Mathematical intuition tells me to ignore the +1, average them to (71.5)^4 and mental math got me 5112.25. I know the answer ends in 1 or 9 because of how our number also originally ends in 1 and the implication this is an integer answer, so I rounded to 5111. What a nice coincidence that was to see it as the answer!
That's a lot of imagination to do, but it's obviously faster than solving it on paper. However you can't use those techniques in a subjective assessment.
Consider: 71.72 = 5112 70.73 = 5110 We are looking for a positive X for which X² = 70.71.72.73 + 1 = 5110.5112 + 1. = (5111 - 1)(5111 + 1) + 1 = (5111² - 1²) + 1 = 5111² So X = 5111
As long as the numbers are consecutive, the first term is multiplied by the last term and 1 is added.70*73+1=5111...Another example: square root(94*95*96*97+1).....94*97+1=9119
Hello friends! For any consecutive sequence ABCD expession like sqrt(A*B*C*D+1) should be equal B*C-1 For 70, 71, 72, 73 we have 71*72-1 = 5112-1 = 5111 For example for 170, 171, 172, 173 we have 171*172-1 = 29412-1 = 29411 🙃
That seems a very involved and algebraic approach for a purely arithmetical question. My first thought was difference of two squares, using the average 71.5. Then I decided to avoid fractions inside the square root by multiplying inside by 16 and dividing outside by 4. Then it all just came out nicely as a simple straightforward chain of = signs, with no need for subsidiary calculations, new notation, substitutions or whatever. Here goes: √(70 x 71 x 72 x 73 + 1) = 0.25√(140 x 142 x 144 x 146 + 16) = 0.25√((143 - 3)(143 - 1)(143 + 1)(143 + 3) + 16) = 0.25√((143² - 3²)(143² - 1²) + 16) = 0.25√((143⁴ - (9 + 1)143² + 9 + 16) = 0.25√((143⁴ - (2)(5)143² + 25) = 0.25√((143² - 5)²) = 0.25(143² - 5) = 0.25(20449 - 5) = 20444/4 = 5111 Of course, you need to use the difference of two squares and binomial expansion identities (a - b)(a + b) ≡ a² - b² (a + b)(a + b) ≡ a² + 2ab + b², and you need to be able to square 143 without a calculator 1 4 3 1 4 3 X ------------- 4 2 9 5 7 2 0 1 4 3 0 0 --------------------- 2 0 4 4 9 So the multiplication is a bit harder than yours, but I think avoiding two layers of subsidiary notation and substitutions makes that worth while.
I just put on my Graphics calculator and in seconds the correct answer. This is the 21st century so when in Rome do as the Romans do(or people of this century). Nostalgia is nice but not efficient. Besides AI will probably doing most of this type of work anyway.
I smiled at that small addition in the end: "... and you need to be able to square 143 without a calculator". I imagined a lazy uneducated person who came all the way to the end just to see: "Multiplication! What kind of a nonsense solution is that? How can you expect me to multiply 3 digit numbers! What I am, a genius?" )))
@@Pilot_engineer_19 It's not about getting the answer, it's about teaching your brain to think. You need to make not so obvious logical steps in order to solve a task at hand. So yes, you can use your calc if you like. But you will miss the very point of that task then.
The product of two adjacent numbers is 2 greater than the product of the numbers on either side of it 71x72=70x73+2 70x73x71x72+1=70x73x(70x73+2)+1 =(70x73)^2+2x(70x73)+1 =(70x73+1)^2 So answer is 70x73+1=5111
when you get x² + 3x + 1, you can simplify it further as this: = x² + 2x + x + 1 (since 3x = 2x + x) = x² + 2x + 1 + x (reshuffling the last x and 1) = (x² + 2x + 1) + x (no change, just adding bracket to show next calculation) = (x + 1)² + x (since (x² + 2x + 1) = (x+1)(x+1)) now substitute x with 70, you get, = (70 + 1)² + 70 = (71)² + 70 = 5041 + 70 = 5111 Hence answer: Note, answer focuses on minimal numbers computation, so if you computed in x² + 3x + 1 itself, its not wrong, but it could have been further simplified!
70*(70+1)*(70+2)*(70+3)=70^4+6*70^3+11*70^2+6*70=26122320+1= 26122321; sqr (26122321); I make from right to left groups (21;23;12;26; so I got sqr (26’12’23’21) and by hand calculate sir of this large number. Answer is 5111. I did all calculation without calculated how was taught at the school. The answer is correct; I checked one by calculator. P.S. I used multiplications binomial and trinomial using Table, where on vertical line I wrote first, binomial and on horizontal line; and on vertical line-trinomial, on horizontal line-binomial; and collected like terms. Since I have known 7^2=49 and 7^3=343, algorithm multiplication by 11; rest procedure was easy.
