Unusual Dice Game
During a game of dice, a spectator decided to keep track of the rolls. He decided to track 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
Can you provide the numbers rolled on each of the five rolls?
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
Can you provide the numbers rolled on each of the five rolls?
posted by Zaux at
7:39 PM
10 Comments:
1st roll-2 and 5=10
2nd roll-3 and 5=15(+5)
3rd roll-3 and 3=9(-6)
4th roll-4 and 5=20(+11)
5th roll-2 and 6,or,3 and 4=12(-8)
gotta love brute force
and brute force is the WINNER ... you got it Knightmare
anyone have an algebraic solution?
that's above my head but i bet it has something to do with primes
I tried but no luck .... and the published solution to this one only has the dice rolls ... no back up info.
1) X
2) X+5
3) X+5-6=X-1
4) X-1+11=X+10
5) X+10-8=X+2
Rolls must be a product 1 to 36
No product may be a prime >6 (as would require one die to be 1 and the other die >6)
All products must contain >6 must have both of 2,3,4,5,6 as a factor
Analyze X+10
X+10<=36 thus X<=26, 26 is impossible so <=25
X<=25
Analyze X-1
Smallest result is 1
X-1>=1 , X>=2
X is between 2 and 25
2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25
FOR X:
Eliminate primes >6
2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25
Eliminate numbers >6 without both of 2,3,4,5,6 as a factor
2,3,4,5,6,8,9,10,12,15,16,18,20,22,24,25
FOR X+5:
Eliminate primes >6
__X: 2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25
X+5: 7,8,9,10,11,13,14,15,17,19,20,21,23,25,26,27,29,30
Result:
__X:,3,4,5,9,10,15,16,20,21,22,25
X+5: ,8,9,10,14,15,20,21,25,26,27,30
Eliminate numbers >6 without both of 2,3,4,5,6 as a factor
__X:,3,4,5,9,10,15,16,20,21,22,25
X+5: ,8,9,10,14,15,20,21,25,26,27,30
Result:
__X:,3,4,5,10,15,20,25
X+5: 8,9,10,15,20,25,30
FOR X-1:
__X:,3,4,5,10,15,20,25
X-1: 2,3,4,9,14,19,24
Eliminate primes >6
Result:
__X:,3,4,5,10,15,25
X-1: 2,3,4,9,14,24
Eliminate numbers >6 without both of 2,3,4,5,6 as a factor
Result:
__X:,3,4,5,10,25
X-1: 2,3,4,9,24
FOR X+10:
___X:,3,4,5,10,25
X+10:13,14,15,20,30
Eliminate primes >6
Result:
___X:,4,5,10,25
X+10:14,15,20,30
Eliminate >6 without both of 2,3,4,5,6 as a factor
___X:,5,10,25
X+10: 15,20,30
FOR X+2:
___X:,5,10,25
X+2: 7,12,27
Eliminate primes >6
Result:
___X:10,25
X+2: 12,27
Eliminate >6 without both of 2,3,4,5,6 as a factor
Result:
___X:10
X+2:12
X=10
1) 10=X
2) 15=X+5
3) 9=X+5-6=X-1
4) 20=X-1+11=X+10
5) 12=X+10-8=X+2
Answer:
1st roll) 10, 2*5
2nd roll) 15, 3*5
3rd roll) 9, 3*3
4th roll) 20, 4*5
5th roll) 12, 2*6 or 3*4
Cam
thanks Cam ... now I don't feel so bad about not being able to solve it algebraically
1st roll 2x5=10
2nd roll 3x5=15
3rd roll 3x3=9
4th roll 4x5=20
5th roll 3x4=12
Couldn't work any other way than maths with a hammer.
Ah, just seen Cams algebraic approach....
Didn't think to eliminate primes and others...
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.
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