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

m (Solution 1)
(One intermediate revision by one other user not shown)
Line 44: Line 44:
  
 
~harungurcan
 
~harungurcan
 +
 +
==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=633
 +
 +
~hsnacademy
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/bVA6Sx4mbdM
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=13|num-a=15}}
 
{{AMC8 box|year=2023|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:10, 27 October 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

Video Solution by Dr. David

https://youtu.be/bVA6Sx4mbdM

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png