Somehow I’m much more interested in the approach than the solution. I am not a math expert just interested but I always seem to be completely oblivious to the way how to! Guess I’ll watch some more then. …. Btw , 50 years old and haven’t had math for some 30-35 years. It still does interest me though
About the same with me. I'm 42. Listening to these videos at this age makes me realize why I struggled in high school. His explanation is great but I feel it's probably not too different from how my teachers used to explain. It's just that back then my young brain couldn't comprehend it and couldn't keep my mind on the task. The moment i got confused and he'd too many questions my mind wondered off.
It's been 26 years since I took a formal math content class. This past year I volunteered to teach algebra and geometry to high school students. A lot of it was like falling off a bike, you never forget. But some of it was surprisingly difficult. Still other parts I had never learned. So far, my favorite day was teaching derivatives... I had never learned them, and didn't understand them while I prepped for the class, but somehow doing it on the board caused something to click in my brain. Good times.
As long as the numbers are consecutive, the first term is multiplied by the last term and 1 is added.70*73+1=5111...Another example: square root(94*95*96*97+1).....94*97+1=9119
There is actually a decent trick for squaring those big numbers. Now you gotta know the tens trick to be decent for numbers like _4 or _6. All it is is whenever you have something like 72^2, it's just 70^2+70+71+71+72 so it'd be equal 4900+70+142+72=5184
Haven't watched the vid. 71=x, then we get x*(x+2)*(x^2-1)+1 then expand it. Then try to approximate this with (x^2+a*x+b)^2. Apparently, a=+1. Then b has to be +1 or -1 (turns to be -1). Those are boring problems, since we know that the answer exists. Real science is when you don't know whether there's an answer
That's nice but too long. I mean, you got it right but you wrote "evaluate". I understand that it is allowed to approximate the result. Which you can do by : 1) forget the +1 in the square root 2) replace 70*71*72*73 by (71-1)*(71+1)*(72-1)*(72+1) 3) it becomes (71^2-1)*(72^2-1) 4) again, forget the ones 5) It becomes √(71^2*72^2) 6) so 71*72 7) = 5112 Close enough !
Proof that for any consecutive A,B,C,D the square root of A.B.C.D + 1 is B.C - 1. To prove this we will need to write the product A.B.C.D + 1 in the form of a square so we can take the square root. The idea is to get rid of A and D by writing them in terms of B and C using the fact that A,B,C,D are consecutive. This way B and C will occur in the product twice, i.e. as squares. The additional +1 is needed to turn the whole thing into a perfect square as we will see when we try this. *Proof in detail:* Since A and B are consecutive we have A = (B-1). Since C and D are consecutive we have D = (C+1). Their product is A.D = (B-1)(C+1) = B.C + B - C - 1. Since B and C are also consecutive we have B - C = -1 so that A.D = B.C - 2. Substituting this expression in A.B.C.D gives A.B.C.D = B.C.(B.C - 2). Now write B.C as (B.C - 1) + 1, and write (B.C - 2) as (B.C - 1) - 1. This results in A.B.C.D = ((B.C-1) + 1)(B.C-1) - 1). The expression on the right is of the form (a+b)(a-b) which is equal to (a² - b²) according to a well known formula. Consequently we get A.B.C.D = (B.C-1)² - 1². Adding 1 on both sides gives A.B.C.D + 1 = (B.C - 1)² of which the square root is B.C - 1. *Alternative proof:* Write A, C and D in terms of B A.B.C.D + 1 = (B-1).B.(B+1).(B+2) + 1 = reordering (B-1).(B+1).B.(B+2) + 1 = rewrite the last two factors (B-1).(B+1).((B+1)-1).((B+1)+1) + 1 = using (a-b)(a+b) = a² - b² twice (B²-1).((B+1)²-1) + 1 = write out the product B².(B+1)² - (B+1)² - B² + 2 = write out the second (B+1)² B².(B+1)² - B² - 2B -1 - B² + 2 = B².(B+1)² - 2B² - 2B + 1 = (B.(B+1))² - 2.(B.(B+1)) + 1 = using a²-2a+1 = (a-1)² (B.(B+1) - 1)² of which the square root is B.(B+1) - 1 which is the same as B.C - 1
Took me too long to figure out. For some reason I got stuck when 71*72 was two more than 70*73 instead of one. After cheating and verifying the answer was an integer I understood how to shift a n(n+2) rectangle so you can add a 1x1 block and make a (n+1)^2. Haven't seen more than the thumbnail yet, but if you use formulae instead of drawing the rectangles I'm going to scream
Well, it was easy enough to follow. To simplify; if you have a rectangle with x rows and x+3 columns, you can move one of the columns to a new row and get x+1 rows and x+2 columns with the new row missing two blocks. So you know 70*73=71*72-2. If you have a rectangle with n rows and n+2 columns, you can make it a n+1 square by moving the last column to a new row and adding one block. That's the +1 from the beginning. So the square has a side of 70*73+1=4900+210+1=5110
When you have x=70 the calculation turns out to be so easy and it is even faster to calculate directly OR even have an estimation from 71*72=5112 plus we know the final digit should be 1 or 9 so we have 5111 or 5119 thus we should introduce a more complicated number for x like 2022 then the answer turns out to be 2023X2024-1=4094551
Good solution, but I solved through the factors' symmetry with respect to 71,5. Let me notice, that in the general case, at the timing of 3:35, you need to write a |y+1| instead of y+1.
