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Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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==Solution==
 
==Solution==
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Let m be the sqaure root of the smaller of the two perfect squares. Then, <math>(m-1)^2 - m^2 = (2m+1)(1) = 2m+1 \le 2023</math>. Thus, <math>m \le 1011</math>. So there are 1011 numbers that satisfy the equation. <math>\boxed{\text{B}}</math>
+
Let m be the sqaure root of the smaller of the two perfect squares. Then, <math>(m-1)^2 - m^2 = (2m+1)(1) = 2m+1 \le 2023</math>. Thus, <math>m \le 1011</math>. So there are <math>\boxed{\text{(B)}1011}</math> numbers that satisfy the equation.
 
~andliu766
 
~andliu766

Revision as of 16:13, 15 November 2023

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 sqaure root of the smaller of the two perfect squares. Then, $(m-1)^2 - m^2 = (2m+1)(1) = 2m+1 \le 2023$. Thus, $m \le 1011$. So there are $\boxed{\text{(B)}1011}$ numbers that satisfy the equation. ~andliu766

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