Saturday 21 February 2004 at 11:06 pm | In Articles | Post Comment

Inspired by a test posting on S.O.S. Mathematics CyberBoard

Factorials are fascinating. They are obtained by multiplying the numbers 1, 2, 3, 4, 5 … together and are written using the ! sign. Thus

1! &=&1&=&1 \\
2!&=&1\times2&=&2 \\
3!&=&1\times2\times3&=&6 \\
4!&=&1\times2\times3\times4&=&24 \\
5!&=&1\times2\times3\times4\times5&=&120 \\
&&\cdots \\
10!&=&1\times2\times3\times\ldots\times10&=&3628800 \\

Factorials get large very quickly so 25! = 15 511 210 043 330 985 984 000 000 and as it gets larger there are more and more zeros.

Question: How can you find the number of zeros in 102! without calculating it?
Answer: All you do is count the number of times powers of 5 divide into 102 and add them up. So
102\div5&=&20& \textrm{plus a remainder} \\
102\div5^2&=&4& \textrm{plus a remainder} \\
102\div5^3&=&0& \textrm{plus a remainder - get 0 so no higher powers needed}

Throw away the remainders and we get that there are 20+4=24 zeros in 102!
Now find how many zeros there are in 2004!

Can you show why this method works? Can you generalise it? Have a go, then look for answers on the internet.

Mathematicians will recognise this as a special case of
\textit{The highest power of a prime p dividing n! is }
\sum_{k=1}^\infty \left[\frac{n}{p^k}\right]
which is not difficult to prove by induction.

More on factorials at mathworld

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