The problem here is to find the number of zeros at the end of the number which is the product of the first hundred positive integers. We call this `100 factorial' and write it 100!

For example 4! = 1x2x3x4 = 24 and 5! = 1x2x3x4x5 = 120.

Well done Xinxin of Tao Nan School, Singapore who sent in the solution below in record time and also the Key Stage 3 Maths Club at Strabane Grammar School, N. Ireland and James of Hethersett High School, Norfolk.

The Strabane group said ``We started with a 10 x 10 number square and worked out 2!, 3!, 4! ... *etc.* We quickly realised that the number of zeros at the end of 100! depends on the number of tens appearing within the product which in turn depends on the number of twos and fives'' and Xinxin's solution is re-produced in full
below.

This question basically asks about the number of zeros ending the number `100!'.

Since 2x5 equals 10, the key to answering this question is finding out the number of matches of 2 and 5 occurring in the prime factors of 100!.

Since it is obvious that when 100! is factorised, there are more 2's than 5's. As a result, all the 5's will find matches. Counting the number of 5's gives the number of matches.

First, all the multiples of 5:

5=1x5

10=2x5

15=3x5

20=4x5

...

There are a total of 20 multiples of 5. As a result, we have already found 20 matches, and thus 20 zeros.

However, it is noted that four numbers contribute two 5's to the factors of 100!. They are:

25=5x5

50=2x5x5

75=3x5x5

100=2x2x5x5

As a result, there are, in fact, 24 5's in the factors of 100!.

Thus, 100! ends with 24 zeros.

Done on the 4 ^{th} of November 1998 by:

Li Xinxin

Tao Nan School

Singapore.