Difference between revisions of "2014 AMC 10A Problems/Problem 8"
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==Problem== | ==Problem== | ||
| − | Which of the following | + | Which of the following numbers is a perfect square? |
<math> \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 </math> | <math> \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 </math> | ||
| − | == Solution== | + | == Solution 1== |
Note that for all positive <math>n</math>, we have | Note that for all positive <math>n</math>, we have | ||
| Line 16: | Line 16: | ||
In order for <math>\frac{n+1}{2}</math> to be a perfect square, <math>n+1</math> must be twice a perfect square. From the answer choices, <math>n+1=18</math> works, thus, <math>n=17</math> and our desired answer is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math> | In order for <math>\frac{n+1}{2}</math> to be a perfect square, <math>n+1</math> must be twice a perfect square. From the answer choices, <math>n+1=18</math> works, thus, <math>n=17</math> and our desired answer is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math> | ||
| − | (Solution/ | + | ==Solution 2 (Variation of Solution 1)== |
| + | |||
| + | We take the first option A. \( \frac{14! \cdot 15!}{2} \). This is just equal to \( 14!^2 \cdot \frac{15}{2} \). \( \frac{15}{2} \) is 7.5, and 7.5 is not a perfect square, therefore option A is incorrect. | ||
| + | |||
| + | We can also rule that \( \frac{17}{2} = 8.5 \) and \( \frac{19}{2} = 9.5 \) are both not perfect squares, so option choices C and E are also incorrect. | ||
| + | |||
| + | If we take B. \( \frac{15! \cdot 16!}{2} \), we see that \( 15!^2 \) is a perfect square, but \( \frac{16}{2} = 8 \) is not, so B is incorrect. Our remaining option is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math>. | ||
| + | |||
| + | Proof: | ||
| + | |||
| + | <cmath>\frac{17! \cdot 18!}{2} \Rightarrow 17!^2 \cdot \frac{18}{2} \Rightarrow 17!^2 \cdot 9 \Rightarrow 17!^2 \cdot 3^2.</cmath> | ||
| + | |||
| + | The product of two squares is always a perfect square. | ||
| + | |||
| + | ~Pinotation | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/sa9OON6KXb8 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/uY_Xp8GtXP8 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}} | {{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category: Introductory Number Theory Problems]] | ||
Latest revision as of 22:02, 24 August 2025
Contents
Problem
Which of the following numbers is a perfect square?
Solution 1
Note that for all positive 
, we have
![\[\dfrac{n!(n+1)!}{2}\]](http://latex.artofproblemsolving.com/c/6/c/c6cadcea118b402a6a8d08b9ed33371afc4b43a0.png)
![\[\implies\dfrac{(n!)^2\cdot(n+1)}{2}\]](http://latex.artofproblemsolving.com/0/f/4/0f4a73e6e6c4cc24be518527f91d6a867e8f5b24.png)
![\[\implies (n!)^2\cdot\dfrac{n+1}{2}\]](http://latex.artofproblemsolving.com/2/1/3/2130d4422339b6c4b17ef79fc45d476fffa67ed2.png)
We must find a value of such that 
is a perfect square. Since 
is a perfect square, we must also have 
be a perfect square.
In order for to be a perfect square, 
must be twice a perfect square. From the answer choices, 
works, thus, 
and our desired answer is 
Solution 2 (Variation of Solution 1)
We take the first option A. \( \frac{14! \cdot 15!}{2} \). This is just equal to \( 14!^2 \cdot \frac{15}{2} \). \( \frac{15}{2} \) is 7.5, and 7.5 is not a perfect square, therefore option A is incorrect.
We can also rule that \( \frac{17}{2} = 8.5 \) and \( \frac{19}{2} = 9.5 \) are both not perfect squares, so option choices C and E are also incorrect.
If we take B. \( \frac{15! \cdot 16!}{2} \), we see that \( 15!^2 \) is a perfect square, but \( \frac{16}{2} = 8 \) is not, so B is incorrect. Our remaining option is .
Proof:
![\[\frac{17! \cdot 18!}{2} \Rightarrow 17!^2 \cdot \frac{18}{2} \Rightarrow 17!^2 \cdot 9 \Rightarrow 17!^2 \cdot 3^2.\]](http://latex.artofproblemsolving.com/0/1/e/01e1ca23aaa3e7528b1f60c0104d21f8becc2c15.png)
The product of two squares is always a perfect square.
~Pinotation
Video Solution (CREATIVE THINKING)
https://youtu.be/sa9OON6KXb8
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