The question is easy the only thing is it has long calculation.. Sqroot [(70*71*72*73)+1] Sqroot [(70+1)(70 +2)(70 +3)70 + 1] Let x =70 So, the above expression become: Sqroot [(x+1)(x+2)(x+3)x + 1] Sqroot [x^2 + 3^x +2)(x+3)x + 1] Sqrt[(x^2 + 3^x +2)(x^2+ 3^x) +1] exp-2 Let x^2 + 3^x = y So exp2 becomes Sqrt. [(y+2)y +1] Sqrt [ y^2 +2^y + 1] Sqrt [y^2 + y + y + 1] Sqrt [y(y + 1) + 1(y+1)] Sqrt [ (y+1)^2] = [ y+1] y = x^2 + 3x x = 70 So y become 70^2 + 70x3 = 4900+210= 5110 y = 5110 Final answer is y+1 = 5110+1 = 5111 4900+ 210 5110
the answer is (70)(73)+1. so if we let x=70 z=73 the question q is sqrt(xz(x+1)(z-1) +1)) noting z-x=3 q²= (xz)(xz+z-x-1)+1 = (xz)(xz+2) +1= (xz)^2+2xz+1 =(xz+1)² so q( the answer)= xz+1 =70*73+1 =4900+210+1=5111 this question comes up quite a lot so if you get it as multiple choice you will know the answer is first*last+1
5111. Took me 10 seconds, although I know the trick that the square root of n(n+1)(n+2)(n+3)+1 is n(n+3)+1, because n(n+1)(n+2)(n+3)+1=n(n+3)(n+1)(n+2)+1=(n^2+3n)(n^2+3n+2)+1=(k-1)(k+1)+1=k^2-1+1=k^2, where k=n^2+3n+1=n(n+3)+1.
8 squared is 64, and 7 times 9 is 63. So 70 times 72 is 71 squared minus one, and 71 times 73 is 72 squared minus 1. So we get 71 squared times 72 squared, minus 71 squared, minus 72 squared, plus two.
There's a simple formula for the square root of 4 consecutive numbers multiplied + 1. Here it is: √ABCD + 1 = BC - 1. So 71 x 72 - 1 = 5111. Here's another example: √16 x 17 x 18 x 19 + 1) = 17 x 18 - 1 = 305. Make your own example and give it a try. It works. Cool beans. 😄😄😄
I keep seeing everyone doing this the exact same way. Are you all smoking crack? You need to do no more than basic multiplication. multiply 70 and 73. You can do this in your head. It's 5110. multiple 71 and 72. You can do this in your head or use paper and pencil. It's 5112. Let n = 5111. Why? Because, if you do, here's the quantity under the sq. rt rewritten: square root of [(n-1) * (n+1) +1] Multiply it out. square root of [n^2 + 1n - 1n -1 + 1] Square root of n^2. n was 5111, so that's your answer.
I had Calculus and I am like what the hell I never got this in Algebra 2 What level of math are we talking about I didn’t have anything like this in Pre-calculus After seeing this I feel like a complete fool
I had a quite different idea: I also call the result x, so I can write x = sqrt(70*71*72*73 + 1) Square this equation: x^2 = 70*71*72*73 + 1 Subtract 1: x^2 - 1 = 70*71*72*73 On the left side, use the 3rd binomic formula: (x + 1)*(x - 1) = 70*71*72*73 The two factors (x + 1) and (x - 1) of the left side differ by 2, so I try to build such two factors on the right side, too. Since the factors differ only by 2, the are almost equal, so I try to group the four factors to approximately equal products. I try to achieve this by grouping the largest and lowest factor, and the two factors in between: 70 * 73 = 70 * (70 + 3) = 4900 + 210 = 5110 71 * 72 = (70 + 1)(70 + 2) = 4900 + 140 + 70 + 2 = 5040 + 72 = 5112 Bingo! 5110 and 5112 differ by exactly 2, so x + 1 = 5112 x - 1 = 5110 and x = 5111 and this is the solution to the problem (the sought square root).
It's a factor, with (y + 1)² you multiply each thing inside the parentheses with itself and the other number then add it all so y * y + y *1 + 1 * y + 1*1 = y² + 2y + 1 Sorry if this is didn't help I'm not that good at math haha
That’s the hard way to go about that problem. Multiply (70•71•72•73) + 1 26,122,320 + 1 26,122,321 Then take the square root 5111 You really overthought that one!
(70*71*72*73)+1=26,122,321 Sq root 25,000,000 = 5000 5100^2 = 26,010,000 5110^2 = 26,112,100 5111^2 = 25,122,321 See how easy? As I said, you overthought the problem. Here’s a quote - “If you’re a hammer, you see every problem as a nail” 😎 So my advice is “Don’t let your schooling interfere with your education” 😎 Good Day, Sir
30 секундную писанину умудрились растянуть на 5 минут. Такие видео совсем дурачки обычно не смотрят можно пропустить половину вспомогательных пояснений.
