$7^n+57^m$ and $27^n+4*7^m$ are divisible by 3

Question

While I was helping my daughter with some advanced task from homework, we came to assumption in title. Experiment shows that it is most likely true. But I can't came up with formal proof. Any ideas?

score 4 · Answer 1 · answered Oct 13 '13 at 14:20

4

Hint: $7 = 1 + 2\cdot3$.${}{}{}{}{}$

answered Oct 13 '13 at 14:20

njguliyev

14,473
1
26
43

score 3 · Answer 2 · answered Oct 13 '13 at 14:21

3

Notice that $7\equiv 1 \mod 3$ so the two expressions are $\equiv 0\mod 3$ so your guess is true.

answered Oct 13 '13 at 14:21

score 1 · Answer 3 · answered Oct 13 '13 at 14:21

1

If you know Binomial Expansion then you can expand $7^n$ and $7^m$ using:$$7^n=(1+6)^n$$ $$7^m=(1+6)^m$$ to get the desired result.

answered Oct 13 '13 at 14:21

rnjai

1,774

$7^n+5*7^m$ and $2*7^n+4*7^m$ are divisible by 3

3 Answers3

$7^n+57^m$ and $27^n+4*7^m$ are divisible by 3