Find the last two non zero digits in 2017!

Question

A four digit positive number having all non-zero distinct digits is such that the product of all the digits is least. If the difference of hundreds digit and tens digit is 1.
How many possibilities are there of such number?

Answer 1

last digit of 2017! = last digit of (2^(A))x A! x B! where A is largest multiple 5 of 2017 which is 2015 and B is 2 ie., remainder
= remainder of (2^(403))x 403! x 2! ------ 1 { remainder of 2! is 2 and 2^403 is 8 (powers of 2 are cyclic between 2, 4, 8 & 6} }
For the remainder of 403! (2^(80))x 80! x 3! ------ 2
For the remainder of 80! (2^(16))x 16! x 0! ------ 3
For the remainder of 16! (2^(3))x 3! x 1! ------ 4
=> 8 x 6 x 1 = 48 ===> 8 ( the last digit }.
resubstituting 4's remainder back in 3 and working back we get the last non zero digit of 2017! as 8.

Answer 2

Last two non zero digits in 2017 are, 1 and 7

Comments

Note that the number is 2017 factorial in other words 2017×2016×2015×2014×........×3×2×1 = 2017!