![Megan Thee Stallion - Bigger In Texas [Official Video]](http://i.ytimg.com/vi/QxTkZTLWdo4/mqdefault.jpg)
Mathematical intuition tells me to ignore the +1, average them to (71.5)^4 and mental math got me 5112.25. I know the answer ends in 1 or 9 because of how our number also originally ends in 1 and the implication this is an integer answer, so I rounded to 5111. What a nice coincidence that was to see it as the answer!
That's a lot of imagination to do, but it's obviously faster than solving it on paper. However you can't use those techniques in a subjective assessment.
@@D_T_999 thats what the square route does. You obviously did not understand lol good luck with your studies!
@@alumarunt lmao you did it wrong shut up
@@RebornKaotic hi pal, nice to see you outside the AA meetings, this week don't touch any alcohol and you will get your chip!
can you pls explain how u knew the answer ended in 1 or 9 ? i don’t understand the intuition behind that. what original number ended in 1?
Consider:
71.72 = 5112
70.73 = 5110
We are looking for a positive X for which
X² = 70.71.72.73 + 1
= 5110.5112 + 1.
= (5111 - 1)(5111 + 1) + 1
= (5111² - 1²) + 1
= 5111²
So X = 5111
That's very clever and elegant, congrats! 👍
Nice
As long as the numbers are consecutive, the first term is multiplied by the last term and 1 is added.70*73+1=5111...Another example: square root(94*95*96*97+1).....94*97+1=9119
Hello friends! For any consecutive sequence ABCD expession like sqrt(A*B*C*D+1) should be equal B*C-1
For 70, 71, 72, 73 we have 71*72-1 = 5112-1 = 5111
For example for 170, 171, 172, 173 we have 171*172-1 = 29412-1 = 29411
🙃
That's right!
I solved using using A*D +1 because finding 73*70 was easier. But general idea is the same.
can you prove it?
@@char8169 Hello! Yep!
Prove that sqrt(A*B*C*D+1) = B*C-1 where C-B=1
sqrt((B-1)*B*C*(C+1)+1) = B*C-1
square both parts of equation
(B-1)*B*C*(C+1)+1 = (B*C-1)^2
(B^2-B)*(C^2+C)+1 = B^2*C^2 - 2*B*C + 1
B^2*C^2 + B^2*C - B*C^2 - B*C + 1 = B^2*C^2 - 2*B*C + 1
B^2*C - B*C^2 = -B*C
divide both parts of equation by B*C
B-C = -1
where C-B=1
Proven
Can you prove to A,B,C,D can be every number? This example is for A,A+1,A+2,A+3
That seems a very involved and algebraic approach for a purely arithmetical question. My first thought was difference of two squares, using the average 71.5. Then I decided to avoid fractions inside the square root by multiplying inside by 16 and dividing outside by 4. Then it all just came out nicely as a simple straightforward chain of = signs, with no need for subsidiary calculations, new notation, substitutions or whatever.
Here goes:
√(70 x 71 x 72 x 73 + 1)
= 0.25√(140 x 142 x 144 x 146 + 16)
= 0.25√((143 - 3)(143 - 1)(143 + 1)(143 + 3) + 16)
= 0.25√((143² - 3²)(143² - 1²) + 16)
= 0.25√((143⁴ - (9 + 1)143² + 9 + 16)
= 0.25√((143⁴ - (2)(5)143² + 25)
= 0.25√((143² - 5)²)
= 0.25(143² - 5)
= 0.25(20449 - 5)
= 20444/4
= 5111
Of course, you need to use the difference of two squares and binomial expansion identities
(a - b)(a + b) ≡ a² - b²
(a + b)(a + b) ≡ a² + 2ab + b²,
and you need to be able to square 143 without a calculator
1 4 3
1 4 3 X
-------------
4 2 9
5 7 2 0
1 4 3 0 0
---------------------
2 0 4 4 9
So the multiplication is a bit harder than yours, but I think avoiding two layers of subsidiary notation and substitutions makes that worth while.
Nice
I just put on my Graphics calculator and in seconds the correct answer.
This is the 21st century so when in Rome do as the Romans do(or people of this century). Nostalgia is nice but not efficient. Besides AI will probably doing most of this type of work anyway.
@@Pilot_engineer_19Good grief.
I smiled at that small addition in the end: "... and you need to be able to square 143 without a calculator". I imagined a lazy uneducated person who came all the way to the end just to see: "Multiplication! What kind of a nonsense solution is that? How can you expect me to multiply 3 digit numbers! What I am, a genius?" )))
@@Pilot_engineer_19 It's not about getting the answer, it's about teaching your brain to think. You need to make not so obvious logical steps in order to solve a task at hand. So yes, you can use your calc if you like. But you will miss the very point of that task then.
