AMC 8 · 2015 · #20

Grade 6 algebraarithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

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 bought 12 pairs of socks for a total of

24</span>. Each pair cost <span class="mk-c">

3, or

4, and he bought at least one pair of each price. How many $1 pairs did he buy?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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.

1STEP 1

If all 12 pairs were $1 the bill is <span class="hl-ask">$ 12, but it's $24 — so the pricier socks add <span class="mk-a">$ 12 of extra cost.

12 × 1 = 12, and 24 - 12 = 12 extra dollars needed

2STEP 2

A $3 pair adds$ 2 over baseline, a $4 pair adds$ 3; with y, z their counts, the extra cost is 2y + 3z = 12.

(3-1)=2 y + (4-1)=3 z = 12 → 2y + 3z = 12

3STEP 3

Test z = 1, 2, 3, … in 2y + 3z = 12, keeping only positive whole y: only z = 2, y = 3 works.

z & 2y = 12 - 3z & y ; 1 & 9 & 4.5 (reject: not whole) ; 2 & 6 & 3 (keep) ; 3 & 3 & 1.5 (reject: not whole) ; 4 & 0 & 0 (reject: need z ≥ 1, y ≥ 1)

4STEP 4

With y = 3 and z = 2, the pair-count total x + y + z = 12 gives x = 12 - 3 - 2 = 7.

x + y + z = 12 → x = 12 - 3 - 2 = 7

5STEP 5

Check the original: 7 + 3 + 2 = 12 pairs and 7× $1 + 3×$ 3 + 2× $4 = <span class="mk-a"><span class="hl-ask">$ 24, each price used — matches (D).

Count: 7 + 3 + 2 = 12 ✓ Cost: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ At least one of each: 7, 3, 2 ≥ 1 ✓ → (D) 7

Answer

The average pair cost is

24 ÷ 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.

💡Key takeaway

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.

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