2023 AMC 10B Problems/Problem 9
Contents
[hide]Problem
The numbers and are a pair of consecutive postive squares whose difference is . How many pairs of consecutive positive perfect squares have a difference of less than or equal to ?
Solution 1
Let x 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
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.