God bless Tanzania and whole Africa with Happiness.
Thanks 🙏❤️
Очень красивое решение. Люблю такие примерчики.
Beau petit casse-tête mathématique et très bien expliqué, merci 👍
The product of two adjacent numbers is 2 greater than the product of the numbers on either side of it
71x72=70x73+2
70x73x71x72+1=70x73x(70x73+2)+1
=(70x73)^2+2x(70x73)+1
=(70x73+1)^2
So answer is 70x73+1=5111
when you get
x² + 3x + 1, you can simplify it further as this:
= x² + 2x + x + 1 (since 3x = 2x + x)
= x² + 2x + 1 + x (reshuffling the last x and 1)
= (x² + 2x + 1) + x (no change, just adding bracket to show next calculation)
= (x + 1)² + x (since (x² + 2x + 1) = (x+1)(x+1))
now substitute x with 70, you get,
= (70 + 1)² + 70 = (71)² + 70
= 5041 + 70
= 5111
Hence answer:
Note, answer focuses on minimal numbers computation, so if you computed in x² + 3x + 1 itself, its not wrong, but it could have been further simplified!
(70)² is easier to compute than (71)²
@@moozoo2589 If you are programming, fewer multiplication and operations is better, your algorithm will be more efficient.
@@moozoo2589 71^2=70^2+70+71
So here answer is 4900+211=5111
You can make it.
@@juhwang79 thank you
This video is equals to... enjoyable.
haha nice
God bless Tanzania and whole Africa with Happiness. 💞💕
Nice problem. Well explained as usual.
Thanks again!
70*(70+1)*(70+2)*(70+3)=70^4+6*70^3+11*70^2+6*70=26122320+1= 26122321; sqr (26122321); I make from right to left groups (21;23;12;26; so I got sqr (26’12’23’21) and by hand calculate sir of this large number. Answer is 5111. I did all calculation without calculated how was taught at the school. The answer is correct; I checked one by calculator. P.S. I used multiplications binomial and trinomial using Table, where on vertical line I wrote first, binomial and on horizontal line; and on vertical line-trinomial, on horizontal line-binomial; and collected like terms. Since I have known 7^2=49 and 7^3=343, algorithm multiplication by 11; rest procedure was easy.
That's a clever way at coming to the solution.
Wow
Somehow I’m much more interested in the approach than the solution.
I am not a math expert just interested but I always seem to be completely oblivious to the way how to!
Guess I’ll watch some more then. ….
Btw , 50 years old and haven’t had math for some 30-35 years. It still does interest me though
About the same with me. I'm 42.
Listening to these videos at this age makes me realize why I struggled in high school. His explanation is great but I feel it's probably not too different from how my teachers used to explain. It's just that back then my young brain couldn't comprehend it and couldn't keep my mind on the task. The moment i got confused and he'd too many questions my mind wondered off.
It's been 26 years since I took a formal math content class. This past year I volunteered to teach algebra and geometry to high school students. A lot of it was like falling off a bike, you never forget. But some of it was surprisingly difficult. Still other parts I had never learned. So far, my favorite day was teaching derivatives... I had never learned them, and didn't understand them while I prepped for the class, but somehow doing it on the board caused something to click in my brain. Good times.
Kind of easier to see the product symmetrically:
√ { (x - 3/2)(x - 1/2)(x + 1/2)(x + 3/2) + 1 }
= √ { (x² - 9/4)(x² - 1/4) + 1 }
= √ { x⁴ - (5/2) x² + 9/16 + 1 }
= √ { x⁴ - (5/2) x² + 25/16 }
= √ { (x² - 5/4)² }
= x² - 5/4
So X would be 71.5...
= 5111
Nice trick. No tricky theorems of number theory are needed. Congrats.
Thanks and welcome !
3:30 haven't you forgotten that sqrt(x^2)=abs(x)?
Very cool, thanks for showing.
No problem 👍
Pretty 😊
Thank you very much
👍
As long as the numbers are consecutive, the first term is multiplied by the last term and 1 is added.70*73+1=5111...Another example: square root(94*95*96*97+1).....94*97+1=9119
Nice!
Thank you! Cheers!
There is actually a decent trick for squaring those big numbers. Now you gotta know the tens trick to be decent for numbers like _4 or _6. All it is is whenever you have something like 72^2, it's just 70^2+70+71+71+72 so it'd be equal 4900+70+142+72=5184
Very elegant thank you
Thank you! 😊
I love seeing all the different ways people thought to solve this in the comments
Beautiful
Thank you! Cheers!
Haven't watched the vid. 71=x, then we get x*(x+2)*(x^2-1)+1 then expand it. Then try to approximate this with (x^2+a*x+b)^2. Apparently, a=+1. Then b has to be +1 or -1 (turns to be -1). Those are boring problems, since we know that the answer exists. Real science is when you don't know whether there's an answer
That's nice but too long. I mean, you got it right but you wrote "evaluate". I understand that it is allowed to approximate the result. Which you can do by :
1) forget the +1 in the square root
2) replace 70*71*72*73 by (71-1)*(71+1)*(72-1)*(72+1)
3) it becomes (71^2-1)*(72^2-1)
4) again, forget the ones
5) It becomes √(71^2*72^2)
6) so 71*72
7) = 5112
Close enough !
