Find an integer $r$ with $0 ≤ r ≤ 10$ such that $7^{137 }≡ r (\text{mod} 11)$.

Question

Find an integer r with 0 ≤ r ≤ 10 such that $7^{137}$ ≡ r (mod 11).

So far I have:

$7^{2}$ ≡ 5(mod 11).

$7^{4}$ ≡ $5^2$ ≡ 3(mod 11).

$7^{5}$ ≡ $3*7$ ≡ -1(mod 11).

$7^{135}$ ≡ $(-1)^{27}$ ≡ -1(mod 11).

so $7^{137}=(-1)*(5)≡ -5(mod 11)$.

I feel like I did something wrong but I'm not sure in which part? Can anyone point it out for me?

Check by Fermat's theorem: $7^{10}=1\pmod{11}$, so $7^{137}=7^7=7^5.7^2=-5=6\pmod{11}$. — Chrystomath, Oct 26 '20 at 15:56
So far so good, but note that $-5\equiv6\bmod11$ to find $r$ with $0\le r\le10$. — J. W. Tanner, Oct 26 '20 at 15:53
There's nothing wrong. It remains to note that $-5\equiv6\bmod11$. — Parcly Taxel, Oct 26 '20 at 15:54

score 0 · Answer 1 · answered Oct 26 '20 at 16:03

Your working is fine. You need to end by noting that $-5 \equiv 6 \pmod{11}$ since they asked for a residue between $0$ and $10$.

An alternative approach would be: $7^{137} \equiv (-4)^{137} \equiv -2^{274} \equiv -(2^{5})^{54} \cdot 2^4 \equiv -(32)^{54} \cdot 16 \equiv -(-1)^{54} \cdot 16 \equiv - 16 \equiv -5 \equiv 6 \pmod{11}$

I prefer this approach to reduce the base in magnitude as much as possible to make it easier to take powers. It's much easier to see that $2^5 = 32$, which is one less than a multiple of $11$.

Find an integer $r$ with $0 ≤ r ≤ 10$ such that $7^{137 }≡ r (\text{mod} 11)$.

1 Answers1