Difference between revisions of "2002 AMC 12B Problems/Problem 21"
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\qquad\mathrm{(E)}\ 2002</math> | \qquad\mathrm{(E)}\ 2002</math> | ||
− | == Solution == | + | == Solution 1== |
Since <math>2002 = 11 \cdot 13 \cdot 14</math>, it follows that | Since <math>2002 = 11 \cdot 13 \cdot 14</math>, it follows that | ||
<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
Line 34: | Line 34: | ||
Thus <math>\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}</math>. | Thus <math>\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}</math>. | ||
+ | <math>\begin{array}{lr} | ||
+ | 11, & \text{if\ }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ | ||
+ | 13, & \text{if\ }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ | ||
+ | 14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ | ||
+ | \end{array}</math>. | ||
+ | == Solution 2 == | ||
+ | |||
+ | Find the LCMs of the groups of the numbers. | ||
+ | |||
+ | Notice that the groups are relatively prime. | ||
+ | |||
+ | So <math>a_n=</math>: | ||
+ | |||
+ | 11 if <math>n</math> is a multiple of 182. | ||
+ | |||
+ | 13 if <math>n</math> is a multiple of 154. | ||
+ | |||
+ | 14 if <math>n</math> is a multiple of 143. | ||
+ | |||
+ | When do we see ambiguities (for example: <math>n</math> is a multiple of 11, 13, and 14)? This is only done when <math>n</math> is a multiple of <math>\operatorname{lcm}(11,13,14)=2002</math>. However, since <math>n<2002</math>, this can never happen. | ||
+ | |||
+ | So we have 10 multiples of 182 we have to count (1 to 10 <math>*182</math>), and similarly, 12 multiples of 154, and 13 multiples of 143. The sum is <math>10*11+12*13+13*14=448</math>. Select <math>\boxed{A}</math>. | ||
+ | |||
+ | ~hastapasta | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:52, 8 April 2022
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
. 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.