Missing digits in a factorial

Here is a problem similar to a question which was on brilliant a few days ago:

Given that 37!=13763753091226345046315979581abcdefgh000000037!=13763753091226345046315979581abcdefgh0000000, determine a,b,c,d,e,f,ga,b,c,d,e,f,g and hh.

Details and assumptions: It is expected that people refrain from using calculators and computers.

#NumberTheory #MathProblem #Math

Note by Bruce Wayne
7 years, 6 months ago

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Comments

We know that 37!37! contain no. 55 occur 88- times and 22 occur 3434 no. of times. So we can say that

1010 occur 88 times . bcz 1010is a product of 5×25\times 2 and there are 2626 no. of times 22 remaining.

So we can say that 37!37! terminating with 88 no. of zeros. So h=0h = 0.

Now for calculation of a,b,c,d,e,f,ga,b,c,d,e,f,g , we use divisibility test of 22.

Like gg must be divisible by 22 and ff must be divisible by 222^2 and and ee must be divisible by 232^3

Similarly dd must be divisible by 242^4 and and cc must be divisible by 252^5 and bb must be divisible

by 262^6 and aa must be divisible by 272^7.

jagdish singh - 7 years, 6 months ago

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Can you elaborate a bit more on your technique of using divisibility tests of 2n2^{n} to find the rest of the digits. I can't quite clearly understand what you mean when you say " ee must be divisible by 232^{3}... and so on". ee 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.

Bruce Wayne - 7 years, 6 months ago
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