Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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| + | ==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? | 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? | ||
<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> | ||
| − | ==Solution | + | ==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 <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. | ||
~andliu766 | ~andliu766 | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2023|ab=B|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:32, 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?
Solution
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
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. 
