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

Revision as of 11:44, 16 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 x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$ . Thus, $x \le 1011$ . So there are $\boxed{\text{(B)}1011}$ 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 x be the square root of the larger number, but if you do that, keep in mind that $x=1$ must be rejected, since $(x-1)$ cannot be $0$ .

Solution 3

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

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=qrswSKqdg-Y

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.

@@ Line 6: / Line 6: @@
 ==Solution 1==
-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.
+Let x be the square root of the smaller of the two perfect squares. Then, <math>(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023</math>. Thus, <math>x \le 1011</math>. So there are <math>\boxed{\text{(B)}1011}</math> 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
-==Solution 2==
-Take note that the pairs are defined by their first digit, so we try to find the pair <math>x</math> and <math>x+1</math>. Using the difference of squares factorization, we get the inequality <math>(2x+1) \leq 2023</math>. Following from here, we get <math>x=1011</math>, so there are <math>\boxed{\text{(B) 1011}}</math> numbers that satisfy the equation.
-~darrenn.cp
-~DarkPheonix
 Note from ~milquetoast: Alternatively, you can let x be the square root of the larger number, but if you do that, keep in mind that <math>x=1</math> must be rejected, since <math>(x-1)</math> cannot be <math>0</math>.

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