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Roberto rolls a standard 6-sided die five times, and the product of his rolls is 900. How many different sequences of rolls could there have been? (The order of the rolls matters.)

 Mar 14, 2023
 #1
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To get a product of 900, we need to find the prime factorization of 900: $900=2^2 \cdot 3^2 \cdot 5^2$.

Since we're rolling a standard 6-sided die, the only possible prime factors of each roll are 2, 3, and 5 (which are the prime factors of 6).

We need to distribute the prime factors of 900 among the five rolls in such a way that we get 2, 3, or 5 as the outcome of each roll. Note that each factor can appear zero or more times in each roll.

To do this, we can use the following technique: we represent each roll as a string of three digits, where each digit is either 0, 1, or 2, representing the number of times the factor 2, 3, or 5 appears in that roll. For example, the roll (2, 2, 5) would be represented as the string "220".

Now, we need to count how many such strings of length 15 (5 rolls, each represented by a string of length 3) we can form, where each digit is 0, 1, or 2, and the product of the five rolls is 900.

We can solve this problem using a technique called stars and bars. Imagine we have 15 stars, representing the 15 digits in the 5 rolls. We want to divide these stars into 3 bins, representing the factors 2, 3, and 5. We can do this by placing 2 bars among the stars, which will divide the stars into 3 groups. The number of stars in the first group represents the total number of 2's rolled, the number of stars in the second group represents the total number of 3's rolled, and the number of stars in the third group represents the total number of 5's rolled.

For example, if we roll 2 2 3 3 5, we would represent this as the string "220330550", which corresponds to the stars-and-bars representation "||||*****", where the first two stars are in the first bin (representing the 2's), the next two stars are in the second bin (representing the 3's), and so on.

The number of ways to place 2 bars among 15 stars is ${15+2\choose 2}={17\choose 2}=136$. Therefore, there are 136 different sequences of rolls that could have led to a product of 900.

 Mar 14, 2023
 #2
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15566 , 15656 , 15665 , 16556 , 16565 , 16655 , 23556 , 23565 , 23655 , 25356 , 25365 , 25536 , 25563 , 25635 , 25653 , 26355 , 26535 , 26553 , 32556 , 32565 , 32655 , 33455 , 33545 , 33554 , 34355 , 34535 , 34553 , 35256 , 35265 , 35345 , 35354 , 35435 , 35453 , 35526 , 35534 , 35543 , 35562 , 35625 , 35652 , 36255 , 36525 , 36552 , 43355 , 43535 , 43553 , 45335 , 45353 , 45533 , 51566 , 51656 , 51665 , 52356 , 52365 , 52536 , 52563 , 52635 , 52653 , 53256 , 53265 , 53345 , 53354 , 53435 , 53453 , 53526 , 53534 , 53543 , 53562 , 53625 , 53652 , 54335 , 54353 , 54533 , 55166 , 55236 , 55263 , 55326 , 55334 , 55343 , 55362 , 55433 , 55616 , 55623 , 55632 , 55661 , 56156 , 56165 , 56235 , 56253 , 56325 , 56352 , 56516 , 56523 , 56532 , 56561 , 56615 , 56651 , 61556 , 61565 , 61655 , 62355 , 62535 , 62553 , 63255 , 63525 , 63552 , 65156 , 65165 , 65235 , 65253 , 65325 , 65352 , 65516 , 65523 , 65532 , 65561 , 65615 , 65651 , 66155 , 66515 , 66551 , Total =  120 five-digit such integers.

 Mar 14, 2023

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