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

Revision as of 12:12, 1 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

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 Difference of Squares[1], $n^2-1 = (n+1)(n-1)$ . So:

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

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

Vide Solution 1 by SpreadTheMathLove

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

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

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.

@@ Line 11: / Line 11: @@
 ==Solution==
-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
+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>.
-<math>40^2-1 = (39)(41)</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:
+*<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>
-<math>50^2-1 = (49)(51)</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>.
-<math>60^2-1 = (59)(61)</math>
+~Soupboy0
-<math>70^2-1 = (69)(71)</math>
-<math>80^2-1 = (79)(81)</math>
-<math>90^2-1 = (89)(91)</math>
-Out of these, 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>. four-digit number
-~Soupboy0
 ==Vide Solution 1 by SpreadTheMathLove==
 https://www.youtube.com/watch?v=jTTcscvcQmI

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