{"id":10,"date":"2004-02-21T23:06:00","date_gmt":"2004-02-21T23:06:00","guid":{"rendered":"http:\/\/www.sixthform.info\/maths\/?p=10"},"modified":"2009-11-17T23:44:57","modified_gmt":"2009-11-17T23:44:57","slug":"factorials","status":"publish","type":"post","link":"https:\/\/www.sixthform.info\/maths\/?p=10","title":{"rendered":"Factorials"},"content":{"rendered":"

Inspired by a test posting on S.O.S. Mathematics CyberBoard<\/a> <\/i><\/font><\/p>\n

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

\\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}<\/p>\n

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.<\/p>\n

Question:<\/b> How can you find the number of zeros in 102! without calculating it?
\nAnswer:<\/b> All you do is count the number of times powers of 5 divide into 102 and add them up. So
\n\\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}<\/p>\n

\nThrow away the remainders and we get that there are 20+4=24 zeros in 102!
\nNow find how many zeros there are in 2004!<\/p>\n

Can you show why this method works? Can you generalise it? Have a go, then look for answers on the internet.<\/p>\n

Mathematicians will recognise this as a special case of
\n\\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}}
\nwhich is not difficult to prove by induction.<\/p>\n

More on factorials at mathworld<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

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 Factorials get large very quickly so 25! = 15 511 210 043 330 985 984 000 000 and as it gets […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-10","post","type-post","status-publish","format-standard","hentry","category-articles"],"_links":{"self":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/10","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=10"}],"version-history":[{"count":3,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/10\/revisions"}],"predecessor-version":[{"id":178,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/10\/revisions\/178"}],"wp:attachment":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=10"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=10"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=10"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}