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Difference between revisions of "2023 AMC 10B Problems/Problem 15"

(→‎Solution 3)
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there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math>  (We're not counting 3 2's in 8, 2 3's in 9, etc).
 
there are odd number of 2's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math>  (We're not counting 3 2's in 8, 2 3's in 9, etc).
  
−
There are even number of 3's in <math>2!\cdot3!\cdot4!\cdot5!...16!</math>
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There are even number of 3's in  
 
...etc,
 
...etc,
  
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== Solution 2 ==
 
== Solution 2 ==
 +
Perfect squares have all of the powers in their prime factorization even. To evaluate <math>cdot2!\cdot3!\cdot4!\cdot5!...16!</math> we get the following:
 +
 +
<math>(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^{11}) \times (7^{10}) \times ((2^3)^{9}) \times ((3^2)^8) \times \\ ((5 \times 2)^7) \times (11^6) \times (((2^2) \times 3)^5) \times (13^4) \times ((7 \times 2)^3) \times ((3 \times 5)^2) \times ((2^4)^1)</math>
 +
 +
 +
== Solution 3 ==
  
 
We can prime factorize the solutions:  
 
We can prime factorize the solutions:  
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~aleyang
 
~aleyang
  
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== Solution 3 ==
+
== Solution 4 ==
  
 
First, we note that <math>3! = 2! \cdot 3</math>, <math>5! = 4! \cdot 5, ... 15! = 14! \cdot 15</math>. So, <math>2!\cdot3! ={2!}^2\cdot3 \equiv 3, 4!\cdot5!={4!}^2\cdot5 \equiv 5, ... 14!\cdot15!={14!}^2\cdot15\equiv15</math>. Simplifying the whole sequence and cancelling out the squares, we get <math>3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!</math>. Prime factoring <math>16!</math> and cancelling out the squares, the only numbers that remain are <math>2, 5,</math> and <math>7</math>. Since we need to make this a perfect square, <math>m = 2 \cdot 5 \cdot 7</math>. Multiplying this out, we get <math>\boxed{\text{(C) }  70}</math>.
 
First, we note that <math>3! = 2! \cdot 3</math>, <math>5! = 4! \cdot 5, ... 15! = 14! \cdot 15</math>. So, <math>2!\cdot3! ={2!}^2\cdot3 \equiv 3, 4!\cdot5!={4!}^2\cdot5 \equiv 5, ... 14!\cdot15!={14!}^2\cdot15\equiv15</math>. Simplifying the whole sequence and cancelling out the squares, we get <math>3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!</math>. Prime factoring <math>16!</math> and cancelling out the squares, the only numbers that remain are <math>2, 5,</math> and <math>7</math>. Since we need to make this a perfect square, <math>m = 2 \cdot 5 \cdot 7</math>. Multiplying this out, we get <math>\boxed{\text{(C) }  70}</math>.
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~yourmomisalosinggame (a.k.a. Aaron) & ~Technodoggo (add more examples)
 
~yourmomisalosinggame (a.k.a. Aaron) & ~Technodoggo (add more examples)
  
−
== Solution 4 (Bashy method) ==
+
== Solution 5 (Bashy method) ==
 
We know that a perfect square must be in the form <math>2^{2a_1}\cdot3^{2a_2}\cdot5^{2a_3}...p^{2a_n}</math> where <math>a_1, a_2, a_3, ..., a_n</math> are nonnegative integers, and <math>p</math> is the largest and <math>nth</math> prime factor of our square number.
 
We know that a perfect square must be in the form <math>2^{2a_1}\cdot3^{2a_2}\cdot5^{2a_3}...p^{2a_n}</math> where <math>a_1, a_2, a_3, ..., a_n</math> are nonnegative integers, and <math>p</math> is the largest and <math>nth</math> prime factor of our square number.
  

Revision as of 07:53, 16 November 2023

Problem

What is the least positive integer $m$ such that $m\cdot2!\cdot3!\cdot4!\cdot5!...16!$ is a perfect square?

$\textbf{(A) }30\qquad\textbf{(B) }30030\qquad\textbf{(C) }70\qquad\textbf{(D) }1430\qquad\textbf{(E) }1001$

Solution 1

Consider 2 -- there are odd number of 2's in $2!\cdot3!\cdot4!\cdot5!...16!$ (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 \begin{align*} m \cdot 2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 &\equiv m \cdot 2^8 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8)\\ &\equiv m \cdot 2 \cdot 3 \cdot (2 \cdot 2) \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot (2 \cdot 2 \cdot 2)\\ &\equiv m \cdot 2 \cdot 5 \cdot 7\\ m &= 2 \cdot 5 \cdot 7 = 70 \end{align*}

~Technodoggo ~minor edits by lucaswujc

Solution 2

Perfect squares have all of the powers in their prime factorization even. To evaluate $cdot2!\cdot3!\cdot4!\cdot5!...16!$ we get the following:

$(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^{11}) \times (7^{10}) \times ((2^3)^{9}) \times ((3^2)^8) \times \\ ((5 \times 2)^7) \times (11^6) \times (((2^2) \times 3)^5) \times (13^4) \times ((7 \times 2)^3) \times ((3 \times 5)^2) \times ((2^4)^1)$


Solution 3

We can prime factorize the solutions: A = $2 \cdot 3 \cdot 5,$ B = $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13,$ C = $2 \cdot 5 \cdot 7,$ D = $2 \cdot 5 \cdot 11 \cdot 13,$ E = $7 \cdot 11 \cdot 13,$

We can immediately eliminate B, D, and E since 13 only appears in $13!, 14!, 15, 16!$, so $13\cdot 13\cdot 13\cdot 13$ 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 $7!$ to $16!$ and 14 appears in $7!$ to $16!$. So, there is an odd amount of 7's since there are 10 7's from $7!$ to $16!$ and 3 7's from $7!$ to $16!$, and $10+3=13$ 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 $\boxed{\text{C}}$.

~aleyang

Solution 4

First, we note that $3! = 2! \cdot 3$, $5! = 4! \cdot 5, ... 15! = 14! \cdot 15$. So, $2!\cdot3! ={2!}^2\cdot3 \equiv 3, 4!\cdot5!={4!}^2\cdot5 \equiv 5, ... 14!\cdot15!={14!}^2\cdot15\equiv15$. Simplifying the whole sequence and cancelling out the squares, we get $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 16!$. Prime factoring $16!$ and cancelling out the squares, the only numbers that remain are $2, 5,$ and $7$. Since we need to make this a perfect square, $m = 2 \cdot 5 \cdot 7$. Multiplying this out, we get $\boxed{\text{(C) } 70}$.

~yourmomisalosinggame (a.k.a. Aaron) & ~Technodoggo (add more examples)

Solution 5 (Bashy method)

We know that a perfect square must be in the form $2^{2a_1}\cdot3^{2a_2}\cdot5^{2a_3}...p^{2a_n}$ where $a_1, a_2, a_3, ..., a_n$ are nonnegative integers, and $p$ is the largest and $nth$ prime factor of our square number.

Let's assume $r=m\cdot2!\cdot3!\cdot4!\cdot5!...16!$. We need to prime factorize $r$ 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 $m$. We can do this by finding the number of factors of $2$, $3$, $5$, $7$, $11$, and $13$.

Case 1: Factors of $2$

We first count factors of $2^1$ in each of the factorials. We know there is one factor of $2^1$ each in $2!$ and $3!$, two in $4!$ and $5!$, and so on until we have$8$ factors of $2^1$ in $16!$. Adding them all up, we have $1+1+2+2+...7+7+8=64$.

Now, we count factors of $2^2$ in each of the factorials. We know there is one factor of $2^2$ each in $4!$, $5!$, $6!$, and $7!$, two in $8!$, $9!$, $10$, and $11!$, and so on until we have $4$ factors of $2^1$ in $16!$. Adding them all up, we have $1+1+1+1+2+2+2+2+3+3+3+3+4=28$.

Now we count factors of $2^3$ in each of the factorials. Using a similar method as above, we have a sum of $1+1+1+1+1+1+1+1+2=10$.

Now we count factors of $2^4$ in each of the factorials. Using a similar method as above, we have a factor of $2^4$ in $16$, so there is $1$ factor of $2^4$.

Adding all the factors of $2$, we have $103$. Since $103$ is odd, $m$ has one factor of $2$.

Case 2: Factors of $3$

We use a similar method as in case 1. We first count factors of $3^1$. We obtain the sum $1+1+1+2+2+2+...4+4+4+5+5=50$.

We count factors of $3^2$. We obtain the sum $1+1+1+1+1+1+1+1=8$.

Adding all the factors of $3$, we have $58$. Since $58$ is even, $m$ has $0$ factors of $3$.

Case 3: Factors of $5$

We count the factors of $5^1$: $1+1+1+1+1+2+2+2+2+2+3+3=21$. Since $21$ is odd, $m$ has one factor of $5$.

Case 4: Factors of $7$

We count the factors of $7^1$: $1+1+1+1+1+1+1+2+2+2=13$. Since $13$ is odd, $m$ has one factor of $7$.

Case 5: Factors of $11$

We count the factors of $11^1$: $1+1+1+1+1+1=6$. Since $6$ is even, $m$ has $0$ factors of $11$.

Case 6: Factors of $13$

We count the factors of $13^1$: $1+1+1+1=4$. Since $4$ is even, $m$ has $0$ factors of $13$.

Multiplying out all our factors for $m$, we obtain $2\cdot5\cdot7=\boxed{\text{ (C) }70}$.

~arjken

Video Solution 1 by OmegaLearn

https://youtu.be/fB81j1vbwdM

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

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