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# Help appreciated

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Catherine rolls a standard 6-sided die five times, and the product of her rolls is 900. How many different sequences of rolls could there have been? (The order of the rolls matters.) Here's what I've tried so far: I listed the prime factorization of 900, which is 2*2*3*3*5*5. Using this, I tried to construct sets of five integers and see how many ways we can reorder them. Currently I can only think of the sets {4, 3, 3, 5, 5} and {2, 6, 3, 5, 5}. Could anyone please help?
May 10, 2023

#1
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There are 3 combinations as follows:

(1, 5, 5, 6, 6) ==5!/2!2! ==30 permutations
(2, 3, 5, 5, 6) ==5!/2!    == 60 permutations
(3, 3, 4, 5, 5) ==5! /2!2! ==30 permutations

Total ==30 + 60 + 30 == 120 different sequences of rolls.

May 10, 2023
#4
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Ahhh yes, I failed to consider the combinations with 1s in them. Thank you so much, and I now understand.

seventythree  May 12, 2023
#2
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To find the number of different sequences of rolls that could have resulted in a product of 900, we need to consider the prime factorization of 900.

The prime factorization of 900 is:

900 = 2² * 3² * 5²

We can distribute these prime factors among the five rolls of the die. Since the order of the rolls matters, we can think of each roll as a "slot" where we can place one of the prime factors.

Let's consider the number of ways to distribute the prime factor 2² = 4 among the five rolls. We can have:

4 in the first roll, 0 in the second, third, fourth, and fifth rolls.

3 in the first roll, 1 in the second, 0 in the third, fourth, and fifth rolls.

2 in the first roll, 2 in the second and 0 in the third, fourth, and fifth rolls.

2 in the first roll, 1 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

1 in the first roll, 1 in the second, 1 in the third, and 2 in the fourth and fifth rolls.

Now, let's consider the number of ways to distribute the prime factor 3² = 9 among the remaining slots. We can have:

9 in the first roll, 0 in the second, third, fourth, and fifth rolls.

8 in the first roll, 1 in the second, 0 in the third, fourth, and fifth rolls.

7 in the first roll, 2 in the second and 0 in the third, fourth, and fifth rolls.

6 in the first roll, 3 in the second, 0 in the third, fourth, and fifth rolls.

6 in the first roll, 2 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

5 in the first roll, 4 in the second, 0 in the third, fourth, and fifth rolls.

5 in the first roll, 3 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

4 in the first roll, 4 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

4 in the first roll, 3 in the second, 2 in the third, and 0 in the fourth and fifth rolls.

3 in the first roll, 3 in the second, 2 in the third, and 1 in the fourth and fifth rolls.

Finally, for the remaining prime factor 5² = 25, we have only one possibility:

25 in the first roll, 0 in the second, third, fourth, and fifth rolls.

By multiplying the number of possibilities for each prime factor, we find the total number of different sequences of rolls:

5 * 10 * 1 = 50

Therefore, there are 50 different sequences of rolls that could have resulted in a product of 900.

May 10, 2023
#3
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I think answer 1 is the correct one.  :)

May 10, 2023