Does it help to know that 6 - 5 = 1 ?

Algebra Level 3

It is given that x , y ∈ Z + x,y\in\Bbb{Z^+} and they satisfy the equation given below:

∏ k = 0 5 ( 5 2 k + 6 2 k ) = 6 x − 5 y . \large\prod_{k=0}^5\left(5^{2^k}+6^{2^k}\right)=6^x-5^y.

What is the value of x + y ? x+y?


Note: This problem is not original. It is adapted from a question posted on Math SE.

512 128 256 32 14 20 64 54

View wiki
Prasun Biswas by Prasun Biswas , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Marvin Chong
Feb 7, 2020

I don't think my explanation is the most elegant nor quickest, but here it goes:

Open the LHS of the equation, which should get you (5+6)( 5 2 5^{2} + 6 2 6^{2} )...( 5 32 5^{32} + 6 32 6^{32} ).

If you multiply both sides by (5-6), the LHS will 'telescope' into 5 64 5^{64} - 6 64 6^{64} (as according to the difference of two squares). Thus, 5 64 5^{64} - 6 64 6^{64} = 5-6( 6 x 6^{x} - 5 y 5^{y} ).

Then, divide both sides by ( 6 x 6^{x} - 5 y 5^{y} ):

5 64 − 6 64 6 x − 5 y \frac{5^{64}-6^{64}}{6^x-5^y} = 5-6

=> - 5 64 − 6 64 5 y − 6 x \frac{5^{64}-6^{64}}{5^y-6^x} = -1

For this equation to be true, 5 64 − 6 64 5 y − 6 x \frac{5^{64}-6^{64}}{5^y-6^x} = 1

Therefore, the numerator and the denominator must have the same value. Naturally, y= 64 \boxed{64} and x= 64 \boxed{64} .

Thus, x + y = 128 \boxed{128}

0 Helpful 1 Interesting 0 Brilliant 0 Confused
Priyesh Pandey
May 11, 2015

yes indeed 6-5 is the hint

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Sualeh Asif
Apr 23, 2015

Hint:

Use the identity ( a − b ) ( a + b ) = a 2 − b 2 (a-b)(a+b)= a^2 -b^2 Let a = 6 ; b = 5 a=6 ;b=5 .An interesting terminating sequence comes up where each term joins to create a new term.

p.s. I'll post a complete solution tomorrow.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Please do that! Don't forget to generalize ∏ k = 0 n ( a 2 k − b 2 k ) \displaystyle \prod_{k=0}^n \bigg (a^{2^k} - b^{2^k} \bigg) .

I think there's a bit of typo in the Challenge Master note. Anyway, here's a hint for you to start on the generalization of this problem:

∏ k = 0 n ( a 2 k + b 2 k ) = { a − b a − b ⋅ ∏ k = 0 n ( a 2 k + b 2 k ) , a ≠ b 2 n + 1 ⋅ ∏ k = 0 n ( a 2 k ) , a = b \large\prod_{k=0}^n\left(a^{2^k}+b^{2^k}\right)=\begin{cases}\displaystyle\frac{a-b}{a-b}\cdot\prod_{k=0}^n\left(a^{2^k}+b^{2^k}\right)~,~a\neq b\\~\\ \displaystyle 2^{n+1}\cdot\prod_{k=0}^n\left(a^{2^k}\right)~,~a=b\end{cases}

Case-1 can be solved using the "terminating sequence / pattern" you mentioned.

Case-2 can be solved using the finite GP sum formula and a nice (trivial) relation involving product and sum notation.

Prasun Biswas - 6 years, 1 month ago

Log in to reply

Good problem.

shivamani patil - 6 years ago

CAN YOU SOLVE IT. I AM NEW HERE AND WANT TO LEARN ABOUT THIS EXPRESSION

Nikhil Raj - 4 years ago

Can you solve it??

Loreto Contreras Sandoval - 1 year, 10 months ago

Log in to reply

He did. The hint he posted is the crux of the solution; please don't wait to be spoon feeded a solution: just multiply both sides of the equation given in the question by 1 = 6 − 5 1=6-5 and use the hint.

Prasun Biswas - 1 year, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...