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

m (Protected "2025 AMC 8 Problems/Problem 23" ([Edit=Allow only administrators] (expires 17:59, 29 January 2025 (UTC)) [Move=Allow only administrators] (expires 17:59, 29 January 2025 (UTC))))
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The 2025 AMC 8 is not held yet. Please do not post false problems.
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How many four-digit numbers have all three of the following properties?
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(I) The tens and ones digit are both 9.
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(II) The number is 1 less than a perfect square.
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(III) The number is the product of exactly two prime numbers.
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==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</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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<math>40^2-1 = (39)(41)</math>
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<math>50^2-1 = (49)(51)</math>
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<math>60^2-1 = (59)(61)</math>
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<math>70^2-1 = (69)(71)</math>
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<math>80^2-1 = (79)(81)</math>
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<math>90^2-1 = (89)(91)</math>
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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
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~Soupboy0

Revision as of 20:40, 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.


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 $n$ 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

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