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Difference between revisions of "2025 AMC 8 Problems/Problem 23"
 
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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 24: Line 24:
  
 
~Soupboy0
 
~Soupboy0
 +
 +
~ Edited by [[User:Aoum|Aoum]]
 +
 +
==Solution 2==
 +
Condition 2 states that the number is <math>1</math> less than a perfect square, so the smallest four-digit perfect square is <math>32^2 = 1024</math> and the greatest four-digit perfect square is <math>99^2 = 9801</math>. Condition 1 states that the tens and ones digit are both 9, so the number must be <math>1</math> less than a perfect square with tens and ones digits of <math>0</math>. Possible values are:
 +
 +
*<math>40^2 - 1 = 1599</math>,
 +
*<math>50^2 - 1 = 2499</math>,
 +
*<math>60^2 - 1 = 3599</math>,
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*<math>70^2 - 1 = 3899</math>,
 +
*<math>80^2 - 1 = 6399</math>, and
 +
*<math>90^2 - 1 = 8099</math>.
 +
 +
Condition 3 states that the number is the product of exactly two prime numbers. Applying the divisibility test for threes, we find that <math>1599</math>, <math>2499</math>, and <math>6399</math> are divisible by 3. This leaves us with <math>3599</math>, <math>3899</math>, and <math>8099</math>. The prime factorization for the three remaining possibilities are as follows:
 +
 +
*<math>3899 = 59 \times 61</math>,
 +
*<math>4899 = 3 \times 23 \times 71</math>, and
 +
*<math>8091 = 3^2 \times 29 \times 31</math>.
 +
 +
Only <math>3899</math> meets the third condition, being the product of exactly two prime numbers. Therefore, only <math>1</math> four-digit number has all three of the stated conditions, so the answer is <math>\boxed{\textbf{(B)\ 1}}</math>.
 +
 +
~ [[User:Aoum|Aoum]]
  
 
==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==

Latest revision as of 21:08, 21 February 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 is $60^2-1 = (59)(61)$, so the answer is $\boxed{\text{(B)\ 1}}$.

~Soupboy0

~ Edited by Aoum

Solution 2

Condition 2 states that the number is $1$ less than a perfect square, so the smallest four-digit perfect square is $32^2 = 1024$ and the greatest four-digit perfect square is $99^2 = 9801$. Condition 1 states that the tens and ones digit are both 9, so the number must be $1$ less than a perfect square with tens and ones digits of $0$. Possible values are:

  • $40^2 - 1 = 1599$,
  • $50^2 - 1 = 2499$,
  • $60^2 - 1 = 3599$,
  • $70^2 - 1 = 3899$,
  • $80^2 - 1 = 6399$, and
  • $90^2 - 1 = 8099$.

Condition 3 states that the number is the product of exactly two prime numbers. Applying the divisibility test for threes, we find that $1599$, $2499$, and $6399$ are divisible by 3. This leaves us with $3599$, $3899$, and $8099$. The prime factorization for the three remaining possibilities are as follows:

  • $3899 = 59 \times 61$,
  • $4899 = 3 \times 23 \times 71$, and
  • $8091 = 3^2 \times 29 \times 31$.

Only $3899$ meets the third condition, being the product of exactly two prime numbers. Therefore, only $1$ four-digit number has all three of the stated conditions, so the answer is $\boxed{\textbf{(B)\ 1}}$.

~ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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