# Puzzles

## Not Roman numerals

The letters \(I\), \(V\) and \(X\) each represent a different digit from 1 to 9. If

$$VI\times X=VVV,$$

what are \(I\), \(V\) and \(X\)?

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For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.

If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.

Now that we know that \(VI\) is 37, we see that

$$37\times X=333,$$

and so \(X=9\) and the final solution is

$$37\times9=333.$$

## Backwards fours

If A, B, C, D and E are all unique digits, what values would work with the following equation?

$$ABCCDE\times 4 = EDCCBA$$

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EDCCBA is a multiple of four, so A is even. A cannot be more than 2, as otherwise EDCCBA would have more digits. So A is 2.

E must therefore be 8 or 9 (as 4 times B is less than E) and 3 or 8 (as 4 times E ends in 2). Therefore E is 8.

Carrying on like this, we find:

$$219978\times4=879912$$

## 10 December

How many zeros does 1000! (ie 1000 × 999 × 998 × ... × 1) end with?

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The number of 0s at the end of 1000! will be equal to the number of times 10 is a factor of it.
For each 10 that is a factor, there will be a 5 and a 2 that are factors. Therfore the number of 0s will be equal to the number of times
5 is a factors of 100! (because 5 is larger than 2).

Therefore the number of 0s at the end of 1000! will be equal to
\(\left\lfloor\frac{1000}{5}\right\rfloor+\left\lfloor\frac{1000}{25}\right\rfloor+\left\lfloor\frac{1000}{125}\right\rfloor+\left\lfloor\frac{1000}{625}\right\rfloor\).
This is **249**.

## 17 December

What is the smallest number, n, such that n! ends with 50 zeros?

## One hundred factorial

How many zeros does \(100!\) end with?

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The number of zeros at the end of a number is the same as the number of 10s in the product that makes the number. Each of these 10s is made by multiplying 5 by 2.

There will be more even numbers than multiples of 5 in \(100!\), so the number of 5s will tell us how many zeros the number ends in.

In \(100!\), there will be 20 multiples of 5 and 4 multiples of \(5^2\). This means that \(100!\) will end in 24 zeros.

#### Extension

How many zeros will \(n!\) end in?