Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Video Solution by Math-X (Smart and Simple))
m (→‎Solution 1)
Line 11: Line 11:
 
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 <math>20</math> nickels and <math>20</math> dimes to bring our total cost to <math>7.10 - 3.00 = 4.10</math>. However, when we try to use quarters, the <math>25</math> cents don’t fit evenly, so we have to give back <math>15</math> cents in order to make the quarter amount <math>4.25</math>. The most efficient way to do this is to give back a  <math>10</math>-cent (dime) stamp and a <math>5</math>-cent (nickel) stamp to have <math>38</math> stamps (coins) used so far. Now, we just use <math>\frac{425}{25} = 17</math> quarters to get a grand total of <math>38 + 17 = \boxed{\textbf{(E)}\ 55}</math>.
 
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 <math>20</math> nickels and <math>20</math> dimes to bring our total cost to <math>7.10 - 3.00 = 4.10</math>. However, when we try to use quarters, the <math>25</math> cents don’t fit evenly, so we have to give back <math>15</math> cents in order to make the quarter amount <math>4.25</math>. The most efficient way to do this is to give back a  <math>10</math>-cent (dime) stamp and a <math>5</math>-cent (nickel) stamp to have <math>38</math> stamps (coins) used so far. Now, we just use <math>\frac{425}{25} = 17</math> quarters to get a grand total of <math>38 + 17 = \boxed{\textbf{(E)}\ 55}</math>.
  
−
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
+
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, InterstellerApex
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:46, 13 January 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 (quaters).

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, InterstellerApex

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)

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

See Also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_14&oldid=210649"