AMC 8 · 2023 · #14
Easy mode Grade 4Problem
Picture Nicolas mailing a package to his friend Anton, who loves collecting stamps. Nicolas wants to cover the package with as many stamps as he can.
He has three kinds of stamps in his drawer:
- 5-cent stamps
- 10-cent stamps
- 25-cent stamps
He has exactly 20 of each kind.
The postage he needs to pay is exactly $7.10. (That's 7 dollars and 10 cents. One dollar is 100 cents, so $7.10 is 710 cents.)
He has to pick stamps that add up to exactly $7.10. What is the largest number of stamps he can use?
Pick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Three stamp values: $5$ cents, $10$ cents, $25$ cents; Exactly $20$ stamps of each value are available; The total postage must be exactly $\$7.10 = 710$ cents; Total value of all $60$ stamps: $20\cdot 5+20\cdot 10+20\cdot 25 = 100+200+500 = 800$ cents; Answer choices: (A) $45$, (B) $46$, (C) $51$, (D) $54$, (E) $55$
Unknowns: The greatest number of stamps Nicolas can use whose total value is exactly $710$ cents
Understand
Restated: 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?
Givens: Three stamp values: $5$ cents, $10$ cents, $25$ cents; Exactly $20$ stamps of each value are available; The total postage must be exactly $\$7.10 = 710$ cents; Total value of all $60$ stamps: $20\cdot 5+20\cdot 10+20\cdot 25 = 100+200+500 = 800$ cents; Answer choices: (A) $45$, (B) $46$, (C) $51$, (D) $54$, (E) $55$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #6 Guess & Check
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.
Execute — Answer: E
3.OA.C.7 Step 1 - First gather the total.
- Each denomination contributes (number)$\times$(value), and the three contributions add to the value of the whole collection.
- Also count the total number of stamps.
💡 Multiplying a small whole number by 5, 10, or 25 is Grade 3 multiplication fluency.
2.MD.C.8 Step 2 - Apply the complement flip (Tool #16).
- The collection is worth $800$ cents but only $710$ cents will be mailed, so the leftover (unused) stamps must be worth exactly $800-710=90$ cents.
- Maximizing stamps USED is the same as minimizing stamps LEFT OUT.
💡 Subtracting one money amount from another to find what's left over is a Grade 2 money word-problem skill.
4.NBT.B.6 Step 3 - Now solve the easier subproblem: build $90$ cents using the FEWEST stamps.
- To use few stamps, use the biggest values first.
- Take as many $25$-cent stamps as fit into $90$: $\lfloor 90/25\rfloor = 3$, contributing $75$ cents and leaving $15$ cents to fill.
💡 Dividing $90$ by $25$ with a remainder (or just thinking "how many 25s fit?") is the Grade 4 quotient-and-remainder idea.
2.MD.C.8 Step 4 - Fill the remaining $15$ cents with the fewest stamps.
- Use one $10$-cent stamp, leaving $5$ cents, then one $5$-cent stamp.
- So the leftover pile is $3$ twenty-fives, $1$ ten, and $1$ five — five stamps in total.
- We have at least one of each in stock (we own $20$ of each), so this combination is allowed.
💡 Combining nickels, dimes, and quarters to make a target amount is Grade 2 coin-money work.
4.OA.A.3 Step 5 - Subtract the unused count from the total.
- The greatest number of stamps used is $60-5=55$.
- This matches choice (E).
💡 Wrapping the multi-step plan into a single subtraction is the Grade 4 multi-step word-problem standard.
4.OA.A.3 Step 6 - Verify directly with Tool #6 (Guess & Check).
- Use $20-1=19$ five-cent stamps, $20-1=19$ ten-cent stamps, and $20-3=17$ twenty-five-cent stamps.
- Add their count and their value.
💡 Plug the answer back in and check both the count and the value — Grade 4 multi-step verification.
3.OA.C.7 First gather the total. Each denomination contributes (number)$\times$(value), a 2.MD.C.8 Apply the complement flip (Tool #16). The collection is worth $800$ cents but on 4.NBT.B.6 Now solve the easier subproblem: build $90$ cents using the FEWEST stamps. To us 2.MD.C.8 Fill the remaining $15$ cents with the fewest stamps. Use one $10$-cent stamp, l 4.OA.A.3 Subtract the unused count from the total. The greatest number of stamps used is 4.OA.A.3 Verify directly with Tool #6 (Guess & Check). Use $20-1=19$ five-cent stamps, $2 Review
Reasonableness: 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.
Alternative: Tool #6 (Guess & Check) alone also works: try the largest choice $55$ first. If it is achievable, we are done. To hit $55$ stamps summing to $710$ cents while staying inside the $20$-per-type cap, set up nickels=$a$, dimes=$b$, quarters=$c$ with $a+b+c=55$ and $5a+10b+25c=710$. Subtracting five times the first from the second gives $5b+20c=435$, i.e. $b+4c=87$. Trying $c=17$ gives $b=19$ and then $a=19$ — all $\le 20$, so $55$ is achievable. Choice (E) confirmed.
CCSS standards used (min grade 4)
2.MD.C.8Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies (Reasoning about combinations of nickel, dime, and quarter stamp values to reach an exact money target (both the $710$-cent postage and the $90$-cent unused total).)3.OA.C.7Fluently multiply and divide within 100 (Computing $20\times 5$, $20\times 10$, $20\times 25$, and $3\times 25$ to tally stamp values.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing $\lfloor 90/25\rfloor = 3$ with remainder $15$ when deciding how many quarter stamps fit into the $90$-cent leftover.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Chaining the complement flip, the greedy fill, the count subtraction $60-5$, and the final verification into one coherent solution.)
⭐ This AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know!
⭐ This AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know!
More like this
Same archetype — closest grade level first.