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Missing digits in a factorial

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.

Note by Bruce Wayne
3 years, 11 months ago

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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 \(10\)is a product of \(5\times 2\) 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 \(2^2\) and and \(e\) must be divisible by \(2^3\)

Similarly \(d\) must be divisible by \(2^4\) and and \(c\) must be divisible by \(2^5\) and \(b\) must be divisible

by \(2^6\) and \(a\) must be divisible by \(2^7\).

Jagdish Singh - 3 years, 11 months ago

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

Bruce Wayne - 3 years, 11 months ago

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