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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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==Solution 1==
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<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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 +
==Solution 1 (BRUTE FORCE)==
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How many values could be on the first day? Only <math>2</math> dollars. The second day, you can either add <math>3</math> dollars, or double, so you can have <math>5</math> dollars, or <math>4</math>. For each of these values, you have <math>2</math> values for each. For <math>5</math> dollars, you have <math>10</math> dollars or <math>8</math>, and for <math>4</math> dollars, you have <math>8</math> dollars or \$<math>7</math>. Now, you have <math>2</math> values for each of these. For <math>10</math> dollars, you have <math>13</math> dollars or <math>20</math>, for <math>8</math> dollars, you have <math>16</math> dollars or <math>11</math>, for <math>8</math> dollars, you have <math>16</math> dollars or <math>11</math>, and for <math>7</math> dollars, you have <math>14</math> dollars or <math>10</math>.
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On the final day, there are 11, 11, 16, and 16 repeating, leaving you with <math>8-2 = \boxed{\textbf{(D)\ 6}}</math> different values.
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~ cxsmi (minor formatting edits)
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==Solution 2 (Slightly less BRUTE FORCE)==
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Continue as in Solution 1 to get <math>7</math>, <math>8</math>, or <math>10</math> dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by <math>2</math> or adding <math>3</math>) from here is if <math>7+3=10\cdot 2</math> or <math>7+3=8\cdot 2</math> which both aren't true. Hence our answer is <math>3\cdot2=\boxed{\textbf{(D)\ 6}}</math>.
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 +
~ Sahan Wijetunga
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 +
==Video Solution 1 (easy to digest) by Power Solve==
 +
https://youtu.be/16YYti_pDUg?si=5kw0dc_bZwASNiWm&t=121
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 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/BaE00H2SHQM?si=6wjacdxeAgtpc0fW&t=1762
 +
~Math-X
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 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=V-xN8Njd_Lc
 +
 
 +
~NiuniuMaths
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=L83DxusGkSY
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 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=alDY4yhaEEg
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==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=791
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==See Also==
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{{AMC8 box|year=2024|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 22:57, 17 February 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 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

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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