2023 AMC 10B Problems/Problem 9
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
such numbers. The answer is
.
~Aopsthedude
Solution 3
The difference of 2 consecutive positive integer squares is always an odd number (proved from above). Therefore, the total number of cases is \(\left\lfloor \frac{2023}{2} \right\rfloor\) =
~Pinotation
Solution 4
Notice that the difference of two consecutive squares can be represented as
. Using the Difference of Squares Formula, we simplify to get
The number given in the problem was 2023, and thus we find that
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination