Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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==Solution 1== | ==Solution 1== | ||
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. | 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. | ||
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~andliu766 | ~andliu766 | ||
Revision as of 17:39, 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 1
Let m be the sqaure root of the smaller of the two perfect squares. Then, 
. Thus, 
. So there are 
numbers that satisfy the equation.
~andliu766
