## Tuesday, 23 November 2010

### Solution to last week's problem

What will be the first date from now which when written as XX/XX/XXXX (day/month/year) contains eight different digits?

That was the challenging little problem thrown at the students on University Challenge last week. It's not straightforward! Obviously (mathematicians use this word when we can't bothered to explain our thinking) we have to start by getting the earliest possible year, then the earliest month in that year and finally the earliest day in that month.

My immediate thought was that the earliest possible year, using four different digits, would be 2013. But 2013 will not work because the month must use either 0 or 1 for its leading digit, so we can't use both of these in the year. I then tried 2034. But the problem with that is that the leading digit of the day can be only 0, 1, 2 or 3, so I now realised that we need at least two of the digits 0, 1, 2 and 3 for the first digits in the day and month. So, I then went for the first possible year that uses only two of the digits, 0, 1, 2 and 3: 2045.

I was confident this would work! But, it doesn't. The only month that does not use 0 or 2 is November which repeats the digit 1! So, we can't have both 0 and 2 in the year.

OK, so surely the solution must be 2145? No, that doesn't work either! The month would have to use the 0. So, having now used 0, 1 and 2, the day would have to be 'thirty-something'. But the only possibilities are 30 or 31, which are not available because we have used the 0 and the 1.

Welcome to the year 2345! This works. We can use the earliest month in that year, June (06) and the earliest day in that month (17). So, the solution is 17 June, 2345. But how on earth were those University-Challenged students supposed to work this out on the spot?