2023 AMC 8 Problems/Problem 14


Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$ in postage? (Note: The amount $$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$

Solution 1

Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$-cent (like nickels), $10$-cent (like dimes), and $25$-cent (like quarters).

If we want to have the highest number of stamps, we have to have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our total cost to $7.10 - 3.00 = 4.10$. However, when we try to use quarters, the $25$ cents don’t fit evenly, so we have to give back $15$ cents in order to make the quarter amount $4.25$. The most efficient way to do this is to give back a $10$-cent (dime) stamp and a $5$-cent (nickel) stamp to have $38$ stamps (coins) used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{\textbf{(E)}\ 55}$.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Solution 2

The value of his entire stamp collect is $8$ dollars. To make $$7.10$ with stamps, he should remove $90$ cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many $25$ cent stamps as possible as they have the greatest denomination. He can remove at most $3$ of these stamps. He still has to remove $90-25\cdot3=15$ cents worth of stamps. This can be done with one $5$ and $10$ cent stamp. In total, he has $20\cdot3=60$ stamps in his entire collect. As a result, the maximum number of stamps he can use is $20\cdot3-5=\boxed{\textbf{(E)}\ 55}$.


~MathFun1000 (Rewrote for clarity and formatting)

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See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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