Problem 20 · 2015 AMC 8
Medium
Algebra & Patterns
system-of-equationsparity-mod
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?
Show answer
Answer: D — 7 pairs.
Show hints
Hint 1 of 2
Three unknowns but only two equations — that feels underdetermined, but the counts must be whole positive numbers, and that's enough. Notice a appears with the same coefficient (1) in both the count equation and the cost equation: subtract them and a vanishes.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtracting leaves 2b + 3c = 12 with b, c ≥ 1. Since 2b is even and 12 is even, 3c must be even, so c is even — parity nails it down.
Show solution
Approach: eliminate a by subtracting, then use parity
- Counts: a + b + c = 12. Cost: a + 3b + 4c = 24. Both have a single a, so subtract to erase it: 2b + 3c = 12.
- Parity: 2b and 12 are even, forcing 3c — hence c — to be even. With at least one of each type, 0 < c < 4, so c = 2.
- Back-substitute: 2b = 6 ⇒ b = 3, then a = 12 − 3 − 2 = 7.
- Why this transfers: when a variable shares the same coefficient across two equations, subtract to delete it; then parity/divisibility usually finishes off the few whole-number possibilities.
Mark:
· log in to save