Evaluate $12^{23} \pmod {73}$

Question

$12^{23} \equiv x \pmod{73}$

What is the smallest possible value of x?

Having such a big exponent, it is difficult to use calculator to calculate. May I know is there any simpler way to do so?

Calculate it by repeated squaring and occasional multiplication by $12$, reducing mod $73$ at each step. Surely asked and answered previously on this site, though finding it may not be so easy. — Gerry Myerson, Mar 10 '13 at 10:00
Perhaps it's a duplicate of http://math.stackexchange.com/questions/119374/modular-exponentiation --- have a look! — Gerry Myerson, Mar 10 '13 at 10:03

score 6 · Accepted Answer · edited Mar 10 '13 at 10:18

6

Since $12^2=144\equiv-2\pmod{73}$, we have $12^{12}\equiv(-2)^6\equiv64\equiv-9\pmod{73}$ and $12^{10}\equiv(-2)^5\equiv-32\pmod{73}$. Now $12^{23}=12^{12}12^{10}12\equiv(-9)(-32)12$...

edited Mar 10 '13 at 10:18

Aang

14,672

answered Mar 10 '13 at 10:08

Dennis Gulko

15,640

score 2 · Answer 2 · answered Mar 11 '13 at 03:54

2

$\rm mod\ 73\!:\ \ 12^{24} \equiv (-2)^{12}\equiv (-8)^4 \equiv (-9)^2 \equiv \color{#C00}8$

Therefore $\, 12^{23}\! \equiv \dfrac{12^{24}}{12\ \ }\! \equiv \dfrac{\color{#C00}8}{12}\equiv \dfrac{48}{72} \equiv \dfrac{48}{-1} \equiv 25 $

answered Mar 11 '13 at 03:54

Math Gems

19,574

score 0 · Answer 3 · edited Mar 10 '13 at 10:21

0

$12^2\equiv(-2)\pmod{73}$

$12^{16}\equiv(-2)^{8}\pmod{73}\equiv37\pmod{73}$

$12^6\equiv(-8)\pmod{73}$

$12^{23}\equiv(12)(-8)(37)\pmod{73}\equiv-48\pmod{73}$

$\implies 12^{23}\equiv25\pmod{73}$

edited Mar 10 '13 at 10:21

Aang

14,672

answered Mar 10 '13 at 10:18

Inceptio

7,881

Evaluate $12^{23} \pmod {73}$

3 Answers3