2023 AMC 10B Problems/Problem 15
Contents
- 1 Problem
- 2 Video Solution by MegaMath
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5 (Bashy method)
- 8 Solution 6 (Answer Choices)
- 9 Solution 7 (Fastest Intuition)
- 10 Solution 8(Pretty Quick similar to solution 7)
- 11 Video Solution 2 by OmegaLearn
- 12 Video Solution 3 by SpreadTheMathLove
- 13 Video Solution 4 by paixiao
- 14 Video Solution
- 15 See also
Problem
What is the least positive integer such that is a perfect square?
Video Solution by MegaMath
https://www.youtube.com/watch?v=le0KSx3Cy-g&t=28s
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 &sravya_m18
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 .
~Stead (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
Solution 6 (Answer Choices)
We see that all the answer choices are divisible by except for , and we also notice that the answer choices have , , , or as a prime factor.
Testing, we see that , , have an even power in the product, so we have that all the other answer choices will not work.
Therefore we just have .
Solution 7 (Fastest Intuition)
Notice that you can add the factor to the expression to make it
Every consecutive pair can be simplified to (divide out ).
Resulting is the product
This is easily simplified to when dividing out perfect squares.
This means that must have at minimun a factor of which gives answer choice .
~coolishu
Solution 8(Pretty Quick similar to solution 7)
this can be written as
Thus
Now lets try
thus
we can continue this pattern until one to get
we can remove all the even powers of expression above to get
we can see that all off the numbers are even and multiples of 2,3,5,7
the biggest possible number we should get is 2 times 3 times 5 times 7 which is 210 which means out anwer should be A or C 30 or 70.
Then look all the stuff with 7 which would only consist of
which needs an extra 7 to give it company, so put answer must be divisble by 7 and less than 210
which leaves 70 or C as our answer
.
Note: you can also just remove all pairs of two to get
Video Solution 2 by OmegaLearn
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=sqVY5-h4vfo
Video Solution 4 by paixiao
https://www.youtube.com/watch?v=EvA2Nlb7gi4&t=238s
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
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