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AMC 8 · 1999 · #3

Grade 7 arithmetic
Archetype Small-Domain Constrained Search
fraction-arithmeticsystematic-enumerationmental-arithmetic systematic-enumeration ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic
📏 Medium solution 💡 2 insights

Problem

Which triplet of numbers has a sum NOT equal to 1?

Pick an answer.

(A)
(1/2,1/3,1/6)
(B)
(2,-2,1)
(C)
(0.1,0.3,0.6)
(D)
(1.1,-2.1,1.0)
(E)
(-3/2,-5/2,5)
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Toolkit + CCSS Solution

Understand

Restated: Five triplets of numbers are listed. Four of them have a sum of $1$. Find the one triplet whose sum is NOT equal to $1$.

Givens: Triplet (A): $\left(\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{6}\right)$; Triplet (B): $(2, -2, 1)$; Triplet (C): $(0.1, 0.3, 0.6)$; Triplet (D): $(1.1, -2.1, 1.0)$; Triplet (E): $\left(-\tfrac{3}{2}, -\tfrac{5}{2}, 5\right)$

Unknowns: Which one of the five triplets has a sum that is NOT $1$

Understand

Restated: Five triplets of numbers are listed. Four of them have a sum of $1$. Find the one triplet whose sum is NOT equal to $1$.

Givens: Triplet (A): $\left(\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{6}\right)$; Triplet (B): $(2, -2, 1)$; Triplet (C): $(0.1, 0.3, 0.6)$; Triplet (D): $(1.1, -2.1, 1.0)$; Triplet (E): $\left(-\tfrac{3}{2}, -\tfrac{5}{2}, 5\right)$

Plan

Primary tool: #3 Eliminate Possibilities

The question hands you five labeled candidates and asks which one fails a single, easy-to-check property (sum equals $1$). That is exactly Tool #3 (Eliminate Possibilities): walk through the candidates, compute the sum of each, and cross off any whose sum is $1$. The one that survives the elimination is the answer. No algebra or pattern-hunting is needed — every triplet is just three numbers to add.

Execute — Answer: D

#3 Eliminate Possibilities 5.NF.A.1 Step 1
  • Test (A).
  • Add the fractions using a common denominator of $6$.
$\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{3}{6} + \tfrac{2}{6} + \tfrac{1}{6} = \tfrac{6}{6} = 1$. Sum $= 1$, so (A) is eliminated.

💡 Adding fractions with unlike denominators by rewriting with a common denominator is the Grade 5 standard.

#3 Eliminate Possibilities 7.NS.A.1 Step 2
  • Test (B).
  • Add the signed integers left to right.
$2 + (-2) + 1 = 0 + 1 = 1$. Sum $= 1$, so (B) is eliminated.

💡 Adding a positive and its opposite gives $0$ — the Grade 7 additive-inverse idea.

#3 Eliminate Possibilities 5.NBT.B.7 Step 3
  • Test (C).
  • Add the decimals (each is a tenth).
$0.1 + 0.3 + 0.6 = 1.0$. Sum $= 1$, so (C) is eliminated.

💡 Adding decimals to the hundredths is the Grade 5 standard; here all three numbers already line up at the tenths place.

#3 Eliminate Possibilities 7.NS.A.1 Step 4
  • Test (D).
  • Add the signed decimals.
$1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0$. Sum $= 0$, NOT $1$. (D) survives the elimination.

💡 $1.1$ and $-2.1$ differ in size by exactly $1.0$, so adding them gives $-1.0$, which then cancels the last $+1.0$ to leave $0$.

#3 Eliminate Possibilities 7.NS.A.1 Step 5

Check (E) as a sanity pass, since the answer should be the unique survivor.

$-\tfrac{3}{2} + \left(-\tfrac{5}{2}\right) + 5 = -\tfrac{8}{2} + 5 = -4 + 5 = 1$. Sum $= 1$, so (E) is eliminated. The only triplet left is $\textbf{(D)}$.

💡 Both fractions share denominator $2$, so they combine directly to $-4$, and $-4 + 5 = 1$.

[1] #3 5.NF.A.1 Test (A). Add the fractions using a common denominator of $6$.
[2] #3 7.NS.A.1 Test (B). Add the signed integers left to right.
[3] #3 5.NBT.B.7 Test (C). Add the decimals (each is a tenth).
[4] #3 7.NS.A.1 Test (D). Add the signed decimals.
[5] #3 7.NS.A.1 Check (E) as a sanity pass, since the answer should be the unique survivor.

Review

Reasonableness: Four sums came out to exactly $1$ and one came out to $0$. The problem promises exactly one triplet fails the test, so finding exactly one failure (and four passes) matches the structure perfectly. A quick second look at (D): $1.1 + 1.0 = 2.1$, and $2.1 + (-2.1) = 0$ — same answer either order. The unique mismatch is (D).

Alternative: Tool #16 (Change Focus): instead of computing each sum from scratch, look at each triplet for cancellation. (B) has $2$ and $-2$ that cancel; (E) has $-\tfrac{3}{2} - \tfrac{5}{2} = -4$ that cancels with $+5$ down to $1$. For (D), $1.1$ and $-2.1$ leave $-1.0$, but the remaining number is $+1.0$, which cancels to $0$ — not $1$. So (D) is the only one whose leftover after cancellation is wrong.

CCSS standards used (min grade 7)

  • 5.NF.A.1 Add and subtract fractions with unlike denominators (Computing $\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6}$ in triplet (A) by rewriting all three fractions with denominator $6$.)
  • 5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Computing $0.1 + 0.3 + 0.6$ in triplet (C) as straight decimal addition at the tenths place.)
  • 7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Adding signed numbers in triplets (B), (D), and (E), including the negative decimal $-2.1$ and the negative fractions $-\tfrac{3}{2}, -\tfrac{5}{2}$.)

⭐ When the question is "which one is different?", just test each candidate and cross off the ones that pass. The single triplet that fails is your answer — here, (D) sums to $0$, not $1$.

⭐ When the question is "which one is different?", just test each candidate and cross off the ones that pass. The single triplet that fails is your answer — here, (D) sums to $0$, not $1$.

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