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Difference between revisions of "2024 AMC 8 Problems/Problem 8"
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==Problem==
 
==Problem==
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On Monday Taye has <math>\$2</math>. Every day, he either gains <math>\$3</math> or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, <math>3</math> days later?
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On Monday, Taye has <math>\$2</math>. Every day, he either gains <math>\$3</math> or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, <math>3</math> days later?
  
 
<math>\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7</math>
 
<math>\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7</math>
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~ Sahan Wijetunga
 
~ Sahan Wijetunga
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 +
==Video Solution by Central Valley Math Circle(Goes through the full thought process)==
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https://youtu.be/dC0inhFYaTw
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 +
~mr_mathman
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 +
==Video Solution by Math-X (First fully understand the problem!!!)==
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https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762
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~Math-X
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
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https://youtu.be/ruYzBZYaWSA
 
https://youtu.be/ruYzBZYaWSA
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==Video Solution by WhyMath==
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https://youtu.be/uvizv1hjSps
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=7|num-a=9}}
 
{{AMC8 box|year=2024|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:07, 22 December 2024

Problem

On Monday, Taye has $$2$. Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?

$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$

Solution 1 (BRUTE FORCE)

How many values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$. For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$, and for $4$ dollars, you have $8$ dollars or $$7$. Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$, for $8$ dollars, you have $16$ dollars or $11$, for $8$ dollars, you have $16$ dollars or $11$, and for $7$ dollars, you have $14$ dollars or $10$.

On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{\textbf{(D)\ 6}}$ different values.

~ cxsmi (minor formatting edits)

Solution 2 (Slightly less BRUTE FORCE)

Continue as in Solution 1 to get $7$, $8$, or $10$ dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by $2$ or adding $3$) from here is if $7+3=10\cdot 2$ or $7+3=8\cdot 2$ which both aren't true. Hence our answer is $3\cdot2=\boxed{\textbf{(D)\ 6}}$.

~ Sahan Wijetunga

Video Solution by Central Valley Math Circle(Goes through the full thought process)

https://youtu.be/dC0inhFYaTw

~mr_mathman

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762 

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=4oMd4d0cnZ-nghTe&t=873

~hsnacademy

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/16YYti_pDUg?si=5kw0dc_bZwASNiWm&t=121

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762 

~Math-X

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=alDY4yhaEEg

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=791

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by Dr. David (Tree Method)

https://youtu.be/ruYzBZYaWSA

Video Solution by WhyMath

https://youtu.be/uvizv1hjSps

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

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