Finding the number of zeroes in a factorial value: n! = n * (n – 1) * (n – 2)... 5 * 4 * 3 * 2 * 1 This can also be written as n! = n * (n – 1) * (n –
{"Image DocumentName": "Finding the number of zeroes in a factorial value:", "Image ImageDescription": "n! = n   (n \u2013 1)   (n \u2013 2)   4   3   2   1. This can also be written as. n! = n   (n \u2013 1)   (n \u2013 2)   2   2   3   2   1. Thus in any factorial value the number of 5\u2019s will always be lesser than the number of 2\u2019s and hence the number of zeroes is the number of 5\u2019s in that factorial value. In general, the number of zeroes in n! is given by. where m is the maximum positive integer such that 5m \uf0a3 n and [x] denotes the integral part of x. \u00a9 Mycatstudy.com. All rights reserved.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}