AMC 8 · 2015 · #20

Grade 6 algebraarithmetic

Archetype Small-Domain Constrained Search

linear-diophantinesystems-of-equationssystematic-enumeration systematic-enumerationconvert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var

📏 Medium solution 💡 3 insights

Problem

Ralph went to the store and bought 12 pairs of socks for a total of $24$. Some of the socks he bought cost$ 1 $a pair, some of the socks he bought cost$ $3$ a pair, and some of the socks he bought cost $4$ a pair. If he bought at least one pair of each type, how many pairs of$ 1$ socks did Ralph buy?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Ralph bought $12$ pairs of socks for a total of $\$24$. Each pair cost $\$1$, $\$3$, or $\$4$, and he bought at least one pair of each price. How many $\$1$ pairs did he buy?

Givens: Total number of pairs $= 12$; Total cost $= \$24$; Each pair costs $\$1$, $\$3$, or $\$4$; At least one pair of each price was bought; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$

Unknowns: The number of $\$1$ pairs Ralph bought

Understand

Restated: Ralph bought $12$ pairs of socks for a total of $\$24$. Each pair cost $\$1$, $\$3$, or $\$4$, and he bought at least one pair of each price. How many $\$1$ pairs did he buy?

Plan

Primary tool: #2 Make a Systematic List

Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

There are only three sock prices and a small total of $12$ pairs, so the candidate sets are tiny. Tool #2 (Systematic List) lets us try each possible number of $\$4$ pairs in order and read off the matching $\$3$ and $\$1$ counts. Tool #9 (Easier Related Problem) gives us a shortcut: pretend all $12$ pairs cost $\$1$ — that's only $\$12$, so we need to find $\$12$ of "extra cost" by swapping some $\$1$ pairs for pricier ones. Tool #3 (Eliminate Possibilities) is a safety net: with answer choices A–E giving the $\$1$ count, we can check each candidate against the constraints if needed.

Execute — Answer: D

#9 Solve an Easier Related Problem 3.OA.A.3 Step 1

Set up the "extra cost" idea.
If all $12$ pairs were $\$1$, the bill would be $12 \times \$1 = \$12$. The real bill is $\$24$, so the pricier socks must add $\$24 - \$12 = \$12$ of extra cost on top of the all-$\$1$ baseline.

$12 \times 1 = 12$, and $24 - 12 = 12 \text{ extra dollars needed}$

💡 Comparing a 'pretend all-$\$1$' world to the real total is the Tool #9 move: replace the messy three-price problem with one we can total in our head.

#9 Solve an Easier Related Problem 4.OA.A.3 Step 2

Find each pricier pair's contribution to the extra.
A $\$3$ pair is $\$2$ more than a $\$1$ pair, so it adds $\$2$ extra.
A $\$4$ pair is $\$3$ more, so it adds $\$3$ extra. Let $y$ = number of $\$3$ pairs and $z$ = number of $\$4$ pairs. Then the extra cost equation is $2y + 3z = 12$.

$$\underbrace{(3-1)}_{=2} y + \underbrace{(4-1)}_{=3} z = 12 \;\Rightarrow\; 2y + 3z = 12$$

💡 One equation in two unknowns instead of two equations in three unknowns — a Grade 4 multi-step word-problem reduction.

#2 Make a Systematic List 4.OA.A.3 Step 3

Systematically list every $(y, z)$ with $y \ge 1$ and $z \ge 1$ that satisfies $2y + 3z = 12$.
Go through $z = 1, 2, 3, \ldots$ in order and solve for $y$ each time, keeping only whole-number, positive answers.

$$\begin{array}{c|c|c} z & 2y = 12 - 3z & y \\ \hline 1 & 9 & 4.5 \text{ (reject: not whole)} \\ 2 & 6 & 3 \text{ (keep)} \\ 3 & 3 & 1.5 \text{ (reject: not whole)} \\ 4 & 0 & 0 \text{ (reject: need } z\ge1, y\ge1) \end{array}$$

💡 Ordering by $z$ guarantees we don't miss or double-count cases — that's the whole point of Tool #2.

#2 Make a Systematic List 3.OA.A.3 Step 4

Read off the only surviving case: $y = 3$ pairs of $\$3$ socks and $z = 2$ pairs of $\$4$ socks.
Use the count total to find the number of $\$1$ pairs, $x$.

$$x + y + z = 12 \;\Rightarrow\; x = 12 - 3 - 2 = 7$$

💡 Once two of the three counts are pinned down, the third is just a subtraction from the known total of $12$ pairs.

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

Verify the candidate by checking both totals in the original problem (not the equation we built).

Count: $7 + 3 + 2 = 12$ ✓ \quad Cost: $7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24$ ✓ \quad At least one of each: $7, 3, 2 \ge 1$ ✓ \;\Rightarrow\; \textbf{(D)} \; 7$

💡 Plugging the answer back into the words of the problem is the Tool #3 check — it catches any mis-set-up before we commit.

[1] #9 3.OA.A.3 Set up the "extra cost" idea. If all $12$ pairs were $\$1$, the bill would be $1

[2] #9 4.OA.A.3 Find each pricier pair's contribution to the extra. A $\$3$ pair is $\$2$ more t

[3] #2 4.OA.A.3 Systematically list every $(y, z)$ with $y \ge 1$ and $z \ge 1$ that satisfies $

[4] #2 3.OA.A.3 Read off the only surviving case: $y = 3$ pairs of $\$3$ socks and $z = 2$ pairs

[5] #3 4.OA.A.3 Verify the candidate by checking both totals in the original problem (not the eq

Review

Reasonableness: The average pair cost is $\$24 \div 12 = \$2$, which is close to the cheapest price ($\$1$) and far from the most expensive ($\$4$). That means most of the pairs must be the cheap kind, so the $\$1$ count should be the largest of the three — answer choices (C) $6$, (D) $7$, and (E) $8$ are the only believable ones, and (D) $7$ falls right in the middle and matches our work.

Alternative: Tool #3 (Eliminate Possibilities) straight on the choices: try $x = 7$ ($\$1$ pairs). Then $y + z = 5$ and $3y + 4z = 24 - 7 = 17$. Subtracting $3(y+z)=15$ gives $z = 2$, $y = 3$ — all positive whole numbers, so (D) works. Checking $x = 8$ instead would force $y + z = 4$ and $3y + 4z = 16$, giving $z = 4$, $y = 0$, which violates 'at least one of each'.

CCSS standards used (min grade 6)

3.OA.A.3 Use multiplication and division within 100 to solve word problems (Computing the all-$\$1$ baseline ($12 \times 1 = 12$) and the final subtraction $x = 12 - 3 - 2 = 7$ that pins down the number of $\$1$ pairs.)
4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Reducing the two-condition (count and cost) word problem into the single equation $2y + 3z = 12$ and verifying the final triple $(7, 3, 2)$ satisfies both original totals.)
6.EE.B.6 Use variables to represent numbers and write expressions when solving real-world problems (Naming $y$ for the number of $\$3$ pairs and $z$ for the number of $\$4$ pairs and translating their 'extra cost' contributions into the expression $2y + 3z$.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations (Solving $2y + 3z = 12$ over the positive whole numbers by systematic enumeration, which pinpoints the unique valid pair $(y,z) = (3,2)$.)

⭐ This AMC 8 problem fits in your toolkit: a Grade 6 word-problem-to-equation step plus a tiny systematic list of cases is all it takes to find the answer.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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