Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 10B Problems/Problem 9"

(→‎Solution 2)
(→‎Solution 2)
Line 15: Line 15:
  
 
==Solution 2==
 
==Solution 2==
−
The smallest number that can be expressed as the difference of a pair of consecutive positive squares is <math>3</math>, 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 <math>2023</math> is <math>2023</math>, which is <math>1012^2-1011^2</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> such 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 <math>3</math>, 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 <math>2023</math> is <math>2023</math>, which is <math>1012^2-1011^2</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> such numbers. The answer is <math>\boxed{\text{(B)}1011}</math>.
  
 
~Aopsthedude
 
~Aopsthedude

Revision as of 22:24, 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?

$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

Solution 1

Let m be the square root of the smaller of the two perfect squares. Then, $(m+1)^2 - m^2 = m^2+2m+1-m^2 = 2m+1 \le 2023$. Thus, $m \le 1011$. So there are $\boxed{\text{(B)}1011}$ 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 $m=1$ must be rejected, since $(m-1)$ cannot be $0$.

Solution 2

The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$, which is $2^2-1^2$. 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 $1012^2-1011^2$. Since these numbers are in the form $(x+1)^2-x^2$, which is just $2x+1$. These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\boxed{\text{(B)}1011}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_9&oldid=203779"