Difference between revisions of "2004 AMC 10B Problems/Problem 21"

m (Solution 2)
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Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>.
 
Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>.
===Solution 2===
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===<h>Solution 2<h>===
 
Shift down the first sequence by <math>1</math> and the second by <math>9</math> so that the two sequences become <math>0,3,6,\cdots,6009</math> and <math>0,7,14,\cdots,14028</math>. The first becomes multiples of <math>3</math> and the second becomes multiples of <math>7</math>. Their intersection is the multiples of <math>21</math> up to <math>6009</math>. There are <math>\lfloor \frac{6009}{21} \rfloor=286</math> multiples of <math>21</math>. There are <math>4008-286=\boxed{\textbf{(A)}\ 3722}</math> distinct numbers in <math>S</math>.
 
Shift down the first sequence by <math>1</math> and the second by <math>9</math> so that the two sequences become <math>0,3,6,\cdots,6009</math> and <math>0,7,14,\cdots,14028</math>. The first becomes multiples of <math>3</math> and the second becomes multiples of <math>7</math>. Their intersection is the multiples of <math>21</math> up to <math>6009</math>. There are <math>\lfloor \frac{6009}{21} \rfloor=286</math> multiples of <math>21</math>. There are <math>4008-286=\boxed{\textbf{(A)}\ 3722}</math> distinct numbers in <math>S</math>.
  

Revision as of 21:04, 7 October 2019

Problem

Let $1$; $4$; $\ldots$ and $9$; $16$; $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$?

$\mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007$

Solution

Solution 1

The two sets of terms are $A=\{ 3k+1 : 0\leq k < 2004 \}$ and $B=\{ 7l+9 : 0\leq l<2004\}$.

Now $S=A\cup B$. We can compute $|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|$. We will now find $|A\cap B|$.

Consider the numbers in $B$. We want to find out how many of them lie in $A$. In other words, we need to find out the number of valid values of $l$ for which $7l+9\in A$.

The fact "$7l+9\in A$" can be rewritten as "$1\leq 7l+9 \leq 3\cdot 2003 + 1$, and $7l+9\equiv 1\pmod 3$".

The first condition gives $0\leq l\leq 857$, the second one gives $l\equiv 1\pmod 3$.

Thus the good values of $l$ are $\{1,4,7,\dots,856\}$, and their count is $858/3 = 286$.

Therefore $|A\cap B|=286$, and thus $|S|=4008-|A\cap B|=\boxed{3722}$.

<h>Solution 2<h>

Shift down the first sequence by $1$ and the second by $9$ so that the two sequences become $0,3,6,\cdots,6009$ and $0,7,14,\cdots,14028$. The first becomes multiples of $3$ and the second becomes multiples of $7$. Their intersection is the multiples of $21$ up to $6009$. There are $\lfloor \frac{6009}{21} \rfloor=286$ multiples of $21$. There are $4008-286=\boxed{\textbf{(A)}\ 3722}$ distinct numbers in $S$.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
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 10 Problems and Solutions

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