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> 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

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