AMC 8 · 2017 · #21

Grade 7 algebralogic

Archetype Small-Domain Constrained Search

paritylogical-deduction caseworksystematic-enumeration ↑ Prerequisites: parity

📏 Medium solution 💡 3 insights

Problem

Suppose $a$ , $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ ?

$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$

Pick an answer.

(A)

(B)

$1 ext{ and }-1$

(C)

$2 ext{ and }-2$

(D)

$0,2, ext{ and }-2$

(E)

$0,1, ext{ and }-1$

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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

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

#9 Solve an Easier Related Problem 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.

$$\dfrac{x}{|x|}=\begin{cases}+1 & \text{if } x>0\\ -1 & \text{if } x<0\end{cases}$$

💡 Absolute value just strips the sign, so dividing a number by its own absolute value leaves only the sign behind.

#2 Make a Systematic List 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.

#3 Eliminate Possibilities 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."

$$\text{Allowed: exactly 1 or exactly 2 of } a,b,c \text{ are negative}$$

💡 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.

#9 Solve an Easier Related Problem 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.

$$\underbrace{(+1)+(+1)+(-1)}_{=1}\ +\ \underbrace{\frac{abc}{|abc|}}_{=-1}\ =\ 0$$

💡 The sign of a product is determined by how many factors are negative; here an odd count (one) gives a negative product.

#9 Solve an Easier Related Problem 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).

$$\underbrace{(+1)+(-1)+(-1)}_{=-1}\ +\ \underbrace{\frac{abc}{|abc|}}_{=+1}\ =\ 0 \;\Rightarrow\; \textbf{(A)}$$

💡 Two negatives multiply to a positive, so $(+)(-)(-)$ is positive — and the two cases conspire to give the same total.

[1] #9 6.NS.C.7 Solve the easier sub-problem $\frac{x}{|x|}$ for any nonzero real number $x$. If

[2] #2 6.NS.C.5 List all $2^3=8$ possible sign patterns for $(a,b,c)$ in a systematic order — by

[3] #3 6.NS.C.5 Eliminate impossible patterns. If all three are positive, then $a+b+c>0$, contra

[4] #9 7.NS.A.2 Evaluate Family 1: two positives and one negative. The three sign terms sum to $

[5] #9 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.5 Understand 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.7 Understand 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.2 Apply 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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