AMC 8 · 2017 · #21

Grade 7 algebralogic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

paritylogical-deduction caseworksystematic-enumeration ↑ Prerequisites: parity

📏 Medium solution 💡 3 insights

Problem

Three nonzero real numbers a, b, c satisfy a+b+c=0. We are asked which values the expression a/|a|+b/|b|+c/|c|+abc/|abc| can take. Each piece x/|x| is just the sign of x (either +1 or -1), so the whole expression is a sum of four ± 1 terms whose actual value depends on how the signs of a, b, c line up under the constraint a+b+c=0.

Pick an answer.

(A)

(B)

$1\text{ and }-1$

(C)

$2\text{ and }-2$

(D)

$0,2,\text{ and }-2$

(E)

$0,1,\text{ and }-1$

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

The scary-looking expression collapses once we solve the easier sub-question x/|x|=? for any nonzero x (Tool #9). Each term is just a sign, +1 or -1. After that, we want to know which sign-patterns of (a,b,c) are even allowed: there are only 2³=8 patterns, so we list them systematically (Tool #2) and use the constraint a+b+c=0 to eliminate the all-positive and all-negative cases (Tool #3). Two sign-pattern families survive, and by symmetry we only have to evaluate the expression once per family.

1STEP 1

The easier piece x/|x| is +1 when x is positive and -1 when x is negative — so every term is just +1 or -1, the variable's sign.

x/|x|= +1 & if x ＞ 0 ; -1 & if x ＜ 0

2STEP 2

List all 8 sign patterns of (a,b,c), ordered by how many are negative, then filter them with the constraint a+b+c=0.

(+,+,+), (+,+,-), (+,-,+), (-,+,+), (+,-,-), (-,+,-), (-,-,+), (-,-,-)

3STEP 3

All-positive sums too high and all-negative too low, so both are impossible — leaving exactly 1 or 2 negatives among a, b, c.

Allowed: exactly 1 or exactly 2 of a,b,c are negative

4STEP 4

Family 1 (two positives, one negative): the three signs give 1+1-1=1, and abc is negative so its term is -1 — total 0.

(+1)+(+1)+(-1)=1 + abc/|abc|=-1 = 0

5STEP 5

Family 2 (one positive, two negatives): signs give 1-1-1=-1, abc positive so +1 — total 0 again. Same both ways, so answer (A).

(+1)+(-1)+(-1)=-1 + abc/|abc|=+1 = 0 → (A)

Answer

Concrete test of Family 1: pick a=1, b=1, c=-2 (sum is 0). Expression =1+1+(-1)+-

\frac{2}{2}

=1+1-1-1=0. Concrete test of Family 2: pick a=2, b=-1, c=-1 (sum is 0). Expression =1+(-1)+(-1)+

\frac{2}{2}

=1-1-1+1=0. Both checks give 0, matching choice (A). The deeper reason it always cancels: in Family 1 the three signs sum to +1 and the product-sign is -1, and in Family 2 the three signs sum to -1 and the product-sign is +1 — opposite signs always cancel.

💡Key takeaway

This AMC 8 problem only needs Grade 7 sign-of-a-product reasoning with positive and negative numbers you already know!

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