Very nice 👍
Thank you 👍
Proof that for any consecutive A,B,C,D the square root of A.B.C.D + 1 is B.C - 1.
To prove this we will need to write the product A.B.C.D + 1 in the form of a square so we can take the square root.
The idea is to get rid of A and D by writing them in terms of B and C using the fact that A,B,C,D are consecutive.
This way B and C will occur in the product twice, i.e. as squares.
The additional +1 is needed to turn the whole thing into a perfect square as we will see when we try this.
*Proof in detail:*
Since A and B are consecutive we have A = (B-1).
Since C and D are consecutive we have D = (C+1).
Their product is
A.D = (B-1)(C+1) = B.C + B - C - 1.
Since B and C are also consecutive we have B - C = -1 so that
A.D = B.C - 2.
Substituting this expression in A.B.C.D gives
A.B.C.D = B.C.(B.C - 2).
Now write B.C as (B.C - 1) + 1, and write (B.C - 2) as (B.C - 1) - 1.
This results in
A.B.C.D = ((B.C-1) + 1)(B.C-1) - 1).
The expression on the right is of the form (a+b)(a-b) which is equal to (a² - b²) according to a well known formula.
Consequently we get
A.B.C.D = (B.C-1)² - 1².
Adding 1 on both sides gives
A.B.C.D + 1 = (B.C - 1)²
of which the square root is B.C - 1.
*Alternative proof:*
Write A, C and D in terms of B
A.B.C.D + 1 =
(B-1).B.(B+1).(B+2) + 1 =
reordering
(B-1).(B+1).B.(B+2) + 1 =
rewrite the last two factors
(B-1).(B+1).((B+1)-1).((B+1)+1) + 1 =
using (a-b)(a+b) = a² - b² twice
(B²-1).((B+1)²-1) + 1 =
write out the product
B².(B+1)² - (B+1)² - B² + 2 =
write out the second (B+1)²
B².(B+1)² - B² - 2B -1 - B² + 2 =
B².(B+1)² - 2B² - 2B + 1 =
(B.(B+1))² - 2.(B.(B+1)) + 1 =
using a²-2a+1 = (a-1)²
(B.(B+1) - 1)²
of which the square root is
B.(B+1) - 1
which is the same as
B.C - 1
Nice! Thank you!
Thank you so much!
Took me too long to figure out. For some reason I got stuck when 71*72 was two more than 70*73 instead of one. After cheating and verifying the answer was an integer I understood how to shift a n(n+2) rectangle so you can add a 1x1 block and make a (n+1)^2. Haven't seen more than the thumbnail yet, but if you use formulae instead of drawing the rectangles I'm going to scream
Well, it was easy enough to follow. To simplify; if you have a rectangle with x rows and x+3 columns, you can move one of the columns to a new row and get x+1 rows and x+2 columns with the new row missing two blocks. So you know 70*73=71*72-2. If you have a rectangle with n rows and n+2 columns, you can make it a n+1 square by moving the last column to a new row and adding one block. That's the +1 from the beginning. So the square has a side of 70*73+1=4900+210+1=5110
@@jollyjoker6340 wow really nice approach
Best and very terse explanation@@jollyjoker6340
Calculator foreva! )))))
Молодец! Отлично! Зачёт!
Thanks and Welcome 🙏❤️
@@learncommunolizer Желаю Успехов!
When you have x=70 the calculation turns out to be so easy
and it is even faster to calculate directly OR even have an estimation from 71*72=5112 plus we know the final digit should be 1 or 9 so we have 5111 or 5119
thus we should introduce a more complicated number for x like 2022
then the answer turns out to be 2023X2024-1=4094551
Good solution, but I solved through the factors' symmetry with respect to 71,5.
Let me notice, that in the general case, at the timing of 3:35, you need to write a |y+1| instead of y+1.
Ну во времена когда не было калькуляторов классная штука
The question is easy the only thing is it has long calculation..
Sqroot [(70*71*72*73)+1]
Sqroot [(70+1)(70 +2)(70 +3)70 + 1]
Let x =70
So, the above expression become:
Sqroot [(x+1)(x+2)(x+3)x + 1]
Sqroot [x^2 + 3^x +2)(x+3)x + 1]
Sqrt[(x^2 + 3^x +2)(x^2+ 3^x) +1] exp-2
Let x^2 + 3^x = y
So exp2 becomes
Sqrt. [(y+2)y +1]
Sqrt [ y^2 +2^y + 1]
Sqrt [y^2 + y + y + 1]
Sqrt [y(y + 1) + 1(y+1)]
Sqrt [ (y+1)^2] =
[ y+1]
y = x^2 + 3x
x = 70
So y become
70^2 + 70x3 = 4900+210= 5110
y = 5110
Final answer is
y+1
= 5110+1
= 5111
4900+ 210
5110
the answer is (70)(73)+1. so if we let x=70 z=73
the question q is sqrt(xz(x+1)(z-1) +1)) noting z-x=3
q²= (xz)(xz+z-x-1)+1 = (xz)(xz+2) +1= (xz)^2+2xz+1 =(xz+1)²
so q( the answer)= xz+1 =70*73+1 =4900+210+1=5111
this question comes up quite a lot so if you get it as multiple choice you will know the answer is first*last+1
👏👏👏
Nice abstract solution/pattern but it seems for this problem a simple hand-calc is faster.
