AMC 10 · 2002 · #21
Grade 6 number-theoryPick an answer.
Almost every term is 0, so the sum is really three small counting jobs glued together — that is Tool #7 (Identify Subproblems): count how many n trigger each of the three rules, multiply each count by its value, and add. Before adding, though, the three sets have to be checked for overlap, because a single n obeying two rules would be both double-counted and ambiguously defined; Tool #12 (Draw a Venn Diagram) is the habit of asking what lies in the intersections, and here the intersections turn out to be empty. Tool #2 (Make a Systematic List) then does each count: 'divisible by both' means 'divisible by the least common multiple', so each rule fires exactly on the multiples of one fixed number, and the multiples of m up to 2001 are simply m, 2m, 3m, …
Replace each pair by one number
Divisible by two coprime numbers means divisible by their product: 182, 154, 143.
Two divisibility demands at once collapse into one demand about the least common multiple.
Two divisibility demands at once collapse into a single demand about their least common multiple.
▸ Why?
The numbers on two multiple lists at once are exactly the multiples of the first number where the lists agree.
▸ Why?
The three groups never overlap in range, so their counts can simply be added at the end.
Check the three cases cannot overlap
Two rules could clash only at a multiple of 2002, which the range excludes.
The only numbers that could belong to two groups are multiples of all three factors, and the very first one sits just outside the range.
6.NS.B.4Draw A Venn DiagramCount each group of multiples
Each divides 2002 exactly, so the counts in range are 10, 12, 13.
Since each spacing divides 2002 evenly, the count is one less than 2002 ÷ m — the last multiple lands exactly on the forbidden 2002.
4.OA.B.4Make A Systematic ListWeight each count and add
Weighting and adding gives 110 + 156 + 182 = 448, choice (A).
Only 35 terms out of 2001 are nonzero, so the whole sum is three products added together.
4.OA.A.3Identify SubproblemsWhen a sum is almost all zeros, find the few places where something happens, make sure those places do not overlap, and then just count and multiply.
- Replace each pair by one number
- Check the three cases cannot overlap
- Count each group of multiples
- Weight each count and add