You may also like

problem icon

N000ughty Thoughts

Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!

problem icon

Fac-finding

Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.

problem icon

Binomial Coefficients

An introduction to the binomial coefficient, and exploration of some of the formulae it satisfies.

Factorial

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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.