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Number Theory Level 1

$6^{ n } + 8^{ n }$ is divisible by 7 if:

n

is a prime number

n

is any real number

n

is an odd number

n

is an even number

3 solutions

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Chew-Seong Cheong
Dec 8, 2014

$6^n+8^n = (7-1)^n+(7+1)^n \equiv (-1)^n+1^n \pmod {7}$

$(-1)^n + 1^n = 0$ when $\boxed {n \space is \space odd }$ .

surely it should be $(7-1)^{n} +(7+1)^{n}$ not $(7-1)^{n}+(8-1)^{n}$ as $(8-1)^{n}≠8^{n}$

Brett Hartley - 6 years, 6 months ago

Thanks. Yes you are right. I have edited it.

Chew-Seong Cheong - 6 years, 6 months ago

I didn't really understood how u got -1^n+1^n

Sachin Arora - 6 years, 6 months ago

Sorry, I didn't explain it. How it works is as follows.

$8^n = (7+1)^n$

$=7^n + n(7^{n-1}) +\frac {n(n-1)}{2}(7^{n-2})+...+ \left (\begin {matrix} n \\ r \end {matrix} \right) (7^r)+...+ \frac {n(n-1)}{2}(7^2) + n(7) + 1$

We note that except for the last term, $1$ , all terms of the expression has $7$ as a factor and therefore are divisible by $7$ . Therefore $8^n \equiv 1 \pmod {7}$ . Similarly, $6^n \equiv (-1)^n \pmod {7}$ .

Chew-Seong Cheong - 6 years, 6 months ago

Thank you very much

Sachin Arora - 6 years, 6 months ago

use binomial newton... :)

Aswad Hariri Mangalaeng - 6 years, 6 months ago

Hariom Tiwari
Dec 8, 2014

(7-1)^n + (7+1)^n Using Binomial Expansion: (x-1)^n + (x+1)^n The last terms (independent of 7) in the expansion of (x-1)^n and (x+1)^n are (-1)^n and 1^n respectively. If n is even then remainder left on division by 7 will be 2. If n is odd then the last terms will cancel each other out and hence the given term is divisible by seven when n is odd

Richard Levine
Dec 15, 2014

I did this without modulo, although I prefer Chew-Seong's answer. Though it is trivial to answer the multiple choice by solving for n=2, I decided to see what I could do without modulo arithmetic.

$6^n$ + $8^n$ =7z. Since $6^n$ and $8^n$ are both even numbers, their sum is even. Therefore, $6^n$ + $8^n$ = 2*7z = 14z. Dividing by 14, we get ( $6^n$ + $8^n$ ) / 14 = z.

This can be generalized as ( $x^n$ + $y^n$ ) / (x+y) = z. Since divisible by 7 means no remainder, this will only occur when (x+y) divides evenly. That only occurs when n is odd, because $x^n$ + $y^n$ = (x+y)(...) only when n is odd. When n is even, we cannot factor it evenly. This is a bit more general case, but it seems to answer the question.

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