Factorials

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

\begin{array}{llllr} 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 \\ &&\cdots \end{array}

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
\begin{array}{llrl} 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} \end{array}

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
\par\boxed{\begin{displaymath} \textit{The highest power of a prime p dividing n! is } \sum_{k=1}^\infty \left[\frac{n}{p^k}\right] \hspace{0.5em} \end{displaymath}}
which is not difficult to prove by induction.

More on factorials at mathworld

No Comments yet »

RSS feed for comments on this post. TrackBack URI

Leave a comment

XHTML: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>

Powered by WordPress with Pool theme design by Borja Fernandez.
Entries and comments feeds. Valid XHTML and CSS. ^Top^