apply sophisticated math theorems or knowledge of special functions. For example, I was stuck with some questions that required the Lambert function W(x) at first, which is not included in the high school syllabus in my home country, and I never heard of it before.
I just realised it was a FOIL rectangle and hence the sides are integers (a + c) (b + d). I did NOT consider the cases of 1 and 437 because it seemed too far fetched to be true. After splitting 437 = 19 x 23 I realised whatever the case it must be one or the other (because you can flip a rectangle) so 19 + 23 = 42
By prime factorization we have 2⁶, 2³×11, 2³×3×5 and 5×3×11.
So i realized that since 3×5×11 is odd then we must have 2 odd numbers because of the conditions in the problem but 2⁶ is even therefore we must also have 2 even numbers in the list.
Now let a and b be even, c and d odd .The trick part in my method is to realize that a=b=2³ but why?
Okay so since 2⁶ is a power of 2 then of course a and b are powers of 2 but a,b2² because we've already showed a,b are even and c,d are odd so take for example 2³×11 if a or b is 2² then in 2³×11 we would still have a factor of two remaining which means that c or d is 22, contradiction because 22 is even.
Since we get 2²
I took a different approach looking at the factors of these numbers, which proved to be a faster way to solve this specific problem.
It is easy to see that 88 is 2³×11. Since the only other number containing the factor 11 is 165 and 165 is odd, the common factor must be 11 (and not 22, 44 or 88).
After that, it only takes seconds to find the factors.
The first two factors come from 88 and are 8 and 11.
The third factor is 165 : 11 = 15.
The fourth factor is 64 : 8 = 8.
Surprisingly, we wouldn't even need to know the fourth number; we could calculate it with the only combination of factors that have only been used once*, i.e. 8×15 = 120.
Therefore, the sum is 8+8+11+15 = 42.
But then again, we should have known this from the start since 42 is the answer to life, the universe and everything! 😛
Your approach is more rigorous, though, and less reliant to find these easy to isolate factors such as the 11 in this example.
But then, I'm wondering whether there would be only one solution if we had more combinations, probably not. I suspect, but haven't proofed, that (a+b)(c+b)=x would have more than one exact possible value for a+b and c+d respectively if x were not a product of two primes. So maybe whenever there is a specific solution, my approach does work in all cases? Only a weak conjecture that would have to be shown to be true.
By the way, why do you us Z⁺ for the positive integers as opposed to N?
Those two are identical, from what I know, so it doesn't matter but using N would be the more natural (no pun intended!) choice.
Not a criticism, just curious. ☺
Awesome channel, by the way. If I still were tutoring math students, I guess I could use some of your problems as useful teaching tools.
Love you channel!
PS: I also love how you seem to radiate a natural positiveness with your smile and soothing voice. I usually don't notice these things too much but they stand out with you!
* well, since 8 accounts for two of the factors, it needs to show up four times
10:58 And if a+c=19 -> Then either: a=8 and c=11, or a=11 and c=8.
Also, means that either: b=8 and d=15, or b=15 and d=8.
I just looked for common factors in all. 88 was most telling, as it's 8*11.
Oh looky here, 165 is 11*15.. So we got 8, 11, and 15.
120 is 15*8 and 64 is 8*8, so everything just fell into place.
Factors are 8, 8, 11, 15, adding them up is 42. QED. 😂
I'm so glad that I figured that out in like, 5 seconds
If 64 is, let's say, ab then it's probably 8*8
Also, we have 88, which proves our conjecture, as 8*11 equals 88 so one of the numbers is gonna be 8
And we have 120, which is 8*x where x is obviously 15. So we get 8+8+11+15.
It took me a little bit longer but I thought the same way
I just thought 88 gives me 11 and 8; 165 and 11 give me 15, 120 and 15 give me another 8. 8+8+11+15=42. Then I fast forwarded. 😮
Let me tell you my way.
we have ab, bc, cd and da and it does not matter which number is a or b or c or d. That means If I know one of these number is equal to something, I can think like it is a. So lets start.
64 = 2⁶ = 2³.2³
88 = 8.11 = 2³. 11
120 = 8.15 = 2³ . 3.5
165 = 11.15 = 11 . 3.5
So I think that you saw there are some relationships between others.
for example lets take 11 as our number because its prime. We can see 11 at 88 and 165. And there is no other number divides both of them together except 11.
So I can say :
a = 11
and if a=11, than I can say than one of the numbers (in this case I'll take ab) must be 88 and so I can say b = 8
and also I have to say other number (which is da) must be 165 and that means d = 15
So I got there 3 things :
a = 11
b = 8
c =
d = 15
so thats gonna be really easy to find c because we know bc and cd must 120 and 64 and we know that both have 2³ as I showed before. So that means c = 8 and all are done.
a = 11
b = 8
c = 8
d = 15
and lets take the sum :
11 + 8 + 8 + 15 = 42
Have a nice day 😊❤
42:
64 - 2^6 - ab
88 - 2^3 x 11 - ad
120 - 2^3 x 3 x 5 - bc
165 - 3 x 5 x 11 - cd
a+b+c+d
=2^3 + 2^3 + 15 + 11
=8 + 8 + 15 + 11
= 42
Hm prime factor decomposition + trial & error combination. Knowing that (a+c)(b+d) has to be sorted out, lead me to 8, 11 & 8, 15 pair. And suddenly all adds up correctly. Since having the sum, does not exactly mean, you have a solution - could have been impossible as well, f.e. take 65 and 87 instead. Would sum up fine, but not leading to a*b=65 sortof.
here is how I solved it
without loss of generality we can assume a
88=11*8 165=11*15 120=15*8 64=8*8
gdp(8,15)=1, so one of the numbers is 11, so another is 15, so another is 8, so another is 8
11+15+8+8=42
64=8•8, 88=8•11, 120=8•15, 165=15•11
let a=8, b=8, c=11, and d=15
ab=64, ad=120, bc=88, and cd=165
a+b+c+d=8+8+11+15=16+26=42
Nice... I did prime factoring 64, 88, 120, 165.
