Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> | There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math> | ||
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== Solution 2 == | == Solution 2 == | ||
Revision as of 17:13, 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
...etc,
So, we original expression reduce to
Solution 2
We can prime factorize the solutions:
A = 
B = 
C = 
D = 
E = 
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 
and
14 appears in to .
So, there is an odd amount of 7's since there are 10 7's from to and 3 7's from to , and 
which is odd. So we need to multiply by 7 to get a perfect square. 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)
