2019 AIME II Problems/Problem 4
Problem 4
A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is , where
and
are relatively prime positive integers. Find
.
Solution
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are outcomes).
Case 1 (easy): Four 5's are rolled. This has probability of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For , let
equal the number of outcomes after rolling the die
times, with the property that the product is a square. Thus,
as 1 and 4 are the only possibilities.
To find given
(where
), we observe that if the first
rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives
outcomes. Otherwise, the first
rolls do not multiply to a perfect square (
outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first
rolls is
where
and
are not both even, then we observe that if
and
are both odd, then the last roll must be 6; if only
is odd, the last roll must be 2, and if only
is odd, the last roll must be 3. Thus, we have
outcomes in this case, and
.
Computing ,
,
gives
,
, and
. Thus for Case 3, there are 157 outcomes. For case 2, we multiply by
to distribute the two 5's among four rolls. Thus the probability is
-scrabbler94
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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