Solution for Cheryl's birthday

It is 16 July! Below is the explanation of this very pleasing problem.

Cheryl has told A and B that her birthday is one of these ten dates:

May 15, 16, 19

June 17, 18

July 14, 16

August 14, 15, 17

She has told A the month and B the number of the day, and they both know this.

A says, 'I do not know C's birthday.

This tells us nothing, because whatever month he had been told A would not be able yet to deduce the birthday.

But B would know C's birthday if he had been told that the day was, say, 19: because 19 occurs in the list of options only in May. Likewise B would know the birthday if had been told that the day was 18: because this occurs only in June.

So, then A says:

'... but I know for sure that B does not know either.'

A would not be able to say this, if the month he had been told was May or June, because for each of these two months there is the possibility that B actually knows the birthday already.

From this B (and we) can deduce that the month of C's birthday is neither May nor June.

B replies: 

'At first I did not know C's birthday, but I do now.'

This means that the number of the day (which B knows) must occur in either July or August but NOT in both these months. For example, if B knew the day was 17 then he would know it had to be August 17. Likewise, B would know the birthday if the day was 16 (July) or 15 (July).

A has deduced all this as well! He knows that the day is 15, 16 or 17. But A also knows the month. And he then replies:

'Now I know as well!

Now A could only know the birthday if the month (which he knows) has only one of these three options in the list: 15, 16 or 17. That month is July.

This is how we (and B presumably now) know that Cheryl's birthday must be 16 July.

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?

Mathematical Problem 4: Solution

This is the solution to the problem set on 9 April.

We observed that if a daughter is born in '66 and her mother in '44 then in the year that the mother turns 66, the daughter turns 44.

Was this an unusual coincidence? Of course not! It happens for any two people! It would be fun to try this as a little investigation with some primary school children.

For example, if A was born in 1957 and B was born in 1971, then in the year that A turns 71 B turns 57. Why? Because the age of A minus the age of B (once they have both had their birthdays) is always equal to the birth year of B minus the birth year of A. If A is born 14 years earlier than B then B is always 14 years younger than A. So when A's age equals B's birth year (ignoring the 1900) then B's age (14 less than this) must equal A's birth year.

Why had I never noticed that before?!

There is a complication if they are born in different centuries, – this requires the older person to pass 100 before the phenomenon occurs. For example, Catherine was born in 1966 and her son, Jack, was born in 2001. So, the year Jack turns 66, Catherine will turn 101!

Mathematical Problem 3: Solution

This is the solution to the problem set on 1 April 2010. This was about the metric/imperial curiosity that 10 feet 3 inches converts to 3 metres 10 centimetres, rounded down to the nearest 5 cm. I wondered if there were other lengths like this where you just swap over the two numbers to convert from feet and inches to metres and centimetres. For reference purposes 1 metre is about 3.28084 feet.

Well, the obvious one to try next is 20 feet 6 inches – which works, because it converts to 6 m 20 cm, when rounded down to the nearest 5 cm. But 30 feet 9 inches does not work, because of the rounding error; it actually rounds down to 9 m 35 cm.

It's not too difficult to check all the possibilities. Because of the instruction to round down to the nearest 5 cm, we need check only imperial lengths that start 0 feet, 5 feet, 10 feet, 15 feet, and so on. Then because the number of inches cannot exceed 11, the number of metres cannot exceed 11. So we do not need to check lengths beyond 11 metres 95 cm, which is under 40 feet.

So, I've checked all the possibilities and all I can offer you are these two results:

10 ft 3 ins = 3 m 10 cm and 20 ft 6 ins = 6 m 20 cm

(both metric lengths rounded down to the nearest 5 cm).

Mathematical Problem 1: Solution

Here's a solution to Mathematical Problem 1, posted on 13 March 2010.

With 2 players, A wins a match of 29 rounds when A has won 15 rounds. Even if A loses all the remaining rounds B will only have won 14.

The principle is that A would still win even if all the remaining games were to be won by B (A's closest rival).

With 3 players, A wins when a is greater than b added to the number of rounds still to be played (assuming B has the second highest number of wins so far). The number of rounds still to be played is 29 – (a + b + c). So, algebraically:

a > b + [29 – (a + b + c)] which simplifies to 2a > 29 – c.

With 4 players, A wins when

a > b + [29 – (a + b + c + d)] which simplifies to 2a > 29 – (c + d).

The general solution , with any number of players and n rounds, is that A wins when

a > b + [n – (a + b + c + d + e + ...)] which simplifies to 2a > n – (c + d + e + ...)

The surprise in this result is that b does not appear in the simplified rule! Provided B has won more rounds than C, D, E and so on, the actual number that B has won does not make any difference to when A is declared the winner.