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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.)


This post is under “Logic” and has 5 respond so far.
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5 Responds so far- Add one»

  1. 1. Wizard of Oz Said:

    10, 15, 9, 20, 12
    Can’t see another solution.

  2. 2. Wizard of Oz Said:

    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?

  3. 3. Chris Said:

    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.

  4. 4. slavy Said:

    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…

  5. 5. slavy Said:

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

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