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Help I need to count this

 

Catherine rolls a standard 6-sided die five times, and the product of her rolls is .  How many different sequences of rolls could there have been?  (The order of the rolls matters.)

 Mar 4, 2023
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We can use prime factorization to count the number of sequences of rolls that have a given product.

The prime factors of the product of five rolls must be chosen from the set {2, 3, 5}, since these are the prime factors of the numbers on a standard 6-sided die. Specifically, we need to choose the number of 2's, 3's, and 5's that appear in the prime factorization of the product.

Since the product of the rolls is at most 6^5 = 7776, it follows that the largest possible prime factor in the prime factorization of the product is 5. Therefore, we can have at most one factor of 5 in the prime factorization.

Next, consider the number of factors of 2 and 3 in the prime factorization. If the product is even, then there must be at least one factor of 2 in the prime factorization. If the product is a multiple of 3, then the sum of the digits of the product must be a multiple of 3, which means that there must be at least one factor of 3 in the prime factorization. If the product is a multiple of 4, then there must be at least two factors of 2 in the prime factorization. If the product is a multiple of 6, then there must be both a factor of 2 and a factor of 3 in the prime factorization.

Therefore, we can count the number of sequences of rolls that have a given product as follows:

- If the product is odd, then there is exactly one choice for the factor of 5. For each choice of the factor of 5, there are 2^5 choices for the factors of 2 (each of the five rolls can be either even or odd), and 3^5 choices for the factors of 3 (each of the five rolls can be either a multiple of 3 or not). Therefore, there are 1 x 2^5 x 3^5 = 1458 sequences of rolls.
- If the product is a multiple of 2 but not 4, then there are exactly two choices for the factors of 2 (either two of the rolls are even, or all five rolls are even). For each choice of the factors of 2, there is exactly one choice for the factor of 5 (since the product is not a multiple of 5), and 3^5 choices for the factors of 3. Therefore, there are 2 x 1 x 3^5 = 1458 sequences of rolls.
- If the product is a multiple of 4 but not 8, then there are exactly three choices for the factors of 2 (either two of the rolls are even and the other three are odd, or three of the rolls are even and the other two are odd, or all five rolls are even). For each choice of the factors of 2, there is exactly one choice for the factor of 5 (since the product is not a multiple of 5), and 3^5 choices for the factors of 3. Therefore, there are 3 x 1 x 3^5 = 4374 sequences of rolls.
- If the product is a multiple of 8 but not 16, then there are exactly four choices for the factors of 2 (either three of the rolls are even and the other two are odd, or four of the rolls are even and the other one is odd, or two of the rolls are even and the other three are divisible by 3, or all five rolls are even). For each choice of the factors of 2, there is exactly one choice for the factor of 5 (since the product is not a multiple of 5), and 3^5 choices for the factors of 3. Therefore, there are 4 x 1 x 3^5 = 5824 sequences of rolls.
- If the product is a multiple of 16 but not 32, then there are exactly five choices for the factors of 2 (either four of the rolls are even and the other one is odd, or all five rolls are even). For each choice of the factors of 2, there is exactly one choice for the factor of 5 (since the product is not a multiple of 5), and 3^5 choices for the factors of 3. Therefore, there are 5 x 1 x 3^5 = 7280 sequences of rolls.
- If the product is a multiple of 32, then all five rolls must be even. Therefore, there is exactly one choice for the factors of 2, one choice for the factor of 5, and 3^5 choices for the factors of 3. Therefore, there are 1 x 1 x 3^5 = 243 sequences of rolls.

Therefore, the total number of different sequences of rolls is:

1458 + 145

 Mar 5, 2023

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