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Difference between revisions of "2023 AMC 10B Problems/Problem 9"

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~andliu766
 
~andliu766
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Minor corrections by ~milquetoast
 
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Alternatively, you can let m be the square root of the larger number, but if you do that, keep in mind that <math>m=1</math> must be rejected, since <math>(m-1)</math> cannot be <math>0</math>.
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Note from ~milquetoast: Alternatively, you can let m be the square root of the larger number, but if you do that, keep in mind that <math>m=1</math> must be rejected, since <math>(m-1)</math> cannot be <math>0</math>.
  
 
==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}}

Revision as of 22:03, 15 November 2023

Problem

The numbers 16 and 25 are a pair of consecutive postive 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

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

~andliu766

Minor corrections by ~milquetoast

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

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

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