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

Revision as of 22:11, 29 January 2025

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

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

Revision as of 22:11, 29 January 2025

Solution

Vide Solution 1 by SpreadTheMathLove

@@ Line 29: / Line 29: @@
 ~Soupboy0
+==Vide Solution 1 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=jTTcscvcQmI