Given that 38! = 523,022,617,466,601,111,760,007,224,100,abc,def,gh0,000,000, determine, with a minimum of arithmetical effort, the digits a, b, c, d, e, f, g, and h. No calculators or computers allowed!

when we express 38! interms of its prime factors, we get
(2^35)*(3^17)*(5^8)*(7^5)*(11^3)*(13^2)*(17^2)*(19^2)*23*29*31*37
now by clubbing eight 5s and eight 2s we get 8 zeros at the end . so h=0

taking the remaining terms & regrouping them
[2^27]*[3^17]*[(7^5)*(19^2)]*[(11^3)*(13^2)*37]*[(17^2)*23*29*31]
now expanding each group & considering only the last seven digits in the expansion (since we need only 7 digits before the zeros)

[(4217728)*(9140163)]*[(6067327)*(8322743)]*(5975653)
now multiply consecutive terms & take only last 7 digits....   continue this process until you are left with only one number
[(1409664)*(3312961)]*(5975653)
(1855104)*(5975653)
9782912

so a=9, b=7, c=8, d=2, e=9, f=1, g=2, h=0

Posted by suman on 2005-02-02 21:52:22