AMC 8 · 2011 · #8

Grade 3 counting
systematic-enumerationsequences-arithmeticparity systematic-enumerationpattern-recognition ↑ Prerequisites: multi-digit-arithmetic
📏 Short solution 💡 2 insights
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Problem
Bag A holds three chips marked 1, 3, 5 and Bag B holds three chips marked 2, 4, 6. One chip is drawn from each bag and the two numbers are added. How many different sums are possible?

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
9

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Table

There are only 3 × 3 = 9 ways to pick one chip from each bag, so Tool #11 (Make a Table) is the cleanest way to lay every sum out in a 3 × 3 addition grid — nothing can be missed. Tool #1 (Find a Pattern) then explains the structure of the answer: odd + even is always odd, and the sums are evenly spaced from the smallest (1+2=3) to the largest (5+6=11), so the distinct sums must be 3, 5, 7, 9, 11.

1STEP 1

Build a 3 × 3 addition table: Bag A values down the side, Bag B across the top, each cell the sum a + b.

+ & 2 & 4 & 6 ; 1 & 3 & 5 & 7 ; 3 & 5 & 7 & 9 ; 5 & 7 & 9 & 11
2STEP 2

The table's nine sums are 3, 5, 7, 5, 7, 9, 7, 9, 11; collapsing the repeats leaves 3, 5, 7, 9, 11.

{3, 5, 7, 5, 7, 9, 7, 9, 11} ⟶ {3, 5, 7, 9, 11}
3STEP 3

The set 3, 5, 7, 9, 11 has 5 members, so there are 5 different possible sums.

|{3, 5, 7, 9, 11}| = 5 → (B)
Answer
5
Every chip in Bag A is odd and every chip in Bag B is even, so every sum is odd. The smallest sum is 1+2=3 and the largest is 5+6=11. The odd numbers from 3 to 11 are 3, 5, 7, 9, 11 — exactly 5 values. Each of these is reachable (for instance 3=1+2, 5=1+4, 7=1+6, 9=3+6, 11=5+6), so the count of 5 is tight: no gaps and no extras. Answer (B) is consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 3 skills — make an addition table, spot the pattern, and count — that you already know!