2002 AMC 12B Problems/Problem 21
Contents
Problem
For all positive integers less than , let
Calculate .
Solution 1
Since , it follows that
Thus . .
Solution 2
Find the LCMs of the groups of the numbers.
Notice that the groups are relatively prime.
So :
11 if is a multiple of 182.
13 if is a multiple of 154.
14 if is a multiple of 143.
When do we see ambiguities (for example: is a multiple of 11, 13, and 14)? This is only done when is a multiple of $\DeclareMathOperator*{\lcm}{11, 13, 14}=2002$ (Error compiling LaTeX. Unknown error_msg). However, since , this can never happen.
So we have 10 multiples of 182 we have to count (1 to 10 ), and similarly, 12 multiples of 154, and 13 multiples of 143. The sum is . Select .
~hastapasta
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.