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

Latest revision as of 17:58, 19 December 2023

Problem

The numbers $16$ and $25$ are a pair of consecutive positive 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 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 $1012^2-1011^2$ . 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 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

https://youtu.be/DK--SMnDSr0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

@@ Line 1: / Line 1: @@
 ==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?
+The numbers <math>16</math> and <math>25</math> are a pair of consecutive positive squares whose difference is <math>9</math>. How many pairs of consecutive positive perfect squares have a difference of less than or equal to <math>2023</math>?
 <math>\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017</math>
@@ 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
+Note from ~milquetoast: Alternatively, you can let <math>x</math> 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>.
 ==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>. 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>.
-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.
+~Aopsthedude
-~darrenn.cp
+==Video Solution by Math-X (First understand the problem!!!)==
-~DarkPheonix
+https://youtu.be/EuLkw8HFdk4?si=qr7LZahoIbDMBxvq&t=1848
-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>.
+~Math-X
-==Solution 3==
-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
 ==Video Solution 1 by SpreadTheMathLove==
 https://www.youtube.com/watch?v=qrswSKqdg-Y
+==Video Solution==
+https://youtu.be/DK--SMnDSr0
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Video Solution by Interstigation==
+https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
 ==See also==
 {{AMC10 box|year=2023|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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