AMC 8 · 2007 · #15

Grade 6 logic

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

if-then-reasoninglogical-deductioninterval-arithmetic caseworkcontradiction-elementary ↑ Prerequisites: interval-arithmeticif-then-reasoning

📏 Medium solution 💡 1 insight

Problem

Three positive numbers satisfy 0 ＜ a ＜ b ＜ c. Of the five inequalities listed, which one can NEVER be true?

Pick an answer.

(A)

$a + c < b$

(B)

$$a \cdot b < c$$

(C)

$a + b < c$

(D)

$$a \cdot c < b$$

(E)

$$\dfrac{b}{c} = a$$

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

The question asks which choice is impossible, so Tool #3 (Eliminate Possibilities) is the direct fit: rule out every choice that can happen, and the lone survivor is the answer. To rule a choice OUT of the suspect list, we use Tool #6 (Guess and Check) — pick concrete numbers that satisfy 0 ＜ a ＜ b ＜ c AND the choice's inequality. To rule a choice IN as impossible, we use the basic rules of inequalities: adding a positive number makes things larger. We are deliberately avoiding Tool #13 (Algebra) because no equation needs solving — small numerical experiments and one transitivity step are enough.

1STEP 1

Test (B) a · b ＜ c: take a = 1, b = 2, c = 5 — then a · b = 2 ＜ 5, so (B) can happen and is eliminated.

a = 1, b = 2, c = 5 → a · b = 2 ＜ 5 = c ✓

2STEP 2

Test (C) a + b ＜ c: take a = 1, b = 2, c = 10 — then a + b = 3 ＜ 10, so (C) can happen and is eliminated.

a = 1, b = 2, c = 10 → a + b = 3 ＜ 10 = c ✓

3STEP 3

Test (D) a · c ＜ b: whole numbers fail, so try a = $\frac{1}{2}$ , b = 1, c = $\frac{3}{2}$ — then a · c = $\frac{3}{4}$ ＜ 1, so (D) can happen and is eliminated.

a =

\frac{1}{2}

, b = 1, c =

\frac{3}{2}

→ a · c =

\frac{3}{4}

＜ 1 = b ✓

4STEP 4

Test (E) $\frac{b}{c}$ = a: pick a = $\frac{1}{2}$ , c = 4, set b = 2 — then $\frac{b}{c}$ = $\frac{2}{4}$ = $\frac{1}{2}$ = a, so (E) can happen and is eliminated.

a =

\frac{1}{2}

, b = 2, c = 4 →

\frac{b}{c}

\frac{2}{4}

\frac{1}{2}

= a ✓

5STEP 5

Only (A) remains. Since b ＜ c and a ＞ 0, adding a keeps it larger, giving b ＜ a + c — the exact opposite of (A), so (A) is impossible.

b ＜ c_given and a ＞ 0_given → b ＜ c ＜ c + a = a + c → b ＜ a + c → (A)

Answer

a + c ＜ b

Double-check (A) with a numerical try. Pick a = 1, b = 2, c = 3: then a + c = 4, which is greater than b = 2. Pick a =

\frac{1}{10}

, b =

\frac{1}{2}

, c = 1: then a + c = 1.1, again greater than b = 0.5. No matter how small a is, the inequality a + c ＜ b keeps failing because c alone is already larger than b and we are only adding to it. The four eliminated choices each came with a concrete witness, so the lone survivor is indeed (A).

💡Key takeaway

Adding a positive number always makes things bigger — so a + c can never sink below b. Spot that one rule and this AMC 8 inequality puzzle is decided.

More like this