# Puzzles

## 20 December

The integers from 2 to 14 (including 2 and 14) are written on 13 cards (one number per card). You and a friend take it in turns to take one of the numbers.

When you have both taken five numbers, you notice that the product of the numbers you have collected is equal to the product of the numbers that your friend has collected.
What is the product of the numbers on the three cards that neither of you has taken?

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Because 13 and 11 are prime and larger than 7, no-one can have taken them, as their is no way the other person's product could then be the same. Once 13 and 11 are discarded,
the prime factorisation of the product of all the other cards is \(2^{11}\times3^5\times5^2\times7^2\). In order to be shared to give the same product, the powers must be even, and so the other card not taken must be 6 (as 2 and 3 are the numbers with odd powers).

Therefore the product of the three cards is **858**.

## 15 December

There are 5 ways to make 30 by multiplying positive integers (including the trivial way):

Today's number is the number of ways of making 30030 by multiplying.

## 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?