Difference between revisions of "2025 AMC 8 Problems/Problem 23"
| (5 intermediate revisions by 4 users not shown) | |||
| Line 11: | Line 11: | ||
==Solution== | ==Solution== | ||
| − | + | The <code>Condition II</code> perfect square must end in "<math>00</math>" because <math>...99+1=...00</math> (<code>Condition I</code>). Four-digit perfect squares ending in "<math>00</math>" are <math>{40, 50, 60, 70, 80, 90}</math>. | |
| − | <math>40^2-1 = (39)(41)</math> | + | <code>Condition II</code> also says the number is in the form <math>n^2-1</math>. By ''Difference of Squares''[https://en.wikipedia.org/wiki/Difference_of_two_squares], <math>n^2-1 = (n+1)(n-1)</math>. So: |
| + | *<math>40^2-1 = (39)(41)</math> | ||
| + | *<math>50^2-1 = (49)(51)</math> | ||
| + | *<math>60^2-1 = (59)(61)</math> | ||
| + | *<math>70^2-1 = (69)(71)</math> | ||
| + | *<math>80^2-1 = (79)(81)</math> | ||
| + | *<math>90^2-1 = (89)(91)</math> | ||
| − | <math> | + | On this list, the only number that is the product of <math>2</math> prime numbers is <math>60^2-1 = (59)(61)</math>, so the answer is <math>\boxed{\text{(B)\ 1}}</math>. |
| − | + | ~Soupboy0 | |
| − | + | ==Video Solution 1 by SpreadTheMathLove== | |
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| − | + | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | |
| + | https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539 | ||
| + | ~hsnacademy | ||
| − | + | ==Video Solution by Thinking Feet== | |
| + | https://youtu.be/PKMpTS6b988 | ||
| − | + | ==Video Solution by Dr. David== | |
| + | https://youtu.be/jn-qIwv57nQ | ||
| − | + | ==A Complete Video Solution with the Thought Process by Dr. Xue's Math School== | |
| − | == | + | https://youtu.be/aEPPwMIQ52w |
| − | |||
| − | |||
| − | |||
| − | https://youtu.be/ | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2025|num-b=22|num-a=24}} | {{AMC8 box|year=2025|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 02:16, 3 February 2025
Contents
Problem
How many four-digit numbers have all three of the following properties?
(I) The tens and ones digit are both 9.
(II) The number is 1 less than a perfect square.
(III) The number is the product of exactly two prime numbers.
Solution
The Condition II perfect square must end in "
" because 
(Condition I). Four-digit perfect squares ending in "" are 
.
Condition II also says the number is in the form 
. By Difference of Squares[1], 
. So:
On this list, the only number that is the product of 
prime numbers is 
, so the answer is 
.
~Soupboy0
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539 ~hsnacademy
Video Solution by Thinking Feet
Video Solution by Dr. David
A Complete Video Solution with the Thought Process by Dr. Xue's Math School
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
