Competition · AMC preparation · step 4 of 4

AMC 10 · 2023B · #11

Grade 6 arithmetic
linear-diophantinesystematic-enumerationlinear-equations-two-varbound-inequality-then-enumerate systematic-enumerationcaseworkbound-inequality-then-enumerate ↑ Prerequisites: linear-equations-two-varsystematic-enumeration
📏 Medium solution 💡 3 insights
Problem
Suzanne withdraws 800 from the bank. The teller hands it over using only 20, 50, and 100 bills, with at least one bill of every kind. Count the number of different stacks (combinations of bill counts) she could receive.

Pick an answer.

(A)
45
(B)
21
(C)
36
(D)
28
(E)
32

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

Tool #2 (Systematic List) is the natural fit because the question literally asks "how many collections" — that triggers an organized count. Tool #9 (Easier Problem) helps: dividing the equation by 10 shrinks 20x + 50y + 100z = 800 to 2x + 5y + 10z = 80, the same problem with smaller numbers. To avoid listing all 21 by hand, Tool #5 (Pattern) catches that the count of (x, y) pairs for each fixed z is an arithmetic sequence — its sum is the answer.

1STEP 1

Shrink the money equation

Divide 20x + 50y + 100z = 800 by 10 to get the same solutions with smaller numbers.

2x + 5y + 10z = 80, x, y, z ≥ 1
2STEP 2

Fix the largest bill first

Fix z (the number of 100s); then 2x + 5y = 80 - 10z forces y to be even.

Fix z. Need 2x + 5y = 80 - 10z with x ≥ 1, y ≥ 1, y even.
3STEP 3

Count the pairs for each z

With y = 2b, x = 5(8 - z - b), so each z gives exactly 7 - z valid pairs.

Number of pairs for fixed z: max(7 - z, 0).
4STEP 4

Add the counts down the list

Add the counts for z = 1 through 6: 6 + 5 + 4 + 3 + 2 + 1 = 21.

6 + 5 + 4 + 3 + 2 + 1 = 21
5STEP 5

Match against the choices

21 matches choice (B); 45, 36, 28, 32 fit no natural miscount of the cases.

Answer = 21 → (B)
Answer
21
Sanity-check the boundary cases. When z = 1 (one 100), the remaining 700 must come from 20s and 50s with at least one of each: 20x + 50y = 700 with x, y ≥ 1. Listing: y ∈ {2, 4, 6, 8, 10, 12} (must be even, and 50y ≤ 680), giving 6 collections — matches the 7 - 1 = 6 formula. Spot-check z = 6 (six 100s, 200 left): 20x + 50y = 200, y ∈ {2}, x = 5 — exactly 1 collection, matches 7 - 6 = 1. Total 21 is small enough to feel right for a "three denominations, 800 total" setup.
💡Key takeaway

This AMC 10 problem only needs Grade 6 "equivalent equations and variables" you already know — once you divide the dollar amounts by 10 and fix the number of 100 bills, each case turns into a tidy little count, and the cases add up to 21.

  • Shrink the money equation
  • Fix the largest bill first
  • Count the pairs for each z
  • Add the counts down the list
  • Match against the choices

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