Here is a problem similar to a question which was on brilliant a few days ago:
Given that 37!=13763753091226345046315979581abcdefgh0000000, determine a,b,c,d,e,f,g and h.
Details and assumptions: It is expected that people refrain from using calculators and computers.
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2^{34}
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We know that 37! contain no. 5 occur 8− times and 2 occur 34 no. of times. So we can say that
10 occur 8 times . bcz 10is a product of 5×2 and there are 26 no. of times 2 remaining.
So we can say that 37! terminating with 8 no. of zeros. So h=0.
Now for calculation of a,b,c,d,e,f,g , we use divisibility test of 2.
Like g must be divisible by 2 and f must be divisible by 22 and and e must be divisible by 23
Similarly d must be divisible by 24 and and c must be divisible by 25 and b must be divisible
by 26 and a must be divisible by 27.
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Can you elaborate a bit more on your technique of using divisibility tests of 2n to find the rest of the digits. I can't quite clearly understand what you mean when you say " e must be divisible by 23... and so on". e is a single digit number. So I can't comprehend your divisibility tests. I can sense what you are saying but probably you haven't expressed it in the right way. Correct me if I am wrong.