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

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==Video Solution by Magic Square==
 
==Video Solution by Magic Square==
 
https://youtu.be/-N46BeEKaCQ?t=4280
 
https://youtu.be/-N46BeEKaCQ?t=4280
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==Video Solution by Interstigation==
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https://youtu.be/1bA7fD7Lg54?t=1078
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=13|num-a=15}}
 
{{AMC8 box|year=2023|num-b=13|num-a=15}}
 
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{{MAA Notice}}

Revision as of 21:44, 16 February 2023

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

Most stamps make $7.10.$ You have $20$ of each coin, nickles, dimes and quarters.

If we want to have the most amount of stamps we have to have the most amount of smaller value coins. 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 give back a dime and a nickel to have $38$ 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 $20\cdot3-5=\boxed{\textbf{(E)}\ 55}$.

~pianoboy

~MathFun1000 (Rewrote for clarity and formatting)

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/1bA7fD7Lg54?t=1078

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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