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Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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==Problem==
 
==Problem==
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The numbers <math>16</math> and <math>25</math> are a pair of consecutive postive squares whose difference is <math>9</math>. How many pairs of consecutive positive perfect squares have a difference of less than or equal to <math>2023</math>?
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The numbers <math>16</math> and <math>25</math> are a pair of consecutive positive squares whose difference is <math>9</math>. How many pairs of consecutive positive perfect squares have a difference of less than or equal to <math>2023</math>?
  
 
<math>\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017</math>
 
<math>\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017</math>
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==Solution 1==
 
==Solution 1==
  
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Let x be the square root of the smaller of the two perfect squares. Then, <math>(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023</math>. Thus, <math>x \le 1011</math>. So there are <math>\boxed{\text{(B)}1011}</math> numbers that satisfy the equation.  
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Let <math>x</math> be the square root of the smaller of the two perfect squares. Then, <math>(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023</math>. Thus, <math>x \le 1011</math>. So there are <math>\boxed{\text{(B)}1011}</math> numbers that satisfy the equation.  
  
 
~andliu766
 
~andliu766
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==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=qrswSKqdg-Y
 
https://www.youtube.com/watch?v=qrswSKqdg-Y
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 +
==Video Solution==
 +
 +
https://youtu.be/DK--SMnDSr0
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 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution by Interstigation==
 +
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
 +
 
==See also==
 
==See also==
 
{{AMC10 box|year=2023|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2023|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:57, 2 November 2024

Problem

The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$. How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$?

$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

Solution 1

Let $x$ be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$. Thus, $x \le 1011$. So there are $\boxed{\text{(B)}1011}$ numbers that satisfy the equation.

~andliu766

A very similar solution offered by ~darrenn.cp and ~DarkPheonix has been combined with Solution 1.

Minor corrections by ~milquetoast

Note from ~milquetoast: Alternatively, you can let $x$ be the square root of the larger number, but if you do that, keep in mind that $x=1$ must be rejected, since $(x-1)$ cannot be $0$.

Solution 2

The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$, which is $2^2-1^2$. The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to $2023$ is $2023$, which is $1012^2-1011^2$. These numbers are in the form $(x+1)^2-x^2$, which is just $2x+1$. These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\boxed{\text{(B)}1011}$.

~Aopsthedude

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/EuLkw8HFdk4?si=qr7LZahoIbDMBxvq&t=1848

~Math-X

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=qrswSKqdg-Y

Video Solution

https://youtu.be/DK--SMnDSr0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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. AMC logo.png

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