Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
The smallest number that can be expressed as the difference of a pair of consecutive positive squares is 3, which is <math>2^2-1^2</math>. The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to 2023 is 2023, which is <math>1012^2-1011^1</math>. Since these numbers are in the form <math>(x+1)^2-x^2</math>, which is just <math>2x+1</math>.These numbers are just the odd numbers from 3 to 2023, so there are <math>[(2023-3)/2]+1=1011</math> numbers. The answer is <math>\boxed{\text{(B)}1011}</math>. | The smallest number that can be expressed as the difference of a pair of consecutive positive squares is 3, which is <math>2^2-1^2</math>. The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to 2023 is 2023, which is <math>1012^2-1011^1</math>. Since these numbers are in the form <math>(x+1)^2-x^2</math>, which is just <math>2x+1</math>.These numbers are just the odd numbers from 3 to 2023, so there are <math>[(2023-3)/2]+1=1011</math> numbers. The answer is <math>\boxed{\text{(B)}1011}</math>. | ||
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~Aopsthedude | ~Aopsthedude | ||
Revision as of 22:22, 15 November 2023
Contents
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 1
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
Note from ~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 .
Solution 2
The smallest number that can be expressed as the difference of a pair of consecutive positive squares is 3, which is . The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to 2023 is 2023, which is . Since these numbers are in the form , which is just .These numbers are just the odd numbers from 3 to 2023, so there are numbers. The answer is .
~Aopsthedude
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.