2017 AMC 8 Problem 21

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To help you master 2017 AMC 8 Problem 21, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2017 AMC8 Problem 21

Solution

Solution 1

Logical Case Analysis

Since a + b + c = 0, the numbers cannot all be positive or all be negative. We can break the problem into the only two possible cases (two positive/one negative, or one positive/two negative) and evaluate the expression for each.

Step 1: Understand the behavior of the absolute value fractions

For any nonzero real number, dividing the number by its absolute value gives 1 if the number is positive, and -1 if the number is negative. We can also split the absolute value of a product into the product of the absolute values.

$$\frac{x}{|x|}=1\text{if}x>0$$

$$\frac{x}{|x|}=-1\text{if}x<0$$

$$\frac{abc}{|abc|}=\frac{a}{|a|}\cdot \frac{b}{|b|}\cdot \frac{c}{|c|}$$

Step 2: Determine the possible signs of the variables

We are given that a, b, and c add up to zero. If they were all positive, their sum would be greater than zero. If they were all negative, their sum would be less than zero. Therefore, there must be a mix of positive and negative numbers. Since there are three numbers, there are only two possibilities.

$$\text{Case 1: Two positive numbers and one negative number.}$$

$$\text{Case 2: One positive number and two negative numbers.}$$

Step 3: Evaluate Case 1 (Two positive, one negative)

Suppose two of the numbers are positive and one is negative. The three individual fractions will evaluate to 1, 1, and -1. Their product will be -1. Adding all four terms together gives zero.

$$1+1+(-1)=1$$

$$1\cdot 1\cdot (-1)=-1$$

$$\text{Total Sum}=1+(-1)=0$$

Step 4: Evaluate Case 2 (One positive, two negative)

Suppose one number is positive and two are negative. The three individual fractions will evaluate to 1, -1, and -1. Their product will be positive 1 because multiplying two negative numbers gives a positive result. Adding all four terms together again gives zero.

$$1+(-1)+(-1)=-1$$

$$1\cdot (-1)\cdot (-1)=1$$

$$\text{Total Sum}=-1+1=0$$

Step 5: Reach the conclusion

In both possible cases, the expression evaluates to exactly 0. There are no other ways to assign signs to a, b, and c such that they sum to zero.

$$0$$

Correct answer: A. $0$
The only possible value is 0, which corresponds to choice A. We can check this by testing a specific boundary case, like a=1.5, b=1.5, c=-3. The expression becomes 1 + 1 + (-1) + (1)(1)(-1) = 2 – 1 – 1 = 0, confirming our result.

Solution 2

Algebraic Pairing Method

Instead of checking cases separately, we can use the fact that at least two numbers must have opposite signs. Grouping these two opposite-sign variables together simplifies the entire expression to zero immediately.

Step 1: Rewrite the expression using sign variables

Let x, y, and z represent the signs of a, b, and c. Each of these variables is either 1 or -1. We can rewrite the original expression in terms of these new variables.

$$x=\frac{a}{|a|},y=\frac{b}{|b|},z=\frac{c}{|c|}$$

$$\text{Expression}=x+y+z+xyz$$

Step 2: Identify a pair with opposite signs

Because a + b + c = 0, the three original numbers cannot all share the same sign. Therefore, we can guarantee that at least two of the numbers have opposite signs. Let’s assume without loss of generality that a and b have opposite signs.

Step 3: Simplify using the properties of the opposite pair

Since a and b have opposite signs, one of x and y is 1 and the other is -1. This means their sum is exactly 0, and their product is exactly -1.

$$x+y=0$$

$$xy=-1$$

Step 4: Substitute the pair properties back into the expression

Rearrange the expression to group the opposite pair together, then substitute the sum and product we just found. The entire expression collapses to zero regardless of the value of z.

$$x+y+z+xyz=(x+y)+z+(xy)z$$

$$(x+y)+z+(xy)z=0+z+(-1)z$$

$$z-z=0$$

Correct answer: A. $0$
The expression evaluates to 0 (Choice A). This algebraic approach mathematically guarantees 0 without needing to split into multiple cases, securely verifying our previous result.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2017 Question 21. These real AMC 8 problems test the same key concept—Algebra. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2023 AMC 8 Problem 9

2002 AMC 8 Problem 6

2026 AMC 8 Problem 1

2026 AMC 8 Problem 2

2026 AMC 8 Problem 4

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Sign Function (Signum):

x is any nonzero real number. The problem is built entirely out of this expression. Recognizing that it simply extracts the sign of the variable is the critical first step to solving the problem.

• Multiplicative Property of Absolute Value:

a, b, and c are real numbers. This property allows us to break the large fraction containing the product 'abc' into the product of three individual sign fractions.

• Zero Sum Constraint:

Applies whenever a sum of real numbers equals zero. This logical constraint limits the possible sign combinations we have to check from eight down to just two.

Key Takeaway: Remember This Strategy for Next Time

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