64 = 2^6
88 = 11*(2^3)
120 = (2^3)*3*5
165 = 3*5*11
Now it is easy to see the numbers of a, b, c and d.
did it in my head using prime factors. Once you know that 88 can only be 8*11, the rest is downhill
It can be 4*22 and 2*44 also
@@proisborn no, because 22 does not work for any of the others.
@@lynnrathbun I was saying 88 can have more factors you did not say comparison to others
neat solution
I see a lot of people using prime factorization and I'm starting to think I'm the only one who actually used your method.
But I didn't solve for a, b, c, and d.
I worked it out a different way, b/d = 8/11 so b and d must be multiples of 8 & 11 respectively, then I tested all possible combinations and found only 1 in which all numbers were integers, in which a + b + c + d = 42, when b = 8, d = 11, a = 8 and c = 15
It's actually easier to figure out what a, b, c and d are and directly compute.
Two of the products have 11 as a factor (88 and 165), one of which is odd and the other only has 2 as the other prime factor, so you know that w.l.g. a = 11 and that means w.l.g. b = 8.
Dividing 165 by 11 gives you another one, d = 15 and you conclude that c = 8.
a is not 11!
ab = 64
bc = 88
cd = 120
da = 165
ab + bc + cd + da = b(a+c) + d(a+c) = 152 + 285 = 437
(b+d)(a+c) = 437 = 19 * 23 or 1 * 437 --> a + c cannot be 1 and b + d cannot be 1 because this would mean one of a, b, c, d would have to be 0
b + d + a + c = 19 + 23 = 42 --- deep thought
Very nice "Good morning!" mental math kata.
Three and five occur twice each as factors, together; then eleven on its own and the other two numbers must be powers of 2 with a product of 64. That's 15 + 11 + 8 + 8 = 42.
64 is 8*8 (2 factors) , divide 88 by 8 get 11 , divide 120 by 8 get 15. Sum 8,8,11,15 get 42
Eu também fiz o cálculo dessa forma. Muito mais fácil.
Im making this comment before I watch the video
I think that the "in some order" is just some red herring when order doesn't matter at all, it's just multiplying and adding
I did it in my head in a couple of minutes. It took me some time to accept that a=b=8, to form the square of 8 that is 64.
ab+ada(b+d)=8(8+11)
bc+cd)=c(b+d)=120+165=15(8+11)
a=8,c=15
b=64/8=8,d=165/15=11
A+b+c+d=8+8+15+11=42
I just drew a rectangle and assigned the 4 products as areas.
11*15 11*8
8*15 8*8
So you get a 15+8 side and an 8+11 side. Semiperimeter is 42.
Genius.
No wait your solution is just insane, how did you even think that? :0
I am so happy i did it correctly
Hey, I was wondering where I could find the shirt you are wearing? A previous link didn't work so I was wondering if there is an official store? Thank you, Sir. Love your videos.
Very nice
Brilliant!
So if you tried to pick a random order of ab, bc, cd, and da setting then equal to any of the 4 numbers to make a system of equations and then solve for a, b, c, and d, shouldn't you get the same sum no matter what order combination you went in? Or would only 1 of them actually work where a, b, c, and d are positive integers while the other combinations have non positive integer values or no solution?
I had the solution quicker than you stated the problem! Way to complicated way to do it. two of the productes are multiples of 11 (8x11 and 15x11) one being 8x8 and the remaining one 15x8. That's all folks!
Excellent!
64 = 8*8
88 = 8*11
120= 8*15
165= 15*11
took me a few seconds to solve lol
Nice explanation 👌👌
Another answer is the negative of those same number, -8, -8, -11, -15, sum becomes -42.
The answer is fine of course,but there is a condition given that all the numbers are positive integers.
Sweetly done - reminded me of 'The Hitchhiker's Guide to the Galaxy'
8,8,11,15 in my head in less than a minute, next
Let's go!
I got the factors 19 & 43 BUT how did you arrive at 8,8, 11, 15 - just a 57 year old that fell in love with math so starting all over again👊👊
It's a great video,
Very nice, I like
Tricky problem
Easy by factorizing.
8:16 19 and 23
Jesus is Lord
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Me @@Zerotoinfinityroad
@@morbrakai8533
So you care,
Idfc
@@Zerotoinfinityroad u asked
Good question Sir... thank You very much...
64 // 88 // 120 // 165
2⁶ // 2³11 // 2³3¹5¹ // 3¹5¹11
2⁶ + 2³3¹5¹ = 2³(2³ + 3¹5¹)
2³11 + 3¹5¹11 = 11(2³ + 3¹5¹)
(2³ + 11)(2³ + 3¹5¹)
a + b + c + d = 2³ + 11 + 2³ + 3¹5¹
*a + b + c + d = 42*