Is $11^{11296}-1$ divisible by 11 and/or 12? It is easy to see that is divisible by 10 as $x^n-1$ has factor $x-1$. I can't see how to apply Fermat's or Eulers theorem of congruence (if applicable)? What would be the approach to test divisibility by 11 and/or 12?
Asked
Active
Viewed 108 times
0
-
1Hint: you don't need those theorems to raise integers congruent to $0$ and $-1$ to arbitrary powers. – Bill Dubuque May 29 '19 at 13:57
-
$11^{11296}$ is divisible by $11$; if $11^{11296}-1$ were divisible by $11$ too, then their difference would be... – J. W. Tanner May 29 '19 at 13:59
-
@J.W.Tanner Note the "using congruence" in the title. – Bill Dubuque May 29 '19 at 14:00
-
Ok @BillDubuque: $11^{11296}\equiv0\pmod{11}$ so $11^{11296}-1\equiv...$ – J. W. Tanner May 29 '19 at 14:01
-
1Do you know the basic Congruence Rules? – Bill Dubuque May 29 '19 at 14:02
1 Answers
3
Divisibility by $11$: it's not, because we know $11^{11296}$ is divisible by $11$, so subtracting $1$ gives a number that is not divisible by $11$.
Divisibility by $12$: $11 \equiv -1$ (mod $12$) so $11^{11296} \equiv (-1)^{11296} $ (mod $12$). This is equivalent to $1$ and therefore subtracting $1$ gives you $0$. This means the number is divisible by $12$ as it is equivalent to $0$ in a mod $12$ system.
J. W. Tanner
- 60,406
neopalm2050
- 76