AMC 8 · 2005 · #5

Grade 4 arithmetic

Archetype Small-Domain Constrained Search

optimization-countingdivisibility-rulesmultiplesmulti-digit-arithmetic optimization-countingsystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticdivisibility-rules

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Problem

Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Soda is sold in packs of $6$, $12$, and $24$ cans. What is the fewest packs you can buy that gives exactly $90$ cans?

Givens: Available pack sizes: $6$, $12$, $24$ cans; Target total: exactly $90$ cans; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $8$, (E) $15$

Unknowns: The minimum total number of packs that sums to exactly $90$ cans

Understand

Restated: Soda is sold in packs of $6$, $12$, and $24$ cans. What is the fewest packs you can buy that gives exactly $90$ cans?

Givens: Available pack sizes: $6$, $12$, $24$ cans; Target total: exactly $90$ cans; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $8$, (E) $15$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

Tool #7 (Identify Subproblems) splits "minimum packs for $90$ cans" into a short chain: peel off as many $24$-packs as fit, then peel off $12$-packs from what's left, then finish with $6$-packs. Each sub-problem is one division-with-remainder, and stacking the largest packs first keeps the pack count as small as possible. Tool #3 (Eliminate Possibilities) then verifies the minimum: the smaller answer choice (A) $4$ packs is impossible, so the candidate from Tool #7 really is the minimum.

Execute — Answer: B

#7 Identify Subproblems 4.OA.A.3 Step 1

Subproblem 1: peel off as many $24$-packs as fit in $90$ cans.
Divide $90$ by $24$ and keep the remainder.

$$90 \div 24 = 3 \text{ R } 18 \;\Rightarrow\; 3 \text{ packs of } 24 = 72 \text{ cans, } 18 \text{ cans left}$$

💡 Division-with-remainder is the Grade 4 "how many groups fit, and what is left over" move.

#7 Identify Subproblems 4.OA.A.3 Step 2

Subproblem 2: peel off as many $12$-packs as fit in the $18$ leftover cans.

$$18 \div 12 = 1 \text{ R } 6 \;\Rightarrow\; 1 \text{ pack of } 12 = 12 \text{ cans, } 6 \text{ cans left}$$

💡 Same move, smaller numbers: one $12$-pack covers all the $12$s in $18$, leaving a multiple of $6$.

#7 Identify Subproblems 3.OA.A.3 Step 3

Subproblem 3: the last $6$ cans are exactly one $6$-pack.

$$6 \div 6 = 1 \text{ R } 0 \;\Rightarrow\; 1 \text{ pack of } 6 = 6 \text{ cans, } 0 \text{ cans left}$$

💡 One $6$-pack closes out the leftover and lands the total at exactly $90$.

#7 Identify Subproblems 3.OA.A.3 Step 4

Add the packs from each subproblem.

$$\text{Total packs} = 3 + 1 + 1 = 5$$

💡 Combine the three subproblem counts to get the candidate minimum.

#3 Eliminate Possibilities 4.OA.B.4 Step 5

Verify with Tool #3: rule out the smaller choice (A) $4$ packs.
Every pack size is a multiple of $6$, so $4$ packs together can hit only the totals $4a + 8b + 16c$ multiplied by $6$, where $a + b + c = 4$ and $a, b, c \geq 0$.
The targets are $a + 2b + 4c = 15$ with $a + b + c = 4$.
Trying $c = 3$: $a + b = 1$ and $a + 2b = 3$, giving $b = 2$, but then $a + b = -1 + 2 \ne 1$.
Trying $c = 4$: total $= 4 \times 24 = 96 \ne 90$.
Every other split undershoots or overshoots; no $4$-pack mix lands on $90$.
So $5$ really is the minimum.

$$5 \text{ packs} \;\Rightarrow\; \textbf{(B)}$$

💡 Pack totals are all multiples of $6$, so we can search small cases by hand and confirm $4$ packs cannot reach $90$.

[1] #7 4.OA.A.3 Subproblem 1: peel off as many $24$-packs as fit in $90$ cans. Divide $90$ by $2

[2] #7 4.OA.A.3 Subproblem 2: peel off as many $12$-packs as fit in the $18$ leftover cans.

[3] #7 3.OA.A.3 Subproblem 3: the last $6$ cans are exactly one $6$-pack.

[4] #7 3.OA.A.3 Add the packs from each subproblem.

[5] #3 4.OA.B.4 Verify with Tool #3: rule out the smaller choice (A) $4$ packs. Every pack size

Review

Reasonableness: Add the chosen packs back: $3 \times 24 + 1 \times 12 + 1 \times 6 = 72 + 12 + 6 = 90$. The total matches exactly, and the pack count is $5$, so choice (B) is consistent. Choices (C) $6$, (D) $8$, and (E) $15$ are all larger and would only be the answer if smaller counts didn't work; we already hit $90$ with $5$ packs, so they can't be minimal. Choice (A) $4$ was ruled out above. Another quick check: $15$ packs of $6$ also gives $90$, which is why (E) appears, but that's the maximum, not the minimum.

Alternative: Tool #6 (Guess and Check): start with the educated guess of $5$ packs and try to push lower. With $4$ packs the largest possible total is $4 \times 24 = 96$ and the next options drop to $3 \times 24 + 1 \times 12 = 84$ or $3 \times 24 + 1 \times 6 = 78$. Nothing with $4$ packs hits $90$ exactly, so $5$ wins, matching (B).

CCSS standards used (min grade 4)

3.OA.A.3 Use multiplication and division within $100$ to solve word problems (Recognizing that each $6$-, $12$-, or $24$-pack is a fixed number of cans, so the total is a sum of small multiplications, and adding the pack counts $3 + 1 + 1 = 5$.)
4.OA.A.3 Solve multistep word problems using the four operations, including problems with remainders (Running division-with-remainder on $90 \div 24 = 3 \text{ R } 18$ and $18 \div 12 = 1 \text{ R } 6$ to peel off the largest packs first.)
4.OA.B.4 Find factors and multiples of a whole number (Noticing that every pack size is a multiple of $6$, which both guides the peel-off order and lets us rule out $4$-pack mixes that cannot equal $90$.)

⭐ When you want the fewest packs that total an exact amount, start with the biggest pack and peel off as many as fit, then move to the next size. That "biggest first" plan turns this AMC 8 problem into three short Grade 4 division-with-remainder steps.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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