Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | Let m be the | + | |
+ | Let m be the square root of the smaller of the two perfect squares. Then, <math>(m+1)^2 - m^2 = m^2+2m+1-m^2 = 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 | ||
+ | Minor corrections by ~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 20:53, 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 square root of the smaller of the two perfect squares. Then, . Thus, . So there are numbers that satisfy the equation.
~andliu766 Minor corrections by ~milquetoast
Alternatively, you can let m be the square root of the larger number, but if you do that, keep in mind that must be rejected, since cannot be .
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.