Prove that 2*(34^n)-3*(23^n)+1 is divisible by 726 for all positive integers n.try to prove that 2*(34^n)-3*(23^n)+1 is divisible by 726 for all positive integers n. I've tried rearranging the...

Prove that 2*(34^n)-3*(23^n)+1 is divisible by 726 for all positive integers n.

try to prove that 2*(34^n)-3*(23^n)+1 is divisible by 726 for all positive integers n.

I've tried rearranging the equation, and experimenting and ive come close but never to a proper answer. Would proof by mathematical induction work? i'm not sure but i think it might have to do with that! :) Thank you for any help offered! I'm just really stuck here! thanks! :)

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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`726=2*3*11^2`

If we can show that 2, 3, and 121 divide the expression, then we know 726 does as well.





The expression is divisible by 2:

`2*34^n-3*23^n+1 -= 0*0^n-1*1^n+1 -= 1+1 -= 0 ("mod" 2)`



The expression is divisible by 3:

`2*34^n-3*23^n+1 -= 2*1^n-0*1^n+1 -= 2+1 -= 0 ("mod" 3)`



The expression is divisible by 121:



`34^n=(33+1)^n=((n),(0))33^n + ((n),(1))33^(n-1) + ... + ((n),(n-1))33 + ((n),(n))`

`23^n=(22+1)^n=((n),(0))22^n + ((n),(1))22^(n-1) + ... + ((n),(n-1))22 + ((n),(n))`


33^2 = 3*3*11*11, which is divisible by 121

Similarly, 33^n and 22^n are divisibly by 121, if n>1

So, all the terms are divisible by 121, except for:

`((n),(n-1))33` , `((n),(n-1))22` , and `((n),(n))`



Thus:


`2*34^n-3*23^n+1 -= 2*( ((n),(n-1))33 + ((n),(n))) -3*( ((n),(n-1))22 + ((n),(n))) +1 ("mod" 121)`

`-=2(n*33+1)-3(n*22+1)+1 = 0 ("mod" 121)`

Thus, the expression is divisible by 121





The expression is divisible by 2, 3, and 121. These are all powers of distinct primes. Thus the expression is divisible by 2*3*121=726

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