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Difference between revisions of "2024 AMC 8 Problems/Problem 8"

(→‎Solution 1 (BRUTE FORCE))
(→‎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>.
 
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 is 11, 11, 16, and 16 repeating, thus 8-4 \$<math>11</math> and \$<math>16</math> repeat leaving you with <math>8-4 = \boxed{\textbf{(B)\ 4}}</math> different values.
+
On the final day, there are 11, 11, 16, and 16 repeating, thus 8-4 \$<math>11</math> and \$<math>16</math> repeat leaving you with <math>8-4 = \boxed{\textbf{(B)\ 4}}</math> different values.
 
   
 
   
 
~ cxsmi (minor formatting edits)
 
~ cxsmi (minor formatting edits)
−
~ corrections
 
  
 
==Solution 2 (Slightly less BRUTE FORCE)==
 
==Solution 2 (Slightly less BRUTE FORCE)==

Revision as of 11:33, 6 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, thus 8-4 $$11$ and $$16$ repeat leaving you with $8-4 = \boxed{\textbf{(B)\ 4}}$ 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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