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 Missing Hexadecimal Digits (Posted on 2011-06-18)
Given that (1C)16! = (3D925Bpqrstuvw00000000000)16 determine, with a minimum of arithmetical effort, the digits p, q, r, s, t, u, v and w. No calculators or computer programs allowed!

 No Solution Yet Submitted by K Sengupta No Rating

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 What is wrong? Comment 2 of 2 |
Just expanding on Charlies post trying to figure out the problem:

3D925Bpqrstuvw00000000000 is 25 hexadecimal digits which seems right.  (1C)16 =(28)10

In decimal 28! = 304888344611713860501504000000
Rather than convert this to hexadecimal lets just get an estimate of the size:  log(304888344611713860501504000000)/log(16) ≈24.486 which indicates 25 digits.  So far so good.  16^.486 ≈ 3.848 which also indicates a first digit of 3.

Now for the zeros at the end:  Since 16 = 2^4 we only need to find the number of factors of 2 and divide by 4 to get the number of zeros.  If the numbers 1 to 28, 14 are even, 7 are also multiples of 4, 3 are also multiples of 8, and 1 is a multiple of 16.  14+7+3+1 = 25.  25/4=6.25 which indicates the number in hexadecimal should end in 6 zeros, not the 11 shown.

It doesnt seem the problem is workable.

 Posted by Jer on 2011-06-21 10:47:33

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