Prove that $108^3-7^3$ is a multiple of $101$

Question

Encountered the next problem: prove that $108^3-7^3$ is the multiple to $101$.

As I understand if $c$ is the multiple to $a$ it means that there exist $b\in\mathbb{N}$ that $ab=c$. So i want to factor out $101$ from the expression, and I don't see a way to do it. I will appreciate any ideas.

Do you know about congruence modulo $n$? – egreg May 24 '16 at 15:57 — egreg, May 24 '16 at 15:57
$108^3-7^3=(108-7)(108^2+108\times 7+7^2)$ – Brian Cheung May 24 '16 at 15:57 — Brian Cheung, May 24 '16 at 15:57

score 5 · Accepted Answer · answered May 24 '16 at 15:57

5

Hint: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

answered May 24 '16 at 15:57

MrYouMath

15,833

You beat me to it. – Dustan Levenstein May 24 '16 at 15:59
Thanks. Now it seems so obvious! – Mr Bubble Hubble May 24 '16 at 16:10
Also check out egreg's method. It is more general than this solution :). – MrYouMath May 24 '16 at 16:12

score 3 · Answer 2 · answered May 24 '16 at 16:09

3

Use the properties of congruences, so you don't have to do any factorization.

Since $108\equiv 7\pmod{101}$, you have $$ 108^3\equiv7^3\pmod{101} $$ which is the same as saying that $108^3-7^3$ is a multiple of $101$.

answered May 24 '16 at 16:09

egreg

238,574

thx, for showing this. Have just red about it - learnt something new :-) – Mr Bubble Hubble May 24 '16 at 18:38
Can you say the law that allows to elevate to 3 a congruence? – José Osorio May 24 '16 at 18:41
@JoséOsorio If $a\equiv b\pmod{k}$ and $c\equiv d\pmod{k}$, then $ac\equiv bd\pmod{k}$. – egreg May 24 '16 at 19:41

Prove that $108^3-7^3$ is a multiple of $101$

2 Answers2