Difference between revisions of "2025 AMC 8 Problems/Problem 23"
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| − | The | + | ==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. | ||
| + | |||
| + | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math> | ||
| + | |||
| + | ==Solution 1== | ||
| + | The ''Condition (II)'' perfect square must end in "<math>00</math>" because <math>...99+1=...00</math> ''Condition (I)''. Four-digit perfect squares ending in "<math>00</math>" are <math>{40, 50, 60, 70, 80, 90}</math>. | ||
| + | |||
| + | ''Condition (II)'' also says the number is in the form <math>n^2-1</math>. By the '''[https://en.wikipedia.org/wiki/Difference_of_two_squares Difference of Squares]''', <math>n^2-1 = (n+1)(n-1)</math>. Hence: | ||
| + | *<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> | ||
| + | |||
| + | On this list, the only number that is the product of <math>2</math> prime numbers (condition <math>3</math>) is <math>60^2-1 = (59)(61)</math>, so the answer is <math>\boxed{\text{(B)\ 1}}</math>. | ||
| + | |||
| + | ~Soupboy0 | ||
| + | |||
| + | ~ Edited by [[User:Aoum|aoum]] | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | ==Video Solution by hsnacademy== | ||
| + | https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539 | ||
| + | |||
| + | ==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 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 11:05, 15 August 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 1
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 the Difference of Squares, 
. Hence:
On this list, the only number that is the product of 
prime numbers (condition 
) is 
, so the answer is 
.
~Soupboy0
~ Edited by aoum
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by hsnacademy
https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
