Difference between revisions of "2023 AMC 8 Problems/Problem 14"

Revision as of 01:10, 21 August 2024

Problem

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 (nickels), $10$ -cent (dimes), and $25$ -cent (quarters).

If we want the highest number of stamps, we must 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 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 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, InterstellerApex, mahika99

Solution 2

The value of his entire stamp collection 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 collection. As a result, the maximum number of stamps he can use is $20\cdot3-5=\boxed{\textbf{(E)}\ 55}$ .

~pianoboy

~MathFun1000 (Rewrote for clarity and formatting)

~vadava_lx (minor edits)

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/vq3voJZ-hvw

~Education, the Study of Everything

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=GHLp1q_7Le68a4rJ&t=2454 ~Math-X note from InterstellerApex: this is wrong, he didn’t choose the correct answer. :(

Animated Video Solution

https://youtu.be/XP_tyhTqOBY

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4280

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=1435

Video Solution by harungurcan

https://www.youtube.com/watch?v=VqN7c5U5o98&t=449s

~harungurcan

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=633

~hsnacademy

@@ Line 44: / Line 44: @@
 ~harungurcan
+==Video Solution (Solve under 60 seconds!!!)==
+https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=633
+~hsnacademy
 ==See Also==
 {{AMC8 box|year=2023|num-b=13|num-a=15}}
 {{MAA Notice}}

2023 AMC 8 (Problems • Answer Key • Resources)
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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