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Posted 04 Nov '11
> Here is another approach to find last non-zero digit of n! Last non-zero digit of 10! = 8 For 20! = 8*8 = 4 This can continue for any number of 10s. For example, Last digit of 70! will be given by 8^7 For last digit of 8^7 = 2^21 =( 2^10)^2 * 2 = 76*2 =2 Ther...
Here is another approach to find last non-zero digit of n!

Last non-zero digit of 10! = 8
For 20! = 8*8 = 4
This can continue for any number of 10s.

For example,

Last digit of 70! will be given by 8^7

For last digit of 8^7 = 2^21
=( 2^10)^2 * 2
= 76*2
=2

Therefore, last digit of 70! is 2.

If you are asked for 73!, then just multiply the above 2 with 71*72*73
or simply 2 with 1*2*3

Therefore, last non zero digit for 73! become 2*6 = 2

Hope this helps.

One correction. If
Last non-zero digit of 10! = 8
then the last non-zero digit of 20! = 8x8x2 (this 2 comes from 20)=8
Last non zero digit of 70!= last non zero digit of (8^7) x2x3x4x5x6x7


http://www.pagalguy.com/forum/quantitative-questions-and-answers/55747-cat-2010-concepts-fundas-tips-5.html