Expert: Pawan Musale - 9/8/2009

Question
My daughter is in 7th grade and come home with this math problem that I cannot help to solve.
1. Find the sum of all positive proper fractions (not necesarily in lowest term) whose numerators are positive integers and whose denominators are positive integers less or equal to 100.
2. Determine the largest positive integer N such that 1005! is divisible by 10 to the power of N.  

Answer
Jasmine,

This question was not asked to me, but still I will try to answer it.

I need to think on 1st question. But I have an answer for 2nd question:

1005! means 1005 factorial which is:

1005! = 1005 * 1004 * 1003 * 1002 * 1001 * 1000 * 999 * ......... * 3 * 2 * 1 ------#1

Now we have to divide this with 10^N.

Now look at expression #1 above. It is clear that 1005! can be divided by maximum 1000 (There is no other number/multiplier greater than 1000 in 1005!)

So 1000 = 10^3

This means that largest positive integer N = 3.

I hope this solves your half problem.

All the best. :)

Regards,
Pawan