AMC 10 · 2003 · #7

Grade 8 algebra
linear-diophantinesystems-of-equationsunit-conversion convert-to-algebraextreme-principle ↑ Prerequisites: linear-equations-two-var
📏 Long solution 💡 3 insights
Problem
A piggy bank holds exactly 100 coins, each a nickel, a dime, or a quarter, and the whole pile is worth 835 cents. A type may be missing entirely. Find how far apart the biggest possible number of dimes and the smallest possible number of dimes are.

Pick an answer.

(A)
0
(B)
13
(C)
37
(D)
64
(E)
83
How to solve
Strategy Convert to Algebra

Two facts are given about the same pile — how many coins and how much money — so tool #13 (Convert to Algebra) turns each fact into one equation and lets them be combined. Tool #4 (Introduce a Variable) names all three counts even though only the dimes are asked about, because the nickels are what makes the count work out and the quarters are what pushes the dime count around. Two equations cannot pin down three unknowns, so there is a whole family of legal coin piles; tool #14 (Extreme Principle) asks only for the two ends of that family, which is all the question needs. Tool #8 (Analyze the Units) does the small but essential first move: work in cents, not dollars, so every number in sight is a whole number.

1STEP 1

Work in cents, not dollars

Working in cents makes every number whole, so the total reads 835.

$8.35 = 835 cents, nickel=5, dime=10, quarter=25
2STEP 2

Name all three counts

Naming the three counts gives a coin equation and a value equation.

a+b+c=100 5a+10b+25c=835 a,b,c ≥ 0 whole numbers
3STEP 3

Get rid of the nickels

Dividing by five and subtracting removes the nickels, leaving b + 4c = 67.

(5a+10b+25c)/5=a+2b+5c=167 ⟹ (a+2b+5c)-(a+b+c)=167-100 ⟹ b+4c=67
4STEP 4

Check the nickels never go negative

Checking back, the nickels come to 33 plus 3c, always positive, so nothing is blocked.

b=67-4c ⟹ a=100-b-c=33+3c ≥ 33 > 0
5STEP 5

Push c to both ends

Pushing the quarters to both ends gives dimes from 3 to 67, a spread of 64, choice (D).

c=0 → b=67 (max); 4c ≤ 67 → c ≤ 16 → b=67-64=3 (min); 67-3=64 → (D)
Answer
64
Both extreme piles can be built and checked. Largest dimes: 33 nickels, 67 dimes, 0 quarters — that is 33+67+0=100 coins worth 165+670+0=835 cents. Smallest dimes: 81 nickels, 3 dimes, 16 quarters — that is 81+3+16=100 coins worth 405+30+400=835 cents. Both are legal, so the spread really is at least 64, and b=67-4c shows no pile can beat either end. The other choices die fast: (A) 0 would mean the dime count is forced, but two different legal piles were just built; (E) 83 would need a pile with 83 dimes or more, yet b=67-4c ≤ 67; (B) 13 and (C) 37 are neither of the two ends.
💡Key takeaway

When two facts cannot pin down three unknowns, boil them down to one relation and then slide the free unknown to each end — the answer lives at the ends, not in the middle.

  • Work in cents, not dollars
  • Name all three counts
  • Get rid of the nickels
  • Check the nickels never go negative
  • Push c to both ends