Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Give the remainder when you divide $3*(16!)+2$ by $17$.

I don't have much to go on, but i'm not asking you to simply give me the answer even though that would be great. Could someone show me where I could learn a method to go about solving this problem, and problems similar.


share|cite|improve this question
up vote 5 down vote accepted

Using Wilson's theorem $16!\equiv -1\pmod {17}$

So, $3\cdot (16!)+2\equiv 3(-1)+2\equiv -1\equiv 16\pmod {17}$

share|cite|improve this answer
Makes sense, but what is the remainder? $16$? – student.llama Oct 16 '12 at 19:44
@student.llama, yes. – lab bhattacharjee Oct 17 '12 at 4:48

The operation of computing the remainder of a division of $a$ by $b$ is called modulo, and written $\text{mod}$. An important property of the modulo operation is that it commutes with addition and multiplication, in a way. You have that $$ \begin{eqnarray} (a+b) \text{ mod } c &=& ((a \text{ mod } c) + (b \text{ mod } c)) \text{ mod } c \\ (a*b) \text{ mod } c &=& ((a \text{ mod } c) * (b \text{ mod } c)) \text{ mod } c \end{eqnarray} $$

Restated in these terms, the term you want to evaluate is $$ 3\cdot16! + 2 \text{ mod } 17$$ With the rules above you can simplify that to $$ (3\cdot(16! \text{ mod } 17) + 2) \text{ mod } 17 $$ Now, to evaluate $16! \text{ mod } 17$, you can either work recursively, i.e. split the product in half, evaluate each side (by using the same splitting technique) and then multiplying the result and applying module again. Or you can use the theorem that $(p-1)! \text{ mod } p = p-1$ if $p$ is a prime. With the second method, you immediate get $$ 3\cdot 16 + 2 \text{ mod } 17 = 16 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.