nkmungila wrote:

Jack has a few sixfaced dice. Each die has numbers 3, 5, and 6 each printed on two faces. He throws all dice and multiplies

the digits displayed on the top surface of each die. If the product he got is 243,000, how many dice does Jack have?

A. 8

B. 11

C. 12

D. 14

E. More than 14

et’s break 243,000 into prime factors:

243,000 = 243 x 1000 = 3^5 x 10^3 = 3^5 x 2^3 x 5^3

We have the prime factorization of 243,000; however, since there are only 3s, 5s, and 6s (but not 2s) on the faces of the dice, the factor 2 must be a factor of 6. Thus, we have to redistribute the factors so that the factorization only contains factors of 3, 5, and 6:

243,000 = 3^5 x 2^3 x 5^3 = 3^2 x 3^3 x 2^3 x 5^3 = 3^2 x 6^3 x 5^3

Since now all the factors are 3s, 5s, and 6s, the sum of the exponents will be the number of dice:

2 + 3 + 3 = 8

Therefore, there are 8 dice.

Note: If you are uncertain about why the number of dice is equal to the sum of the exponents, think of more straightforward outcomes of dice rolls: If the product of the dice rolls were, say, 125, you would know that each of the 3 dice showed a 5. Expressing 125 as 5^3, you see that the exponent 3 tells how many dice were rolled. Similarly, if the product of the dice rolls were, say, 135, then you would factor 135 as 5 x 27 = 5^1 x 3^3, and this would indicate that there were 1 + 3 = 4 dice and they showed 5,3,3,3 on their faces.

Answer: A

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