## Another dice problem

Posted by Chris on April 25, 2012 – 7:59 pm

During a game of dice, a spectator decided to keep track of the rolls. He recorded the product of the rolls. For example: a roll of 4 and 5 is scored as 20.

1st roll score – (we do not know)

2nd roll score – 5 more than the 1st roll score

3rd roll score – 6 less than the 2nd roll score

4th roll score – 11 more than the 3rd roll score

5th roll score – 8 less than the 4th roll score

What numbers were rolled on each of the five rolls? (Sadly, there are two answers.)

April 25th, 2012 at 11:05 pm

10, 15, 9, 20, 12

Can’t see another solution.

April 25th, 2012 at 11:36 pm

Ah, just re-read the question. You want the actual numbers on the dice at ach roll. Here they are:

10 = 2, 5

15 = 3, 5

9 = 3, 3

20 = 4, 5

12 = 3, 4 or 2, 6.

I suppose that the two ways of getting 12 on the last roll are the two answers that you are looking for?

April 26th, 2012 at 6:06 pm

You got it Wiz. As I doubt anyone else is going to bother with this rubbish problem, here’s my prepared solution:

The only possible rolls are 1*(1 to 6), 2*(1 to 6), etc. =>

1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36.

Let x be the number sought. The rolls are: x,x+5,x-1,x+10,x+2.

4th roll: We must check that for each x, that it’s possible to

roll x+10 e.g. if x = 1, then x+10 = 11 isn’t possible.

That leaves x = 2,5,6,8,10,15,20.

2nd roll: Check x+5 is possible. That leaves x = 5,10,15,20.

3rd roll: Check x-1 is possible. That leaves x = 5,10

5th roll: Check x+2 is possible. That leaves x = 10.

So roll 10,15,9,20 and 12. i.e. 2+5,3+5,3+3,4+5 and 2+6 or 3+4.

April 26th, 2012 at 10:56 pm

Hi Chris. Another possible solution is the following: none of the scores should be divisible by 7, and since they all give different reminders modulo 7, then x= 3 or 6 (mod 7). Since x+10 is a valid score, then it is at most 36, thus x is less or equal to 26. This leaves us with only seven candidates for x: 3,6,10,13,17,20,24. 13 and 17 are primes and are automatically out, 3+10, 6+5, 20-1, and 24-1 are also primes (greater than 6), so the only possibility is 10. The rest is the same…

April 26th, 2012 at 11:11 pm

Chris said: I’ve made Slavy’s comment become problem 2013