AMC 8 · 2017 · #21
Grade 7 algebralogicProblem
Suppose a, b, and c are nonzero real numbers, and a+b+c=0. What are the possible value(s) for ∣a∣a+∣b∣b+∣c∣c+∣abc∣abc?
Pick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three nonzero real numbers $a$, $b$, $c$ satisfy $a+b+c=0$. We are asked which values the expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ can take. Each piece $\frac{x}{|x|}$ is just the sign of $x$ (either $+1$ or $-1$), so the whole expression is a sum of four $\pm 1$ terms whose actual value depends on how the signs of $a$, $b$, $c$ line up under the constraint $a+b+c=0$.
Givens: $a$, $b$, $c$ are nonzero real numbers; $a+b+c=0$; The expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$; Answer choices: (A) $0$; (B) $1$ and $-1$; (C) $2$ and $-2$; (D) $0,2,-2$; (E) $0,1,-1$
Unknowns: The set of all possible values the expression can take
Understand
Restated: Three nonzero real numbers $a$, $b$, $c$ satisfy $a+b+c=0$. We are asked which values the expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ can take. Each piece $\frac{x}{|x|}$ is just the sign of $x$ (either $+1$ or $-1$), so the whole expression is a sum of four $\pm 1$ terms whose actual value depends on how the signs of $a$, $b$, $c$ line up under the constraint $a+b+c=0$.
Givens: $a$, $b$, $c$ are nonzero real numbers; $a+b+c=0$; The expression $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$; Answer choices: (A) $0$; (B) $1$ and $-1$; (C) $2$ and $-2$; (D) $0,2,-2$; (E) $0,1,-1$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The scary-looking expression collapses once we solve the easier sub-question $\frac{x}{|x|}=?$ 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 $2^3=8$ 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.
Execute — Answer: A
6.NS.C.7 Step 1 - Solve the easier sub-problem $\frac{x}{|x|}$ for any nonzero real number $x$.
- If $x>0$, then $|x|=x$ and the ratio is $+1$.
- If $x<0$, then $|x|=-x$ and the ratio is $-1$.
- So each of the four terms in our big expression is just the sign of its variable — $+1$ or $-1$ — no calculation needed.
💡 Absolute value just strips the sign, so dividing a number by its own absolute value leaves only the sign behind.
6.NS.C.5 Step 2 - List all $2^3=8$ possible sign patterns for $(a,b,c)$ in a systematic order — by how many of them are negative.
- Then use the constraint $a+b+c=0$ to keep only the patterns that can actually occur.
💡 There are only finitely many sign patterns, so we can just walk through them in order without missing any.
6.NS.C.5 Step 3 - Eliminate impossible patterns.
- If all three are positive, then $a+b+c>0$, contradicting $a+b+c=0$.
- If all three are negative, then $a+b+c<0$, also a contradiction.
- So $(+,+,+)$ and $(-,-,-)$ are out.
- The six survivors split into two families: "two positives and one negative" and "one positive and two negatives."
💡 A sum of three numbers can only equal zero if they have mixed signs — that's what the constraint $a+b+c=0$ is telling us.
7.NS.A.2 Step 4 - Evaluate Family 1: two positives and one negative.
- The three sign terms sum to $1+1+(-1)=1$.
- The product $abc$ has the sign of $(+)(+)(-)=-$, so $\frac{abc}{|abc|}=-1$.
- Total $=1+(-1)=0$.
- By symmetry, it does not matter which of $a,b,c$ is the negative one — the expression's value depends only on how many negatives there are.
💡 The sign of a product is determined by how many factors are negative; here an odd count (one) gives a negative product.
7.NS.A.2 Step 5 - Evaluate Family 2: one positive and two negatives.
- The three sign terms sum to $1+(-1)+(-1)=-1$.
- The product $abc$ has the sign of $(+)(-)(-)=+$, so $\frac{abc}{|abc|}=+1$.
- Total $=-1+1=0$.
- Both surviving families give the same value, $0$, so that is the only possible value of the expression — answer choice (A).
💡 Two negatives multiply to a positive, so $(+)(-)(-)$ is positive — and the two cases conspire to give the same total.
6.NS.C.7 Solve the easier sub-problem $\frac{x}{|x|}$ for any nonzero real number $x$. If 6.NS.C.5 List all $2^3=8$ possible sign patterns for $(a,b,c)$ in a systematic order — by 6.NS.C.5 Eliminate impossible patterns. If all three are positive, then $a+b+c>0$, contra 7.NS.A.2 Evaluate Family 1: two positives and one negative. The three sign terms sum to $ 7.NS.A.2 Evaluate Family 2: one positive and two negatives. The three sign terms sum to $ Review
Reasonableness: 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.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices alone: a single concrete test like $a=1,b=1,c=-2$ already yields $0$, so the answer must include $0$. That rules out (B) and (C) immediately. To pick between (A), (D), (E), try a Family-2 example like $a=2,b=-1,c=-1$, which also gives $0$ — not $\pm 2$ and not $\pm 1$. With both families exhausted and only $0$ appearing, (A) is forced.
CCSS standards used (min grade 7)
6.NS.C.5Understand that positive and negative numbers describe quantities (Listing the $8$ possible sign patterns of $(a,b,c)$ and using the constraint $a+b+c=0$ to rule out the all-positive and all-negative cases.)6.NS.C.7Understand ordering and absolute value of rational numbers (Recognizing that $\frac{x}{|x|}$ for nonzero $x$ equals $+1$ when $x>0$ and $-1$ when $x<0$ — i.e., dividing by $|x|$ extracts the sign.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Determining the sign of the product $abc$ from the signs of its factors (two positives and a negative give a negative product; one positive and two negatives give a positive product).)
⭐ This AMC 8 problem only needs Grade 7 sign-of-a-product reasoning with positive and negative numbers you already know!
⭐ 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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