An algebra problem by A Former Brilliant Member

Algebra Level 3

∏ i = 0 23 ( 1 7 2 i + 1 6 2 i ) = 1 7 x − 1 6 x \large \prod _{i=0}^{23} \left(17^{2^i}+16^{2^i}\right) =17^x-16^x

Solve for an value of x x .


The answer is 16777216.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

X X
Dec 9, 2018

The left side is equal to

( 17 + 16 ) ( 1 7 2 + 1 6 2 ) ( 1 7 4 + 1 6 4 ) . . . ( 1 7 8388608 + 1 6 8388608 ) (17+16)(17^2+16^2)(17^4+16^4)...(17^{8388608}+16^{8388608})

We can times a ( 17 − 16 ) (17-16) to the LHS, so it becomes

( 17 − 16 ) ( 17 + 16 ) ( 1 7 2 + 1 6 2 ) ( 1 7 4 + 1 6 4 ) . . . ( 1 7 8388608 + 1 6 8388608 ) {\color{#3D99F6}(17-16)(17+16)}(17^2+16^2)(17^4+16^4)...(17^{8388608}+16^{8388608})

= ( 1 7 2 − 1 6 2 ) ( 1 7 2 + 1 6 2 ) ( 1 7 4 + 1 6 4 ) . . . ( 1 7 8388608 + 1 6 8388608 ) ={\color{#3D99F6}(17^2-16^2)(17^2+16^2)}(17^4+16^4)...(17^{8388608}+16^{8388608})

= ( 1 7 4 − 1 6 4 ) ( 1 7 4 + 1 6 4 ) ( 1 7 8 + 1 6 8 ) . . . ( 1 7 8388608 + 1 6 8388608 ) ={\color{#3D99F6}(17^4-16^4)(17^4+16^4)}(17^8+16^8)...(17^{8388608}+16^{8388608})

⋮ \vdots

= 1 7 16777216 − 1 6 16777216 =17^{16777216}-16^{16777216}

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