2004 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2004 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

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Concepts:arithmetic sequence least common multiple inclusion-exclusion

Difficulty rating: 1740

21.

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 $S?$

$3722$

$3732$

$3914$

$3924$

$4007$

Solution:

The first sequence is $1 + 3k$ with largest term $6010,$ and the second is $9 + 7j$ with a much larger last term, so the binding limit is $6010.$

A common value has the form $16 + 21m$ (the first shared term is $16,$ spaced by $\mathrm{lcm}(3, 7) = 21$ ). Requiring $16 + 21m \le 6010$ gives $0 \le m \le 285,$ that is $286$ common numbers.

The number of distinct values is $2004 + 2004 - 286 = 3722.$

Thus, the correct answer is A.

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