Find $19^{92} \pmod {92}$

Question

How can I find $19^{92}\pmod{92}$?

I am completely stumped.

I thought of calculating $19^{92} \pmod{23}$ and $19^{92} \pmod{4}$.. ( because $23\cdot4 = 92$).

But I don't know the modulo operation to proceed.

Possible Duplicate: http://math.stackexchange.com/questions/195634/how-do-you-calculate-the-modulo-of-a-high-raised-number and http://math.stackexchange.com/questions/81228/how-do-i-compute-ab-bmod-c-by-hand — , May 08 '15 at 10:58
To obtain $19^{92} \pmod{92}$, type 19^{92} \pmod{92} in math mode. — N. F. Taussig, May 08 '15 at 11:00

score 3 · Accepted Answer · answered May 08 '15 at 10:58

3

$$92=23\times 4\quad\textrm{and}\quad \gcd(23,4)=1$$

Compute the residue modulo $23$ and $4$ first. We'll use Fermat's Little Theorem for the modulo $23$ part since it's a prime moduli.

$$19^{92}\equiv 19^{92~\bmod~22}\equiv 19^4\equiv (361)^2\equiv 16^2\equiv 3\pmod{23}$$

$$19^{92}\equiv (20-1)^{92}\equiv (-1)^{92}\equiv 1\pmod4$$

Use Chinese Remainder Theorem now.

answered May 08 '15 at 10:58

Prasun Biswas

2

${\rm mod}\ \color{#c00}{23}!:,\ 19^4\equiv (-4)^4\equiv 256\equiv \color{#c00}{23}0+\color{#c00}{23}+3\equiv 3\ \ \ $ – Bill Dubuque May 08 '15 at 14:09
@BillDubuque, I'm wondering why I missed such a trivial observation! Anyway, thanks for your comment. :) – Prasun Biswas May 08 '15 at 14:36

score 2 · Answer 2 · answered May 08 '15 at 10:58

2

Using Euler-Fermat, we have

$$19^{\varphi(92)} = 19^{44} \equiv 1 \mod 92$$

Thus $$19^{92} \equiv 19^{92-2\cdot44} = 19^4 \equiv 49 \mod 92$$

The last step can be done using a calculator.

answered May 08 '15 at 10:58

wythagoras

2

The last step can be done using a brain! $\tag{}$ ${\rm mod}\ \color{#c00}{92}!:\ 19^2 \equiv 19(5\cdot 4-1),\equiv, \smash[t]{\overbrace{\color{#c00}{95}}^{\large\color{#c00}3}}\cdot 4-19,\equiv -7.\ \ \ $ Alternatively, $\tag{}$ ${\rm mod}\ \color{#c00}{92}!:\ 19^2\equiv (20!-!1)^2\equiv 4\cdot\underbrace{\color{#c00}{100}}_{\Large\color{#c00} 8}-40+1\equiv -7\ \ \ $ – Bill Dubuque May 08 '15 at 14:33

score 0 · Answer 3 · answered May 08 '15 at 12:05

$$19\equiv-1\pmod4,19^{92}\equiv(-1)^{92}\equiv1\ \ \ \ (1)$$

$$19\equiv-4\pmod{23}\implies19^{92}\equiv(-4)^{92}$$

$(-4)^{92}=4^{92}=(2^2)^{92}=2^{184}$

$184\equiv8\pmod{22}\implies2^{184}\equiv2^8\equiv3\pmod{23}\ \ \ \ (2)$

Apply CRT on $(1),(2)$ to find $$19^{92}\equiv49\pmod{92}$$

3 Answers3