Difference between revisions of "2025 AMC 8 Problems/Problem 23"
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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 | + | 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 |
<math>40^2-1 = (39)(41)</math> | <math>40^2-1 = (39)(41)</math> |
Revision as of 21:40, 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 "", then when
is subtracted from this number, (Condition II) the number will end in "
" (Condition I). Therefore, the number is in the form
, where
(otherwise
won't end in "
" or
won't be
digits). Also, note that
. Therefore,
and
are both prime numbers because of (Condition III). Testing, we get
Out of these, the only number that is the product of prime numbers is
, so the answer is
. four-digit number
~Soupboy0