Catherine rolls a standard $6$-sided die six times. If the product of her rolls is $2000,$ then how many different sequences of rolls could there have been? (The order of the rolls matters.)

falafronk Mar 24, 2024

#1**+1 **

The prime factorization of 2000 is 2^{4}*5^{3}.

Catherine must have rolled 3 5s, to form the 5^{3}.

The 2^{4} is more tricky. Catherine could've rolled 4's or 2's a certain number of times to form the 2^{4}, but the number of rolls to acheive this must be 3, because she rolls a total of 6 rolls.

Consider all the cases:

__#1: The rolls 2, 2 and 2 ^{2} in some order.__

__#2: The rolls 1 2 ^{2} and 2^{2} in some order.__

These are the only ways she coud've rolled a 4.

The number of ways to arrange 2, 2, 2^{2}, 3, 3, 3 is 60

The number of ways to arrange 1, 2^{2}, 2^{2}, 3, 3, 3 is also 60.

__Therefore there could be 60+60 = 120 sequences. __

hairyberry Mar 25, 2024