Difference between revisions of "2025 AMC 8 Problems/Problem 23"
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| + | ==Problem== | ||
How many four-digit numbers have all three of the following properties? | How many four-digit numbers have all three of the following properties? | ||
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(III) The number is the product of exactly two prime numbers. | (III) The number is the product of exactly two prime numbers. | ||
| + | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math> | ||
==Solution== | ==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 | |
| − | |||
| − | 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 | ||
<math>40^2-1 = (39)(41)</math> | <math>40^2-1 = (39)(41)</math> | ||
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~Soupboy0 | ~Soupboy0 | ||
| + | ==Vide Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:14, 30 January 2025
Contents
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.
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
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Thinking Feet
See Also
| 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. 
