MODULAR problem

Question

What will be the remainder when 64! is divided by 71?

Do we need to solve this problem by using MOD theorem or need to expands the factorial?

See also http://math.stackexchange.com/questions/99876/closed-form-for-p-n-pmodp-where-p-is-prime. — lhf, Jun 03 '13 at 14:41

score 4 · Answer 1 · answered Jun 03 '13 at 14:35

Hmm. $71$ happens to be a prime. We have Wilson's theorem telling us that for a prime $p$ we have $$ (p-1)!\equiv-1\pmod{p}. $$ So $70!\equiv-1\pmod{71}$ without further ado. Here $$ 70!=64!\cdot65\cdot66\cdot67\cdot68\cdot69\cdot70\equiv 64!(-6)(-5)(-4)(-3)(-2)(-1)=64!\cdot6!\pmod{71}. $$ Can you do the rest?

score 4 · Answer 2 · answered Jun 03 '13 at 14:35

4

Hint: You can use Wilson's Theorem to get:

$$70! \equiv -1 \mod 71$$

Then use that $70! = 64!\times 65\times...\times70$.

answered Jun 03 '13 at 14:35

Milind

3,944

Key Ideas · Accepted Answer · 2013-06-03T15:06:19.470

2

Hint $\displaystyle\ \ {\rm mod}\ 71\!:\,\ 64! = \frac{70!}{\color{#c00}{70}\cdots \color{#0a0}{65}}\!\!\stackrel{\rm\ Wilson_{\phantom{ I_I}}}\equiv\!\!\!\! \frac{-1}{\color{#c00}{(-1)}\cdots \color{#0a0}{(-6)}}\equiv \frac{-1}{720}\equiv\frac{70}{10}\equiv 7$

edited Jun 03 '13 at 15:06

answered Jun 03 '13 at 15:00

Key Ideas

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MODULAR problem

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