Difference between revisions of "2025 AMC 8 Problems/Problem 23"

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

@@ Line 10: / Line 10: @@
 <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>50^2-1 = (49)(51)</math>
@@ Line 24: / Line 24: @@
 ~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>,
+*<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==
 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
-~hsnacademy
 ==Video Solution by Thinking Feet==

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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