AMC 10 · 2017 · #3

Grade 7 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
interval-arithmeticsign-analysislogical-deduction bound-inequality-then-enumerateguess-and-check ↑ Prerequisites: interval-arithmetic
📏 Medium solution 💡 2 insights
Problem
Three real numbers obey 0 ＜ x ＜ 1, -1 ＜ y ＜ 0, and 1 ＜ z ＜ 2. Among the five listed expressions, find the one that comes out positive for every allowed choice of x, y, and z — not just for some lucky values.

Pick an answer.

(A)
$y+x^2$
(B)
y+xz
(C)
$y+y^2$
(D)
$y+2y^2$
(E)
y+z

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

Tool #3 (Eliminate Possibilities): five choices, only one survives, so killing four is as good as proving the fifth. Tool #14 (Extreme Principle): an expression is positive for all values exactly when its worst case (smallest possible value) is still positive — so I push each range to its low end. Tool #6 (Guess and Check): for a choice that can fail, one concrete set of numbers that makes it negative is enough to throw it out.

1STEP 1

Read the ranges

Necessarily positive means an expression must clear 0 even at its worst case, so I test every choice at the low end of each interval.

0 ＜ x ＜ 1, -1 ＜ y ＜ 0, 1 ＜ z ＜ 2
2STEP 2

Add the bounds for y+z

In choice (E), y and z each exceed their low bounds -1 and 1, so their sum exceeds (-1)+1=0 — positive for every allowed value.

y ＞ -1 and z ＞ 1 → y+z ＞ (-1)+1=0
3STEP 3

Knock out (A) and (B)

Take x near 0 (x=0.1, y=-0.9, z=1.5): x² and xz shrink toward 0, leaving y negative, so y+x²=-0.89 and y+xz=-0.75 — both fail.

x=0.1, y=-0.9, z=1.5: y+x²=-0.89, y+xz=-0.75
4STEP 4

Knock out (C) and (D) by factoring

Factor: (C) y+y²=y(1+y) is negative times positive, so always below 0; (D) y+2y²=y(1+2y) flips sign at y=-1/2, so not always positive.

y+y²=y(1+y) ＜ 0; y+2y²=y(1+2y) changes sign at y=-1/2
5STEP 5

Only (E) survives

(A)–(D) each fail for some allowed values, so the only necessarily positive expression is choice (E).

y+z ＞ 0 always → (E)
Answer
y+z
Spot-check the extremes for y+z: even at the worst corner y→ -1, z→ 1, the sum approaches 0 from above and never dips below it, so y+z stays positive — consistent with choosing (E). Meanwhile the discarded choices each had a concrete failing example, so nothing was thrown out by accident.
💡Key takeaway

An expression is 'always positive' only if even its smallest possible value stays above zero; adding the low ends y ＞ -1 and z ＞ 1 gives y+z ＞ 0, so the answer is (E).

  • Read the ranges
  • Add the bounds for y+z
  • Knock out (A) and (B)
  • Knock out (C) and (D) by factoring
  • Only (E) survives

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