AMC 8 · 2023 · #14

Grade 4 arithmetic

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

optimization-countinglinear-diophantinedivisibility-rules optimization-countingsystematic-enumerationcasework ↑ Prerequisites: multi-digit-arithmeticdivisibility-rules

📏 Medium solution 💡 3 insights

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Problem

Nicolas has 20 each of 5-cent, 10-cent, and 25-cent stamps. He must put on stamps that total exactly 710 cents, and he wants to use as MANY stamps as possible. How many stamps is that?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Trying to BUILD 710 cents out of as many stamps as possible is messy — you keep adding small stamps and lose track. Tool #16 (Change Focus) flips the question: since the whole collection is worth 800 cents, the stamps Nicolas LEAVES OUT must be worth exactly 800-710=90 cents. Maximizing the stamps used is the same as MINIMIZING the stamps NOT used. So the problem becomes: "What is the fewest stamps that add up to 90 cents?" That is a small, clean question. Tool #7 splits it into stages (start with the largest denomination, then refill the gap), and Tool #6 lets us double-check the count at the end.

1STEP 1

Total up the collection: 20 of each at 5, 10, 25 cents makes 800 cents across 60 stamps.

20· 5+20· 10+20· 25=100+200+500=800 cents; 20+20+20=60 stamps

2STEP 2

Flip focus: the collection holds 800 but only 710 ships, so the leftover stamps must be worth exactly 90 cents.

800-710=90 cents of unused stamps

3STEP 3

Make 90 with the fewest stamps: biggest first, so ⌊ $\frac{90}{25}$ ⌋=3 quarters give 75, leaving 15 cents.

3× 25=75; 90-75=15 cents remaining

4STEP 4

Fill 15 with one 10 and one 5, so the leftover pile is 3+1+1=5 stamps — all in stock.

15=10+5; unused stamps=3+1+1=5

5STEP 5

Subtract the leftovers: 60-5=55 stamps used — the maximum, matching choice (E).

60-5=55 → (E)

6STEP 6

Check directly: 19 fives + 19 tens + 17 quarters = 55 stamps worth 95+190+425=710 cents. ✓

19+19+17=55 stamps; 19· 5+19· 10+17· 25=95+190+425=710 cents✓

Answer

Is 55 stamps believable? Nicolas owns 60 stamps worth 800 cents but only needs 710 cents, so he is 90 cents "over" the target. The cheapest way to shed 90 cents is with high-value stamps (a few quarters), which removes the fewest physical stamps. The minimum unused count is 5 stamps, so he uses 60-5=55 — the largest answer choice listed, which fits the "maximize stamps" wording. Smaller choices like 45 or 46 would mean throwing away many MORE stamps for no reason, so they cannot be the maximum.

💡Key takeaway

This AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know!

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