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

## 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 $|S|=4008-|A\cap B|=\boxed{3722}$.

### Solution 2

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 $\lfloor \frac{6009}{21} \rfloor$ multiples of $286$. There are $4008-286=\boxed{\textbf{(A)}\ 3722}$ distinct numbers in $S$.

## See also

 2004 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 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