Yet another way of doing it: 70*71*72*73 + 1 = (70*72)*(71*73) + 1 = (71-1)(71+1)(72-1)(72+1) + 1 = (71²-1)(72²-1) + 1 = 71² 72² - 71² - 72² + 2 = 71² 72² - 71 (72-1) - 72(71+1) + 2 = 71² 72² - 71*72 + 71 - 72*71 - 72 + 2 = 71²72² - 2*71*72 + 1 = (71*72 - 1)². Taking the square root, the result is 71*72 - 1 = 5111. (Or you could further calculate: = (70+1)(73-1) - 1 = 70*73 + 73 - 70 - 2 = 70*73 + 1 = 5111.)
I did it in my head and ended up with 5111 in 2 minutes because l have done mental arithmetic for over 30 years - but nice to see how others do it
5111. Took me 10 seconds, although I know the trick that the square root of n(n+1)(n+2)(n+3)+1 is n(n+3)+1, because n(n+1)(n+2)(n+3)+1=n(n+3)(n+1)(n+2)+1=(n^2+3n)(n^2+3n+2)+1=(k-1)(k+1)+1=k^2-1+1=k^2, where k=n^2+3n+1=n(n+3)+1.
thanks!!
You're welcome!
Good Work
Thanks for the visit
@@learncommunolizer can you Promote mein
Ну да, красиво. Вот только если уж взял всё расписывать, то ✓[(у+1)²]= | у+1 | = у+1 (т.к. у>0).
Thanks 👍
Welcome 👍
Потерял корень -5111.
Число под корнем всегда положительное, просто человек забыл про ОДЗ
8 squared is 64, and 7 times 9 is 63. So 70 times 72 is 71 squared minus one, and 71 times 73 is 72 squared minus 1. So we get 71 squared times 72 squared, minus 71 squared, minus 72 squared, plus two.
Haven't got enough fingers and toes
My neighbours won't let me borrow there's
That sheer is endless
70×73 = 5110
71×72 = 4900+70+70+72=5112
5112*5110+1 = sqrt[(5111+1)(5111-1) + 1] = sqrt(5111^2 - 1 + 1) = 5111
(Using (a+b)(a-b) = a^2 - b^2)
I don't why they felt so satisfying.
Excellent penmanship?
There's a simple formula for the square root of 4 consecutive numbers multiplied + 1. Here it is: √ABCD + 1 = BC - 1. So 71 x 72 - 1 = 5111. Here's another example: √16 x 17 x 18 x 19 + 1) = 17 x 18 - 1 = 305. Make your own example and give it a try. It works. Cool beans. 😄😄😄
wow.Thanks❤️❤️❤️ 🤝
70^2+3(70)+1=
4900+210+1=5111
4900 not 490
Hay quá!
Wow that took you 5 minutes? I did it in 10 seconds with 1 easy trick: a calculator
a = 70
sqrt(a(a+1)(a+2)(a+3) + 1) =
sqrt(a(a + 3) * (a + 1)(a + 2)) + 1) =
sqrt((a^2 + 3a)(a^2 + 3a + 2) + 1) =
t = a^2 + 3a
= sqrt(t^2 + 2t + 1) = t + 1 = a^2 + 3a + 1 = 4900 + 210 + 1 = 5111
1) 70*71*72*73+1=26 122 321
2) 5000^2=25 000 000
3) 5100^2=26 010 000
4) 5110^2=26 112 100
5) 5120^2=26 214 400
6) 5111 or 5119?
7) 5111^2=26 122 321
3m 24s, Ok.
This, ladies and gentlemen, is why calculators were invented
I think it would be easier to just multiply the numbers! 😂
Doing sqrt on 2nd degree of X, a person gets absolute value, not the X itself. Yes, I know, here y is positive, but that should be said.
=sqrt[71*72*(71-1)*(72+1)+1] =sqrt[71*72*(71*72-2)+1] =sqrt[(71*72)^2-2*(71*72)+1] =71*72-1
I keep seeing everyone doing this the exact same way. Are you all smoking crack? You need to do no more than basic multiplication.
multiply 70 and 73. You can do this in your head. It's 5110.
multiple 71 and 72. You can do this in your head or use paper and pencil. It's 5112.
