AMC 8 · 1999 · #6

Grade 1 logic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

logical-deductionif-then-reasoning logical-deductioncasework

📏 Medium solution 💡 3 insights

Problem

Five people — Bo, Coe, Flo, Jo, Moe — have different amounts of money. We are given a handful of "more than / less than" clues that compare some of them in pairs. The question asks who holds the smallest amount.

Pick an answer.

(A)

(B)

Coe

(C)

Flo

(D)

(E)

Moe

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

There are only five candidates for "least", so the fastest path is Tool #3 (Eliminate Possibilities): find one clue that beats each candidate and cross them off. The moment four names are crossed off, the fifth must be the answer. Tool #1 (Draw a Diagram) supports the count by stacking the clues into a single chain M ＜ J ＜ B ＜ F (plus a side note M ＜ C), so the ordering is visible at a glance and nothing gets missed.

1STEP 1

Turn each clue into a ＜/＞ statement with name initials: J ＜ F, B ＜ F, B ＞ M, C ＞ M, M ＜ J, J ＜ B.

J ＜ F, B ＜ F, B ＞ M, C ＞ M, M ＜ J, J ＜ B

2STEP 2

Eliminate Bo: B ＞ M means Bo beats Moe, so Bo is not the smallest.

B ＞ M → Bo is not the least

3STEP 3

Eliminate Coe: C ＞ M shows Coe also beats Moe, so Coe is not the smallest either.

C ＞ M → Coe is not the least

4STEP 4

Eliminate Jo: M ＜ J means Jo has more than Moe, so Jo is not the smallest.

M ＜ J → Jo is not the least

5STEP 5

Eliminate Flo by chaining: F ＞ J and J ＞ M give F ＞ M, so Flo beats Moe too.

F ＞ J and J ＞ M → F ＞ M

6STEP 6

All four others are eliminated, so the bottom of the chain M ＜ J ＜ B ＜ F is the least — Moe, answer (E).

M ＜ J ＜ B ＜ F and M ＜ C → Moe is the least → (E)

Answer

Moe

Plug in any concrete amounts that fit the chain M ＜ J ＜ B ＜ F and M ＜ C — for example M = 1, J = 2, B = 3, F = 4, C = 5 (all different, all clues satisfied). The smallest value is 1, held by Moe. Try another assignment such as M = 10, J = 20, B = 30, F = 40, C = 100: still smallest is Moe. Every clue is satisfied and Moe always ends up at the bottom, so (E) is robust.

💡Key takeaway

If a clue says someone has more money than Moe, that person cannot be the least — and Bo, Coe, Jo, and Flo all get knocked out by exactly that rule, leaving Moe. The only Grade 1 idea you need is "order objects from least to greatest, using the chain rule."

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