Competition · AMC preparation · step 4 of 4

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 the clues into inequalities

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

Rule out Bo

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

B ＞ M → Bo is not the least

3STEP 3

Rule out Coe

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

Rule out Jo

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

Rule out Flo

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

Name the one left

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."

Turn the clues into inequalities
Rule out Bo
Rule out Coe
Rule out Jo
Rule out Flo
Name the one left

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