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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, 8 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 $$2^7$$. · 3 years, 8 months ago

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. · 3 years, 8 months ago