Competition · AMC preparation · step 4 of 4
AMC 8 · 2023 · #14
Grade 4 arithmeticPick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Find the total value
Total up the collection: 20 of each at 5, 10, 25 cents makes 800 cents across 60 stamps.
Multiplying a small whole number by 5, 10, or 25 is Grade 3 multiplication fluency.
3.OA.C.7Identify SubproblemsFlip to the unused stamps
Flip focus: the collection holds 800 but only 710 ships, so the leftover stamps must be worth exactly 90 cents.
Subtracting one money amount from another to find what's left over is a Grade 2 money word-problem skill.
The stamps Nicolas leaves out of the mailing must be worth exactly 90 cents, so using the most stamps is the same as leaving out the fewest.
▸ Why?
Every one of the 60 stamps is either mailed or set aside — none skipped, none counted twice — so the mailed pile and the set-aside pile together rebuild the whole collection: 800 cents and 60 stamps.
▸ Why?
The mailed pile is pinned at 710 cents, so the set-aside pile is whatever is left when 710 is taken from the 800-cent whole: 800 - 710 = 90 cents.
▸ Why?
Finding the leftover part when you already know the whole and one part is just undoing the addition of the two parts, and subtraction is what reverses that addition.
▸ Why?
The two piles always share 60 stamps between them, so every extra stamp that goes on the package is one fewer stamp set aside — the largest mailed count lines up with the smallest set-aside count.
▸ Why?
When two parts must add up to one fixed whole, pushing one part as large as it can go forces the other part to be as small as it can go.
Make 90 cents with fewest stamps
Make 90 with the fewest stamps: biggest first, so ⌊⌋=3 quarters give 75, leaving 15 cents.
Dividing 90 by 25 with a remainder (or just thinking "how many 25s fit?") is the Grade 4 quotient-and-remainder idea.
4.NBT.B.6Identify SubproblemsFill the last 15 cents
Fill 15 with one 10 and one 5, so the leftover pile is 3+1+1=5 stamps — all in stock.
Combining nickels, dimes, and quarters to make a target amount is Grade 2 coin-money work.
2.MD.C.8Identify SubproblemsSubtract the unused stamps
Subtract the leftovers: 60-5=55 stamps used — the maximum, matching choice (E).
Wrapping the multi-step plan into a single subtraction is the Grade 4 multi-step word-problem standard.
4.OA.A.3Change Focus Count The ComplementVerify the count directly
Check directly: 19 fives + 19 tens + 17 quarters = 55 stamps worth 95+190+425=710 cents. ✓
Plug the answer back in and check both the count and the value — Grade 4 multi-step verification.
4.OA.A.3Guess And CheckThis AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know!
- Find the total value
- Flip to the unused stamps
- Make 90 cents with fewest stamps
- Fill the last 15 cents
- Subtract the unused stamps
- Verify the count directly
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