Difference between revisions of "2023 AMC 10B Problems/Problem 9"
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==Problem== | ==Problem== | ||
| − | The numbers 16 and 25 are a pair of consecutive | + | 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> | <math>\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017</math> | ||
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==Solution 1== | ==Solution 1== | ||
| − | 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. | + | Let <math>x</math> 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 | ~andliu766 | ||
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Minor corrections by ~milquetoast | 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 <math>x=1</math> must be rejected, since <math>(x-1)</math> cannot be <math>0</math>. | + | 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== | ==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>. | + | 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>. |
~Aopsthedude | ~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== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=qrswSKqdg-Y | 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== | ==See also== | ||
{{AMC10 box|year=2023|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2023|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest 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 ![$[(2023-3)/2]+1=1011$](http://latex.artofproblemsolving.com/7/1/8/718cfce6795fc720f8574fa5b6efd43daf12c5b2.png)
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. 
