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

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How many values could be on the first day? Only <math>2</math> dollars. On the second day, you can either add <math>3</math> dollars, or double so that 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. On the second day, you can either add <math>3</math> dollars, or double so that 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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There is a total of <math>2</math>x<math>2</math>x<math>2</math>=<math>8</math> combinations, and \$<math>11</math> and \$<math>16</math> repeat. Leaving you with <math>8-2 = \boxed{\textbf{(C)} 6}</math> different values.
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There is a total of <math>2*</math>2*2=8<math> combinations, and \$</math>11<math> and \$</math>16<math> repeat. Leaving you with </math>8-2 = \boxed{\textbf{(C)} 6}$ different values.
  
 
~cxsmi (minor formatting edits)
 
~cxsmi (minor formatting edits)

Revision as of 17:25, 25 January 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. On the second day, you can either add $3$ dollars, or double so that 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 $10$.

There is a total of $2*$2*2=8$combinations, and $11$and $16$repeat. Leaving you with$8-2 = \boxed{\textbf{(C)} 6}$ different values.

~cxsmi (minor formatting edits)

Video Solution 1(easy to digest) by Power Solve

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

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