Competition · AMC preparation · step 4 of 4

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. Each pair cost $1, $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

Find the extra dollars needed

If all 12 pairs were $1 the bill is $12, but it's $24 — so the pricier socks add $12 of extra cost.

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

2STEP 2

Write each pair's extra cost

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

List every possible pair of counts

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)

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

The only way the pricier pairs add exactly 12 of extra cost, using at least one 3 pair and at least one 4 pair, is three 3 pairs together with two $4 pairs.

▸ Why?

Each 3 pair adds 2 of extra and each 4 pair adds 3 of extra, and together those two amounts must fill the whole 12 of extra, so the counts satisfy 2y + 3z = 12.

▸ Why?

The 12 of extra has no other source than the 3 pairs' share plus the 4 pairs' share, so those two shares add back to the full 12.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

A 4 pair sits 3 above the 1 baseline, and z such pairs each add that same fixed 3, so together they contribute z groups of $3.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

Once you fix how many 4 pairs z there are, the number of 3 pairs is forced to y = (12 - 3z) / 2, the only value that keeps the extra at $12.

▸ Why?

Reading y back out of 2y = 12 - 3z means undoing the doubling by halving, which leaves exactly one y.

Inverse operationsSubtraction undoes addition and division undoes multiplication — they reverse each other.

▸ Why?

So only z = 1, 2, 3, 4 are worth testing, and of these only z = 2 leaves a whole number of 3 pairs that is at least one, namely y = 3.

▸ Why?

z counts real pairs, so it is a whole number of at least 1, and its share 3z is only part of the 12 of extra, so it cannot exceed 12 and z can be at most 4.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

For z = 1 and z = 3 the forced y comes out as a half, which cannot count actual pairs, and for z = 4 it comes out as 0, which breaks the 'at least one 3 pair' rule, so only z = 2 survives.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

4.OA.A.3Make A Systematic List

4STEP 4

Find the cheapest count

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 count and the cost

Check the original: 7 + 3 + 2 = 12 pairs and 7×$1 + 3×$3 + 2×$4 = $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.

Find the extra dollars needed
Write each pair's extra cost
List every possible pair of counts
Find the cheapest count
Check the count and the cost

A parent dashboard for the family lives at sensimlab.com.

Find the extra dollars needed

Write each pair's extra cost

List every possible pair of counts

Find the cheapest count

Check the count and the cost

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