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

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==Problem==
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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 <math>5</math>-cent, <math>10</math>-cent, and <math>25</math>-cent stamps, with exactly <math>20</math> of each type. What is the greatest number of stamps Nicolas can use to make exactly <math>\$7.10</math> in postage?
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(Note: The amount <math>\$7.10</math> corresponds to <math>7</math> dollars and <math>10</math> cents. One dollar is worth <math>100</math> cents.)
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<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55</math>
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==Solution 1==
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Let's use the most stamps to make <math>7.10.</math> We have <math>20</math> of each stamp, <math>5</math>-cent (nickels), <math>10</math>-cent (dimes), and <math>25</math>-cent (quarters).
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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 <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 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 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>.
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, InterstellerApex, mahika99
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==Solution 2==
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The value of his entire stamp collection is <math>8</math> dollars. To make <math>\$7.10</math> with stamps, he should remove <math>90</math> cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many <math>25</math> cent stamps as possible as they have the greatest denomination. He can remove at most <math>3</math> of these stamps. He still has to remove <math>90-25\cdot3=15</math> cents worth of stamps. This can be done with one <math>5</math> and <math>10</math> cent stamp. In total, he has <math>20\cdot3=60</math> stamps in his entire collection. As a result, the maximum number of stamps he can use is <math>20\cdot3-5=\boxed{\textbf{(E)}\ 55}</math>.
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~pianoboy
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~MathFun1000 (Rewrote for clarity and formatting)
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~vadava_lx (minor edits)
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/vq3voJZ-hvw
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~Education, the Study of Everything
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==Video Solution by Math-X (Smart and Simple)==
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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. :(
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==Animated Video Solution==
 
==Animated Video Solution==
 
https://youtu.be/XP_tyhTqOBY
 
https://youtu.be/XP_tyhTqOBY
Line 4: Line 35:
 
~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
  
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==Written Solution==
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==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=4280
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=1435
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=VqN7c5U5o98&t=449s
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~harungurcan
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==Video Solution (Solve under 60 seconds!!!)==
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https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=633
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~hsnacademy
  
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Most stamps make <math>7.10. You have 20 of each coin, nickles, dimes and quarters.
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==Video Solution by Dr. David==
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https://youtu.be/bVA6Sx4mbdM
  
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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 29 0imes to bring our total cost to </math>7.10 - 3.00 = 4.10<math>. 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 </math>\frac{425}{25} = 17<math> quarters to get a grand total of </math>38 + 17 = \boxed{\text{(E)}55}$
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==Video Solution by WhyMath==
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https://youtu.be/rNnkVEpXr-A
  
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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==See Also==
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{{AMC8 box|year=2023|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 10:12, 18 November 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

Video Solution by WhyMath

https://youtu.be/rNnkVEpXr-A

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

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