Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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Revision as of 21:02, 18 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 can reduce our original expression to
~Technodoggo ~minor edits by lucaswujc
Solution 2
Perfect squares have all of the powers in their prime factorization even. To evaluate we get the following:
.
Taking all powers we get:
Simplifying again, we finally get:
To make all the powers left even, we need to multiply by which is .
~darrenn.cp
Solution 3
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 since 7 appears in 14 once, 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 4
First, we note that , . So, . 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) & ~Technodoggo (add more examples)
Solution 5 (Bashy method)
We know that a perfect square must be in the form where are nonnegative integers, and is the largest and prime factor of our square number.
Let's assume . We need to prime factorize and see which prime factors are raised to an odd power. Then, we can multiply one factor each of prime number with an odd number of factors to . We can do this by finding the number of factors of , , , , , and .
Case 1: Factors of
We first count factors of in each of the factorials. We know there is one factor of each in and , two in and , and so on until we have factors of in . Adding them all up, we have .
Now, we count factors of in each of the factorials. We know there is one factor of each in , , , and , two in , , , and , and so on until we have factors of in . Adding them all up, we have .
Now we count factors of in each of the factorials. Using a similar method as above, we have a sum of .
Now we count factors of in each of the factorials. Using a similar method as above, we have a factor of in , so there is factor of .
Adding all the factors of , we have . Since is odd, has one factor of .
Case 2: Factors of
We use a similar method as in case 1. We first count factors of . We obtain the sum .
We count factors of . We obtain the sum .
Adding all the factors of , we have . Since is even, has factors of .
Case 3: Factors of
We count the factors of : . Since is odd, has one factor of .
Case 4: Factors of
We count the factors of : . Since is odd, has one factor of .
Case 5: Factors of
We count the factors of : . Since is even, has factors of .
Case 6: Factors of
We count the factors of : . Since is even, has factors of .
Multiplying out all our factors for , we obtain .
~arjken
Video Solution by MegaMath
https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s
~megahertz13
Video Solution 2 by OmegaLearn
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=sqVY5-h4vfo
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.