A solo dice game is played thusly: one each turn, a normal pair of dice is rolled. The score is calculated by taking the product, rather than the sum, of the two numbers shown on the dice.

On a particular game, the score for the second roll is five more than the score for the first; the score for the third roll is six less than that of the second; the score for the fourth roll is eleven more than that of the third; and the score for the fifth roll is eight less than that of the fourth. What was the score for each of these five throws?

BTW, Sorry about last week. It's tough to fill in when you're sick.

## Monday, April 26, 2010

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10, 15, 9, 20, 12

ReplyDeleteI agree with Heather.

ReplyDeleteI went about this by putting everything in terms of the first roll. Therefore:

R1 = ?

R2 = R1 + 5

R3 = R1 - 1

R4 = R1 + 10

R5 = R1 + 2

Creating a multiplication table up to 6 and eliminating all of the numbers in it that are not possible you end up with one solution of:

R1 = 10

putting that into the rest of the equations give you the roll order of:

10, 15, 9, 20, 12

Looks like I agree with both of you. Like Dan, I had to write out all the possibilities and see what could happen. I listed out the possible Round 1 scores (products of two dice: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36). Then I checked each one to see if the other round results were possible. For instance, if Round 1 was 1, then round three has to be 0, which is not possible. If round 1 was 2, then round 2 was 7, also impossible. Keep on going till you eliminate all the other possibilities.

ReplyDeleteThat leaves (as Heather and Dan already said) 10, 15, 9 , 20, 12.

I agree with heather Dan and mike

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