AMC 8 · 2015 · #23

Grade 6 logicarithmetic

Archetype Small-Domain Constrained Search

sequences-arithmeticparityfraction-arithmeticlogical-deduction caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var

📏 Long solution 💡 4 insights

Problem

Tom has twelve slips of paper which he wants to put into five cups labeled $A$ , $B$ , $C$ , $D$ , $E$ . He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$ . The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$ . If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$ , then the slip with $3.5$ must go into what cup?

$\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Tom has $12$ slips with numbers $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4, 4.5$ to drop into five cups $A, B, C, D, E$. Each cup's slip-sum must be an integer, and the five sums must be consecutive integers increasing from $A$ to $E$. A $2$ is already in cup $E$ and a $3$ is already in cup $B$. Which cup must hold the $3.5$?

Givens: Slip multiset: $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4, 4.5$ ($12$ slips total); Each cup's sum must be an integer; The five cup-sums are consecutive integers, increasing from $A$ to $E$; A $2$ is already in cup $E$; a $3$ is already in cup $B$; Answer choices: (A) $A$, (B) $B$, (C) $C$, (D) $D$, (E) $E$

Unknowns: The cup ($A, B, C, D,$ or $E$) that must contain the slip with $3.5$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

Rather than searching all placements at once, Tool #7 (Identify Subproblems) breaks the puzzle into three clean stages: (1) compute the total of all slips, (2) use that total to pin down the five required cup-sums, (3) ask where the $3.5$ can legally go. Once the five target sums are known, the third stage is a textbook Tool #3 (Eliminate Possibilities) sweep — try the $3.5$ in each cup and rule out any cup where the remaining slips cannot complete the required sum. The fixed slips ($2$ in $E$, $3$ in $B$) and the parity rule for the $.5$ slips do most of the eliminating.

Execute — Answer: D

#7 Identify Subproblems 5.NBT.B.7 Step 1

Add all $12$ slip values to get the grand total.
Group equal slips first: three $2$s give $6$, two $2.5$s give $5$, four $3$s give $12$, plus a $3.5$, a $4$, and a $4.5$.

$$6 + 5 + 12 + 3.5 + 4 + 4.5 = 35$$

💡 Grouping like-value slips before adding is the Grade 5 decimal-arithmetic move that keeps the total honest.

#7 Identify Subproblems 6.EE.B.7 Step 2

Translate "five consecutive integers summing to $35$" into the per-cup targets.
Five consecutive integers have an average equal to their middle term, so the middle one is $35 \div 5 = 7$.
The five sums are therefore $5, 6, 7, 8, 9$ — cup $A = 5$, $B = 6$, $C = 7$, $D = 8$, $E = 9$.

$$\text{avg} = 35 \div 5 = 7 \;\Rightarrow\; A=5,\; B=6,\; C=7,\; D=8,\; E=9$$

💡 Using the average to recover consecutive integers is the Grade 6 equation-reasoning shortcut for $n+(n+1)+\dots+(n+4)=35$.

#3 Eliminate Possibilities 5.NBT.B.7 Step 3

Try the $3.5$ in cup $A$ (target $5$).
The slips left over in cup $A$ would have to sum to $5 - 3.5 = 1.5$.
The smallest remaining slip is $2$, so no subset can produce $1.5$.
Cup $A$ is ruled out.

$$5 - 3.5 = 1.5 < 2 = \min(\text{remaining slips}) \;\Rightarrow\; \text{impossible}$$

💡 Comparing the required leftover to the smallest available slip is the fastest way to kill a candidate cup.

#3 Eliminate Possibilities 5.NBT.B.7 Step 4

Try the $3.5$ in cup $B$ (target $6$).
Cup $B$ already holds a $3$, so adding the $3.5$ pushes its running sum to $6.5$, which already exceeds the target $6$.
Cup $B$ is ruled out.

$$3 + 3.5 = 6.5 > 6 \;\Rightarrow\; \text{impossible}$$

💡 When a partial sum already overshoots, no extra slips can rescue it — adding slips can only increase the sum.

#3 Eliminate Possibilities 5.NBT.B.7 Step 5

Try the $3.5$ in cup $C$ (target $7$).
The remaining slips in cup $C$ would have to sum to $7 - 3.5 = 3.5$.
The only way to land on a half-integer is to use exactly one $.5$ slip ($2.5$); pairing it with anything else needs a value of $3.5 - 2.5 = 1$, which doesn't exist.
Using a $2.5$ alone gives only $2.5$.
Cup $C$ is ruled out.

$$7 - 3.5 = 3.5;\; 3.5 - 2.5 = 1 \notin \text{slips} \;\Rightarrow\; \text{impossible}$$

💡 To hit a half-integer total you need an odd number of $.5$ slips — checking which $.5$ combinations are available collapses the search quickly.

