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2004 AMC 10B Problems/Problem 21

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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) \...

Solution

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}.

See also

2004 AMC 10B (ProblemsResources)
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
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