AMC 10 · 2004 · #21
Grade 6 arithmeticPick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The set S is the union of two lists that may overlap, so I picture two circles (Tool #12, Draw a Venn Diagram): the size of the union is the two list sizes added, minus the overlap counted twice. So |S|=2004+2004-(number of shared terms), and the whole problem reduces to counting the shared terms. That count is the real work. A number on both lists leaves the same remainders as each list, so the shared numbers form their own steady list that jumps by the least common multiple of the two step sizes (Tool #5, Look for a Pattern). To pin down where that list must stop, I use the smaller of the two final terms as a ceiling (Tool #14, Extreme Principle), which needs each sequence's term rule so I can find those final terms (Tool #4, Introduce a Variable).
Write a rule and find each last term
Term rules: 3n-2 and 7m+2. Their 2004th terms are 6010 and 14030; first-list terms are 1 mod 3, second-list terms are 2 mod 7.
A term is just the starting number plus one step for every jump already made.
6.EE.A.2Introduce A VariableShared numbers jump by 21
A shared number is 1 mod 3 and 2 mod 7. The first is 16, and holding both remainders means stepping by lcm(3,7)=21: 16, 37, 58, …
To keep both remainders at once you can only step by a multiple of both gaps, and the smallest such step is their least common multiple.
To keep both remainders at once you can only step by a multiple of both gaps, so the step is their least common multiple.
▸ Why?
A step that preserves both remainders must be a common multiple, and the smallest is the least common multiple.
▸ Why?
Only the remainder after dividing decides membership in each list, so nothing else can matter.
Count the shared numbers
Shared numbers stop at the smaller last term 6010: 16+21k ≤ 6010 gives k ≤ 285.4…, so k=0,…,285, that is 286 shared numbers.
A number on both lists cannot outrun whichever list ends first, so the shorter list's last term is the finish line.
6.EE.B.5Extreme PrinciplePool the lists with a Venn diagram
Two circles overlap in those 286 numbers, so 2004+2004 double-counts them: |S|=4008-286=3722, choice (A).
Adding both piles double-counts whatever sits in both, so remove one copy of the overlap.
4.OA.A.3Draw A Venn DiagramPool two lists by adding their lengths and subtracting the overlap; the overlapping numbers form their own list that jumps by the least common multiple of the two step sizes.
- Write a rule and find each last term
- Shared numbers jump by 21
- Count the shared numbers
- Pool the lists with a Venn diagram
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