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.
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.)
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Posted by Jer
on 2017-11-11 22:35:23 |
Really good approximation and a way to get exact
