Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Without using a calculator, how can we solve the following?

  1. How do we find the number of zeros at the end of $600!$
  2. What are the last 3-digits of $171^{172}$?
  3. What is the sum of all positive numbers less than or equal to $61$, which are divisible by $3$ as well as by $5$?
share|cite|improve this question


1) The number of zeroes at the end of the decimal expansion of the natural number $N$ is $k$ if and only if $10^k$ divides $N$ but $10^{k+1}$ does not.

Thus, if you want to know how many zeroes does $600!=600\cdot599\cdot598\cdot\dots\cdot3\cdot2\cdot1$ end with, you need to compute the factors $10$. Since $10=2\cdot5$ and $2<5$ that's the same as counting the factors of $5$ of which there is one every $5$ numbers. But that's not the entire story: the number $25=5^2$ gives so far only a contribution of $1$ to the count, whereas it contains two factors $5$. Thus in order to get the correct answer you need to ...

2) You want $171^{172}\bmod 1000$. Note that $$ 171^{172}\equiv(171^2)^{86}\equiv241^{86}\equiv(241^2)^{43} \equiv81^{32+8+2+1}\equiv81^{32}81^881^281\equiv\cdots. $$ Keep going.

3) A number is divisible by $3$ and $5$ if and only if it is divisible by $15$. Thus, what are the summands less than $61$?

share|cite|improve this answer
! could you explain the all questions by more detail please... – gama Nov 12 '12 at 10:16
gama, do you have anything to contribute, at all? – Gerry Myerson Nov 12 '12 at 11:41
I added some detail, but you should really work out the rest by yourself – Andrea Mori Nov 12 '12 at 11:54
@QAndrea Mori!Thank you so much for your explanation. – gama Nov 14 '12 at 6:17

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.