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Difference between revisions of "2002 AMC 12B Problems/Problem 21"
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\qquad\mathrm{(E)}\ 2002</math>
 
\qquad\mathrm{(E)}\ 2002</math>
  
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== Solution ==
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== 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;\\
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14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\
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\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 ==

Latest revision as of 10:52, 8 April 2022

Problem

For all positive integers $n$ less than $2002$, let

\begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n\ \text{is\ divisible\ by\ }13\ \text{and\ }14;\\ 13, & \text{if\ }n\ \text{is\ divisible\ by\ }14\ \text{and\ }11;\\ 14, & \text{if\ }n\ \text{is\ divisible\ by\ }11\ \text{and\ }13;\\ 0, & \text{otherwise}. \end{array} \right. \end{eqnarray*}

Calculate $\sum_{n=1}^{2001} a_n$.

$\mathrm{(A)}\ 448 \qquad\mathrm{(B)}\ 486 \qquad\mathrm{(C)}\ 1560 \qquad\mathrm{(D)}\ 2001 \qquad\mathrm{(E)}\ 2002$

Solution 1

Since $2002 = 11 \cdot 13 \cdot 14$, it follows that \begin{eqnarray*} a_n =\left\{ \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} \right. \end{eqnarray*}

Thus $\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}$. $\begin{array}{lr} 11, & \text{if\ }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{if\ }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array}$.

Solution 2

Find the LCMs of the groups of the numbers.

Notice that the groups are relatively prime.

So $a_n=$:

11 if $n$ is a multiple of 182.

13 if $n$ is a multiple of 154.

14 if $n$ is a multiple of 143.

When do we see ambiguities (for example: $n$ is a multiple of 11, 13, and 14)? This is only done when $n$ is a multiple of $\operatorname{lcm}(11,13,14)=2002$. However, since $n<2002$, this can never happen.

So we have 10 multiples of 182 we have to count (1 to 10 $*182$), and similarly, 12 multiples of 154, and 13 multiples of 143. The sum is $10*11+12*13+13*14=448$. Select $\boxed{A}$.

~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

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