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.

No vote yet

4 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWe 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 \(2^7\).

Log in to reply

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.

Log in to reply