Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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<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== | ||
| + | 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. \boxed{\text{B}} | ||
Revision as of 15:30, 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?
Solution
Let m be the sqaure root of the smaller of the two perfect squares. Then, 
. Thus, 
. So there are 1011 numbers that satisfy the equation. \boxed{\text{B}}
