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2025 AMC 8 Problems/Problem 23: Difference between revisions

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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>


==Solution==
==Solution 1==
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>.  
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>.  


<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:  
''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>40^2-1 = (39)(41)</math>
*<math>50^2-1 = (49)(51)</math>
*<math>50^2-1 = (49)(51)</math>
Line 21: Line 21:
*<math>90^2-1 = (89)(91)</math>
*<math>90^2-1 = (89)(91)</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>.  
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
~Soupboy0


==Vide Solution 1 by SpreadTheMathLove==
~ Edited by [[User:Aoum|aoum]]
 
==Video Solution 1 by SpreadTheMathLove==
https://www.youtube.com/watch?v=jTTcscvcQmI
https://www.youtube.com/watch?v=jTTcscvcQmI


==Video Solution (A Clever Explanation You’ll Get Instantly)==
==Video Solution by hsnacademy==
https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539
https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539
~hsnacademy


==Video Solution by Thinking Feet==
==Video Solution by Thinking Feet==
https://youtu.be/PKMpTS6b988
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==
==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}}
[[Category:Introductory Number Theory Problems]]

Latest revision as of 10:05, 15 August 2025

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.

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

The Condition (II) perfect square must end in "$00$" because $...99+1=...00$ Condition (I). Four-digit perfect squares ending in "$00$" are ${40, 50, 60, 70, 80, 90}$.

Condition (II) also says the number is in the form $n^2-1$. By the Difference of Squares, $n^2-1 = (n+1)(n-1)$. Hence:

  • $40^2-1 = (39)(41)$
  • $50^2-1 = (49)(51)$
  • $60^2-1 = (59)(61)$
  • $70^2-1 = (69)(71)$
  • $80^2-1 = (79)(81)$
  • $90^2-1 = (89)(91)$

On this list, the only number that is the product of $2$ prime numbers (condition $3$) is $60^2-1 = (59)(61)$, so the answer is $\boxed{\text{(B)\ 1}}$.

~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

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

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

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