#3 Eliminate Possibilities 5.NBT.B.7 Step 6

Try the $3.5$ in cup $E$ (target $9$).
Cup $E$ already has a $2$, so the remaining slips in $E$ must sum to $9 - 2 - 3.5 = 3.5$.
This is the same impossible half-integer requirement we just ruled out in cup $C$.
Cup $E$ is ruled out.

$$9 - 2 - 3.5 = 3.5 \;\Rightarrow\; \text{same dead end as cup } C$$

💡 Recognizing the same leftover sum that just failed elsewhere saves a second elimination pass.

#3 Eliminate Possibilities 5.NBT.B.7 Step 7

Cup $D$ (target $8$) is all that's left.
Verify: $8 - 3.5 = 4.5$, and there is exactly one $4.5$ slip available.
Placing $\{3.5, 4.5\}$ in cup $D$ gives the required sum of $8$.
The $3.5$ must go in cup $D$.

$8 - 3.5 = 4.5 = $ available slip $\;\Rightarrow\; \textbf{(D)}$

💡 After elimination, the only surviving option must be the answer — and a quick existence check confirms it works.

[1] #7 5.NBT.B.7 Add all $12$ slip values to get the grand total. Group equal slips first: three

[2] #7 6.EE.B.7 Translate "five consecutive integers summing to $35$" into the per-cup targets.

[3] #3 5.NBT.B.7 Try the $3.5$ in cup $A$ (target $5$). The slips left over in cup $A$ would have

[4] #3 5.NBT.B.7 Try the $3.5$ in cup $B$ (target $6$). Cup $B$ already holds a $3$, so adding th

[5] #3 5.NBT.B.7 Try the $3.5$ in cup $C$ (target $7$). The remaining slips in cup $C$ would have

[6] #3 5.NBT.B.7 Try the $3.5$ in cup $E$ (target $9$). Cup $E$ already has a $2$, so the remaini

[7] #3 5.NBT.B.7 Cup $D$ (target $8$) is all that's left. Verify: $8 - 3.5 = 4.5$, and there is e

Review

Reasonableness: Sanity-check the full placement implied by cup $D = \{3.5, 4.5\}$. Remaining slips: $2, 2, 2, 2.5, 2.5, 3, 3, 3, 4$ (plus the fixed $2$ in $E$ and $3$ in $B$). Cup $A=5$: $\{2, 3\}$. Cup $B=6$: already has $3$, add $3$ to get $6$. Cup $C=7$: $\{2.5, 2.5, 2\}=7$. Cup $E=9$: already has $2$, add $3$ and $4$ to get $9$. All slips used, all sums match. Answer (D) is consistent.

Alternative: Tool #16 (Count the Complement / parity argument). The four half-integer slips are $2.5, 2.5, 3.5$ — only three. To make every cup-sum an integer, the $.5$ contributions per cup must total an integer, so the $.5$ slips must be partitioned so each cup gets an even number of them (including zero). Three half-slips can only fit as $\{2+1\}$ across two cups or $\{3\}$ in one cup. Trying $\{3\}$ in one cup: that cup's $.5$-slip sum is $7.5$, forcing the cup target to be at least $8$, and only cup $D$ ($=8$) or $E$ ($=9$) can hold all three. Cup $E$ already has the $2$, leaving room for only $7$ more, not $7.5$ even — wait, $2 + 7.5 = 9.5$, miss. So all three half-slips ($2.5+2.5+3.5=8.5$) cannot fit any single cup. Therefore exactly two half-slips go together and one stands alone with an even partner count — and walking through the targets again forces the lone $3.5$ to land in $D$ paired with $4.5$.

CCSS standards used (min grade 6)

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Summing the $12$ slip values to $35$ and computing each cup's leftover requirement (e.g., $7 - 3.5 = 3.5$, $8 - 3.5 = 4.5$) using decimal subtraction.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Setting up $n + (n+1) + (n+2) + (n+3) + (n+4) = 35$ and solving $5n + 10 = 35$ to find $n = 5$, giving the cup targets $5, 6, 7, 8, 9$.)
4.OA.A.3 Solve multistep word problems using the four operations, including interpreting remainders (Running the case-by-case elimination across cups $A, B, C, E$ — each case is a small multistep arithmetic check that either fits the target sum or doesn't.)

⭐ This AMC 8 problem only needs Grade 6 equation reasoning to find the cup sums, plus careful elimination — both skills you already have!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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