Let n = 5111. Why? Because, if you do, here's the quantity under the sq. rt rewritten: square root of [(n-1) * (n+1) +1]
Multiply it out. square root of [n^2 + 1n - 1n -1 + 1]
Square root of n^2. n was 5111, so that's your answer.
While this is all well and dandy, please give me one real life situation where I should calculate without a calculator.
It is about the process and the thinking, not about the solution.
A nice solution but the video is too long. It can be reduced to 1 min.
You missed the second solution of the square root.
Complicou demais.
Formula ad +1 here a=70 , d=73
Gauss until 73 minus Gauss until 70
√701
Usually people rish to rwplace letters with numbers, not theother way around😂
Abs (y+1) ?
I had Calculus and I am like what the hell I never got this in Algebra 2 What level of math are we talking about I didn’t have anything like this in Pre-calculus After seeing this I feel like a complete fool
numerically, by approximation and inspection "5000 give or take a couple of hundred" - good enough for any engineering problem.
Am I the only one who thinks it would be easier to just do all the multiplication long hand?
В столбик посчитать быстрее.
The answer is clearly 42....
I had a quite different idea: I also call the result x, so I can write
x = sqrt(70*71*72*73 + 1)
Square this equation:
x^2 = 70*71*72*73 + 1
Subtract 1:
x^2 - 1 = 70*71*72*73
On the left side, use the 3rd binomic formula:
(x + 1)*(x - 1) = 70*71*72*73
The two factors (x + 1) and (x - 1) of the left side differ by 2, so I try to build such two factors on the right side, too.
Since the factors differ only by 2, the are almost equal, so I try to group the four factors to approximately equal products.
I try to achieve this by grouping the largest and lowest factor, and the two factors in between:
70 * 73 = 70 * (70 + 3) = 4900 + 210 = 5110
71 * 72 = (70 + 1)(70 + 2) = 4900 + 140 + 70 + 2 = 5040 + 72 = 5112
Bingo! 5110 and 5112 differ by exactly 2, so
x + 1 = 5112
x - 1 = 5110
and
x = 5111
and this is the solution to the problem (the sought square root).
How to calculate something that you or your future generation will never calculate in a lifetime.
Sqrt(x^2)=|x|
Exactly! No way to get A mark with such a neglecting, at least at my school and university.
Yes, you will lose marks in a Chinese exam if you do not discuss this.
A Nice Radical Problem: √[(70)(71)(72)(73) + 1] = ?
Let: a = 70
(70)(71)(72)(73) + 1 = (a)(a + 1)(a + 2)(a + 3) + 1 = (a)(a + 3)(a + 1))(a + 2) + 1
= (a^2 + 3a)(a^2 + 3a + 2) + 1 = (a^2 + 3a)^2 + 2(a^2 + 3a) + 1 = (a^2 + 3a + 1)^2
= [a(a + 3) + 1 ]^2= [70(73) + 1]^2 = 5111^2
√[(70)(71)(72)(73) + 1] = √(5111^2) = ± 5111
It comes to something around 5,000 or 6,000
More than 70², anyways
Про модуль мы не слышали?
5111
The answer is wrong! You calculated the answer only when (y+1)>0. You have to calculate it when (y+1)
No. The square root of (y+1)² is equal to |y+1|.
Much easier to just use a calculator
=root{(70×73)×(71×72)+1}
=root{5110×5112+1}
=root{(5111-1)*(5111+1)+1}
=root{5111^2-1^2+1}
=root(5111^2)
=5111
Math makes no sense at all to me!
When would you actually USE this problem in the real world?
Hello why y²+2y+1 is (y+1)²
3min35s
It's a factor, with (y + 1)² you multiply each thing inside the parentheses with itself and the other number then add it all so y * y + y *1 + 1 * y + 1*1 = y² + 2y + 1
Sorry if this is didn't help I'm not that good at math haha
@@seymourcouch-billiter2669
you pretend to be very modest:), thx:)
That's an example for a binomial formula, see en.m.wikipedia.org/wiki/Binomial_theorem for examples and an explanation.
In short, because (a+b)² = a² + 2ab + b²
Just hard try the solution, it's 5111.
That’s the hard way to go about that problem.
Multiply (70•71•72•73) + 1
26,122,320 + 1
26,122,321
Then take the square root
5111
You really overthought that one!
Of oourse you do have to know that square root from memory, as you're not allowed a calculator.
(70*71*72*73)+1=26,122,321
Sq root 25,000,000 = 5000
5100^2 = 26,010,000
5110^2 = 26,112,100
5111^2 = 25,122,321
See how easy? As I said, you overthought the problem. Here’s a quote - “If you’re a hammer, you see every problem as a nail” 😎 So my advice is “Don’t let your schooling interfere with your education” 😎 Good Day, Sir
30 секундную писанину умудрились растянуть на 5 минут. Такие видео совсем дурачки обычно не смотрят можно пропустить половину вспомогательных пояснений.
Does this need some trial and error
Sqrt(x**2) = x and -x !
Перемножить быстрее
Kanka basit bunlar ilkokul seviyesinde
Не вижу простоты