How to solve $30^{37} \mod 77$ without calculator?

Question

I tried doing something like this: $$(30^2)^{15}(30^7)\mod 77$$ but it is not effective, maybe someone knows some tips and tricks to solve this ?

Have you heard about the Chinese remainder theorem? It says that because $77 = 7\cdot 11$, where $7$ and $11$ are coprime, you can focus on calculations modulo $7$ and modulo $11$ separately, and put it together to modulo $77$ afterwards. — Arthur, Jun 06 '17 at 20:29
Hint: $30=2\times 3 \times 5$ and $2$, $3$ and $5$ are coprime to $77$ and $\phi(77)=60$ — Amin235, Jun 06 '17 at 20:31
Related: https://math.stackexchange.com/questions/2214567/modulus-in-number-theory/2214578#2214578 — Ethan Bolker, Jun 06 '17 at 20:36

score 5 · Answer 1 · answered Jun 06 '17 at 20:30

5

Use repeated squaring, for example $30^{32}=((((30^2)^2)^2)^2)^2$ and take mod 77 at each step.

answered Jun 06 '17 at 20:30

Akababa

3,169

thanks, you save my time – Walter White Jun 06 '17 at 20:38
In fact, the method of @Akababa is the Square and Multiply Algorithm for Modular Exponentiation – Amin235 Jun 06 '17 at 20:40
+1 for the reminder that you don't have to factor $N$ to calculate powers modulo $N$ efficiently. – Rob Arthan Jun 06 '17 at 20:53

score 4 · Accepted Answer · answered Jun 06 '17 at 20:41

4

By Fermat Little Theorem $$30^{6} \equiv 1 \pmod{7} \\ 30^{10} \equiv 1 \pmod{11} $$

Therefore $$30^{37}=(30^6)^6 \cdot 30 \equiv 30\equiv 2 \pmod{7} \\ 30^{37} \equiv 30^7\equiv (-3)^7 \equiv -3^4 \cdot 3^3 \equiv 7 \cdot 5 \equiv 2\pmod{11} $$

This shows that $7,11 |30^{37}-2$ and hence $77 | 30^{37}-2$. Thus $$30^{37}\equiv 2 \pmod{77}$$ unless I made a small mistake in the computations.

answered Jun 06 '17 at 20:41

N. S.

134,609

1

I got the same answer (but you beat me by 19 seconds). – TonyK Jun 06 '17 at 20:43
@TonyK I know the feeling, happened to me few times in the past too :) – N. S. Jun 06 '17 at 20:43
@N.S thanks for advice. – Walter White Jun 06 '17 at 20:45

score 2 · Answer 3 · answered Jun 06 '17 at 20:39

when confronted with $f(x) \equiv y \pmod {77}$

consider

$f(x) \equiv y \pmod {7}$ and $f(x) \equiv y \pmod {11}$

Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$

$a^{p-1} \equiv 1 \pmod p$

$30^{37} \equiv 2 \pmod 7\\ 30^{37} = (-3)^7\equiv 2 \pmod {11}$

$30^{37}\equiv 2 \pmod{77}$

score 2 · Answer 4 · answered Jun 06 '17 at 20:47

Do the calculation separately modulo $7$ and modulo $11$.

Modulo $7$:

$30^{37}\equiv 2^{37} \equiv 2$

where using Fermat's Little Theorem $2^{36}=(2^6)^6\equiv 1$.

Modulo $11$:

$30^{37}\equiv 8^{37} \equiv 2^{3×37}=2^{111}\equiv 2$

where using Fermat's Little Theorem $2^{110}=(2^{10})^{11}\equiv 1$.

So $30^{37}\equiv 2 \bmod 7$ and $\bmod 11$, thus the Chinese Remainder Theorem gives

$30^{37}\equiv 2 \bmod 77$

How to solve $30^{37} \mod 77$ without calculator?

4 Answers4