Difference between revisions of "2005 AMC 10B Problems/Problem 22"
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are <math>24 - 8 = | are <math>24 - 8 = | ||
| − | \boxed{\textbf{(C) }16}</math> numbers less than or equal to 24 that satisfy the condition. | + | \boxed{\textbf{(C) }16}</math> numbers less than or equal to <math>24</math> that satisfy the condition. |
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
| + | |||
| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=XQQpjNuOL5E ~David | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:20, 17 September 2023
Problem
For how many positive integers 
less than or equal to 
is 
evenly divisible by 
Solution
Since 
, the condition is equivalent to having an integer value for 
. This reduces, when 
, to having an integer value for 
. This fraction is an
integer unless 
is an odd prime. There are 
odd primes less than or equal to , so there
are 
numbers less than or equal to that satisfy the condition.
Video Solution
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=XQQpjNuOL5E ~David
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
