AMC 10 · 2002 · #10

Grade 7 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

sequences-arithmeticmodular-arithmeticconsecutive-integer-count easier-related-problembound-inequality-then-enumerate ↑ Prerequisites: sequences-arithmetic

📏 Long solution 💡 3 insights

Problem

Take the seven numbers 1, 4, 7, 10, 13, 16, 19. Choose three different ones and add them. Different choices can land on the same total — for example 1+19 and 4+16 already tie. Count how many different totals are possible, not how many ways there are to choose.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Solve an Easier Related Problem

Adding up 35 triples by hand and crossing off repeats would work but invites mistakes, so the plan is to make the numbers smaller before counting anything. Tool #5 (Look for a Pattern) notices that the seven numbers climb by 3, which is the fact the whole problem rests on. Tool #9 (Solve an Easier Related Problem) trades the set {1,4,…,19} for the set {0,1,2,3,4,5,6}, so the question becomes the far easier one of which small sums of three distinct digits are reachable. Tool #14 (Extreme Principle) then finds the smallest and largest reachable sum, drawing a window that the answer cannot exceed. That window is only an upper bound, and a bound is not a count — so tool #2 (Make a Systematic List) does the part that is easy to skip and easy to get wrong: exhibiting an actual choice for every single value inside the window.

1STEP 1

Spot the steps of three

Each number is 3 more than the last, so any three add to a multiple of 3.

1, 4, 7, 10, 13, 16, 19: each step +3; sum of three ≡ 1+1+1 ≡ 0 (mod 3)

2STEP 2

Trade the set for 0 through 6

Writing each as 1+3k trades the set for 0 through 6 and the total for 3+3S.

member = 1+3k, k ∈ {0,1,2,3,4,5,6}; sum = 3 + 3S where S = k₁+k₂+k₃

3STEP 3

One total per S, and S is boxed in

The total grows strictly with S, and S runs from 3 to 15, giving at most 13 values.

sum = 3(S+1) strictly increasing in S; 0+1+2 = 3 ≤ S ≤ 4+5+6 = 15; at most 15-3+1 = 13

4STEP 4

Exhibit every total in the window

Three families of choices realise every multiple of 3 from 12 to 48, so the count is exactly 13, choice (A).

{1+4+t} = {12,15,18,21,24}, {1+m+19} = {24,27,30,33,36}, {s+16+19} = {36,39,42,45,48}; (48-12)/3+1 = 13 → (A)

Answer

Choice (E) 35 is exactly C(7, 3), the number of ways to choose three members — it answers a different question, and the gap between 35 choices and 13 totals is expected, because ties are everywhere once 1+19 = 4+16 = 7+13 = 20. Choices (C) 24 and (D) 30 are themselves totals on the list rather than counts of totals. A structural check confirms 13: the set is symmetric about 10, since replacing each member m by 20-m maps the set onto itself, and that swap turns a total N into 60-N. So the set of reachable totals must be symmetric about 30 — and {12, 15, …, 48} is exactly symmetric about 30, with 12 pairing to 48 and 30 fixed in the middle. A lopsided answer would have signalled a missed case.

💡Key takeaway

Finding the smallest and largest total only draws the window — you still have to show a real choice for every value inside it before that window becomes a count.

Spot the steps of three
Trade the set for 0 through 6
One total per S, and S is boxed in
Exhibit every total in the window

Spot the steps of three

Trade the set for 0 through 6

One total per S, and S is boxed in

Exhibit every total in the window

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