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Talking BIG (Posted on 2017-11-10) Difficulty: 3 of 5
The number (10^(666))! where 666 is the beast number and "!" denotes a factorial is called Leviathan Number.

Given that this number has approximately 6.656×10^(668) decimal digits, evaluate the number of its trailing zeros.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts Really good approximation and a way to get exact Comment 3 of 3 |
If n/4 gives a good upper bound for the number of zeros in n! then maybe it's possible to figure out what to subtract to make it exact.

The number to subtract is 1/4 of 
the number of 1's past the last multiple of 5 +
the number of 5's past the last multiple of 5^2 + 
the number of 5^2's past the last multiple of 5^3 +
etc.

For example, 194 is 4 past 190, 3 5's past 175, 2 25's past 125, and 1 125 past 0.  4+3+2+1=10.  194/4-10/4=46 which is the number of zeros at the end of 194! (38+7+1)

The is equivalent to the sum of digits in the base 5 representation of n.  (194 in base 5 is 1234)

Now for n=10^x, we need powers of 10 expressed in base 5 so we can sum their digits.  These numbers end in n zeros.  Take off the zeros and they look just like powers of 2 expressed in base 5.

We're getting close.  Example with x=6 n=1000000:
2^6 is 64 which is 224 in base 5.  Sum of digits = 8.
1000000/4 - 8/4 = 24998 which is exactly right.  

Problem!  I can't easily write 2^666 in base 5 to sum the digits.  (I went up to 2^25 by hand, it's a slog.  No pattern is apparent in the SOD but the average of the digits is pretty close to 2.)  Now for the good estimate: the average randomly chosen base 5 digit is 2.  The number of base 5 digits in 2^666 is 287.  From this, we estimate the SOD as 2*287=574.  The estimated deficiency is then 574/4=143.5

Charlie gave an exact answer of 143, so this estimate turns out to be very good.  

If someone wants to find the SOD of 2^666 in base 5, please do.  I predict it to be 143*2=286.

(There's a potential problem here.  The SOD of all the powers of 2 from 2 to 25 are all divisible by 4.  286 is not divisible by 4 so I'm wary of my prediction.)




  Posted by Jer on 2017-11-11 22:35:23
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