Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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All perfect squares have a difference that is the sum of the two numbers being squared. Looking at this pattern shows that all odd numbers are the difference of two perfect squares.Then you can divide 2023 by 2 because only half of them are odd numbers and get 1011.5. Since The sequence starts and ends with an odd number you round up to 1012, but you have to subtract one because one and zero do not count because zero is not a positive number. This yields a result of 1012-1, or <math>\boxed{\text{(B)}1011}</math>. | All perfect squares have a difference that is the sum of the two numbers being squared. Looking at this pattern shows that all odd numbers are the difference of two perfect squares.Then you can divide 2023 by 2 because only half of them are odd numbers and get 1011.5. Since The sequence starts and ends with an odd number you round up to 1012, but you have to subtract one because one and zero do not count because zero is not a positive number. This yields a result of 1012-1, or <math>\boxed{\text{(B)}1011}</math>. | ||
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~LIUGRA001 | ~LIUGRA001 | ||
Revision as of 20:57, 2 November 2024
Contents
Problem
The numbers and are a pair of consecutive positive squares whose difference is . How many pairs of consecutive positive perfect squares have a difference of less than or equal to ?
Solution 1
Let be the square root of the smaller of the two perfect squares. Then, . Thus, . So there are numbers that satisfy the equation.
~andliu766
A very similar solution offered by ~darrenn.cp and ~DarkPheonix has been combined with Solution 1.
Minor corrections by ~milquetoast
Note from ~milquetoast: Alternatively, you can let 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 , 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 is , which is . These numbers are in the form , which is just . These numbers are just the odd numbers from 3 to 2023, so there are such numbers. The answer is .
~Aopsthedude
Solution 3
All perfect squares have a difference that is the sum of the two numbers being squared. Looking at this pattern shows that all odd numbers are the difference of two perfect squares.Then you can divide 2023 by 2 because only half of them are odd numbers and get 1011.5. Since The sequence starts and ends with an odd number you round up to 1012, but you have to subtract one because one and zero do not count because zero is not a positive number. This yields a result of 1012-1, or .
~LIUGRA001
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/EuLkw8HFdk4?si=qr7LZahoIbDMBxvq&t=1848
~Math-X
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=qrswSKqdg-Y
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
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.