## Entries are now closed

Winners' names will be printed and online in Imperial 47 in November 2019.

## Ready to test your little grey cells? Imperial’s best minds set the ultimate puzzle challenge.

### 1: Hard

My doctor prescribed some red tablets to be taken: one each day from 1 January this year. They come in packets of ten. Later, my doctor decided that I should also take some green tablets, one each day from 1 February. These come in packs of seven. I am very conscientious about taking my tablets and I was wondering what would be the first date (if any) when I would be opening new packets of red and green tablets on the same day.

Dr Lynda White, Principal Teaching Fellow in Experimental Design, Department of Mathematics

Solution: 22 March

On 31 January I would take the first of a new packet of red tablets, leaving nine in the pack.

Assuming the required date exists, let ‘m’ be the number of packets of red tablets used after the nine tablets in the opened packet have been taken, and let ‘n’ be the number of packets of green tablets used from 1 February up to the day when packs of both colours finish.

Then 9 + 10m = 7n.

Putting n = 1, 2, 3, 4, 5 or 6 does not give an integral value for m, but n = 7 gives m = 4.

So on the 49th day (counting 1 February as day one) I finish both a red pack and a green pack. This will be 21 March. The next day, 22 March, I would open new packets of both colours.

### 2: Very hard

Can you find a nine-digit number using each of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once each, with the following property: the first two digits (reading from the left) form a number divisible by two, the first three digits form a number divisible by three, the first four digits form a number divisible by four etc up to the nine-digit number that is divisible by nine?

Dr Lynda White, Principal Teaching Fellow in Experimental Design, Department of Mathematics

Solution: 381654729

The fifth digit must be 5.

You will need to know that a positive whole number is only divisible by nine if the sum of its digits is divisible by nine, and that a positive whole number is only divisible by three if the sum of its digits is divisible by three.

As each digit is used once, the nine-digit number will always be divisible by nine. This follows as the sum of all nine digits is 45, which is divisible by nine. The sum of the first three digits must be a multiple of three, and so the sum of the fourth, fifth and sixth digits must therefore also be a multiple of three (so that the first six digits form a multiple of three) and the sum of the last three digits must be a multiple of three.

The second digit must be even, the third and fourth digits must give a two-digit number divisible by four, and the sixth, seventh and eighth digits must give a three-digit number divisible by eight. This means that the second, fourth, sixth and eighth digits must consist of 2, 4, 6 and 8 in some order.

There are many ways to attack this problem from hereon, but it is a matter of eliminating arrangements which do not satisfy these requirements. There is no simple test of divisibility by seven, so you must check this at the end, having found one or more nine-digit numbers which satisfy all the other criteria.

### 3: Fiendish

I always had difficulty remembering my PIN number until I noticed the following. My house number has three digits and is different if I write it backwards. If I take the difference between my house number and its reverse I get another three-digit number. If I add this new number to what I get if I reverse it, I get my (four-digit) PIN number. It’s secret because no-one knows where I live. So what is my PIN number?

Professor Jonathan Mestel, Professor of Applied Mathematics, Department of Mathematics