Show 666 divides 1296^n - 666n + 36 for all integers n less than or equal to 1.

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- Apr 9th 2009, 07:53 AMmodi4helpInteger division
Show 666 divides 1296^n - 666n + 36 for all integers n less than or equal to 1.

- Apr 9th 2009, 11:01 AMOpalg
If you mean "... for all

*positive*integers n less than or equal to 1," then the result is true. In fact, there is only one such integer, namely n=1, and you can easily verify that 1296–666+36 = 666.

If you allow n to be zero or negative, or if you meant "... for all integers n*greater*than or equal to 1," then the result is false. - Apr 9th 2009, 11:21 AMmodi4help
Can you explain more how can I show it please.

sorry i wrote the question wrong.

Show 666 divides 1296^n - 666n + 36 for all integers n greater than or equal to 1. - Apr 9th 2009, 11:37 AMOpalg
Sorry, I also made a mistake, in my previous comment. The result

*is*true for all integers n greater than or equal to 1. Here's the reason.

Start by noticing that $\displaystyle 1296 = 36^2$ and $\displaystyle 666=18\times37$. Also, $\displaystyle x^n-1$ is always divisible by $\displaystyle x-1$, and so $\displaystyle 1296^n - 1$ is divisible by $\displaystyle 1296-1 = 35\times37$. Hence $\displaystyle 1296^n - 666n + 36 = (1296^n -1) - 37(18n -1)$ is divisible by 37. But $\displaystyle 1296^n - 666n + 36$ is also divisible by 18. Therefore it is divisible by 666.