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Difference between revisions of "2025 AMC 8 Problems/Problem 23"
 
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==Problem==
 
How many four-digit numbers have all three of the following properties?
 
How many four-digit numbers have all three of the following properties?
  
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(III) The number is the product of exactly two prime numbers.
 
(III) The number is the product of exactly two prime numbers.
  
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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>
  
 
==Solution==
 
==Solution==
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−
 
 
Note that if a perfect square ends in "<math>00</math>", then when <math>1</math> is subtracted from this number, (Condition II) the number will end in "<math>99</math>" (Condition I). Therefore, the number is in the form <math>n^2-1</math>, where <math>n = \{40, 50, 60, 70, 80, 90\}</math> (otherwise <math>n^2-1</math> won't end in "<math>99</math>" or <math>n</math> won't be <math>4</math> digits). Also, note that <math>n^2-1 = (n+1)(n-1)</math>. Therefore, <math>n-1</math> and <math>n+1</math> are both prime numbers because of (Condition III). Testing, we get
 
Note that if a perfect square ends in "<math>00</math>", then when <math>1</math> is subtracted from this number, (Condition II) the number will end in "<math>99</math>" (Condition I). Therefore, the number is in the form <math>n^2-1</math>, where <math>n = \{40, 50, 60, 70, 80, 90\}</math> (otherwise <math>n^2-1</math> won't end in "<math>99</math>" or <math>n</math> won't be <math>4</math> digits). Also, note that <math>n^2-1 = (n+1)(n-1)</math>. Therefore, <math>n-1</math> and <math>n+1</math> are both prime numbers because of (Condition III). Testing, we get
  
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~Soupboy0
 
~Soupboy0
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==Vide Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=22|num-a=24}}
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{{MAA Notice}}

Latest revision as of 19:14, 30 January 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

Note that if a perfect square ends in "$00$", then when $1$ is subtracted from this number, (Condition II) the number will end in "$99$" (Condition I). Therefore, the number is in the form $n^2-1$, where $n = \{40, 50, 60, 70, 80, 90\}$ (otherwise $n^2-1$ won't end in "$99$" or $n$ won't be $4$ digits). Also, note that $n^2-1 = (n+1)(n-1)$. Therefore, $n-1$ and $n+1$ are both prime numbers because of (Condition III). Testing, we get

$40^2-1 = (39)(41)$

$50^2-1 = (49)(51)$

$60^2-1 = (59)(61)$

$70^2-1 = (69)(71)$

$80^2-1 = (79)(81)$

$90^2-1 = (89)(91)$

Out of these, the only number that is the product of $2$ prime numbers is $60^2-1 = (59)(61)$, so the answer is $\boxed{\text{(B)\ 1}}$. four-digit number

~Soupboy0

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

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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