Can you solve this math problem in two ways?

Asked by LostInParadise (32225

) 2 days ago

A while back I asked how the numbers are arranged on calendar dice, which allow you to display any number in the month from 01 to 31. There is a simple solution, but you have to use one face to show both 6 and 9.

One solution to the problem is to have 0, 1 and 2 on both dice and then have 3, 4 and 5 on one die and 6, 7, 8 and 9 on the other one. For calculation purposes is is convenient to treat 6 and 9 as if they each appeared on a different face, giving the second die 7 faces.

Now a natural question to ask is, what is the total number of numbers from 00 to 99 that can be formed by the two dice? There are two ways of solving this.

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1 Answer

Let me give one way of determining the number of numbers that can be determined.
From 00 to 99 is 100 possible values and we can say that there are no other possibilities, Now we can subtract from 100 the number of numbers that cannot be formed and subtract the number of such cases from 100.

A moment’s thought should convince you that both dice having numbers from 3 to 5 is not possible and having two numbers from 6 to 9 is also not possible and that all other cases are possible. There are 3×3=9 in the first case and 4×4=16 for the other.

9+16=25 and 100 – 25 =75 and that is the answer.

Now see if you can determine another way of solving this problem.

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