Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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We can immediately eliminate B, D, and E since 13 only appears in <math>13!, 14!, 15, 16!</math>, so <math>13\cdot 13\cdot 13\cdot 13</math> is a perfect square. | We can immediately eliminate B, D, and E since 13 only appears in <math>13!, 14!, 15, 16!</math>, so <math>13\cdot 13\cdot 13\cdot 13</math> is a perfect square. | ||
Next, we can test if 7 is possible (and if it is not we can use process of elimination) | Next, we can test if 7 is possible (and if it is not we can use process of elimination) | ||
− | 7 appears in <math>7! to 16!</math> | + | 7 appears in <math>7!</math> to <math>16!</math> |
− | 14 appears in <math>14! to 16!</math> | + | 14 appears in <math>14!</math> to <math>16!</math> |
So, there is an odd amount of 7's. Since 30 is not a divisor of 7, our answer is 70 which is <math>\boxed{\text{C}}</math>. | So, there is an odd amount of 7's. Since 30 is not a divisor of 7, our answer is 70 which is <math>\boxed{\text{C}}</math>. | ||
Revision as of 17:44, 15 November 2023
Contents
Problem
What is the least positive integer such that
is a perfect square?
Solution 1
Consider 2,
there are odd number of 2's in (We're not counting 3 2's in 8, 2 3's in 9, etc).
There are even number of 3's in
...
So, original expression reduce to
Solution 2
We can prime factorize the solutions:
We can immediately eliminate B, D, and E since 13 only appears in , so
is a perfect square.
Next, we can test if 7 is possible (and if it is not we can use process of elimination)
7 appears in
to
14 appears in
to
So, there is an odd amount of 7's. Since 30 is not a divisor of 7, our answer is 70 which is
.
~aleyang
Solution 3
First, we note . Simplifying the whole sequence and cancelling out the squares, we get
. Prime factoring
and cancelling out the squares, the only numbers that remain are
and
. Since we need to make this a perfect square,
. Multiplying this out, we get
.
~yourmomisalosinggame (a.k.a. Aaron)