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Missing Digits (Posted on 2004-11-10) Difficulty: 3 of 5
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!

No Solution Yet Submitted by Nick Hobson    
Rating: 2.8000 (5 votes)

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solution.... | Comment 6 of 7 |
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
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