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. 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>. | ||
− | 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- | + | 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. |
~ cxsmi (minor formatting edits) | ~ cxsmi (minor formatting edits) |
Revision as of 10:31, 6 February 2024
Contents
[hide]- 1 Problem
- 2 Solution 1 (BRUTE FORCE)
- 3 Solution 2 (Slightly less BRUTE FORCE)
- 4 Video Solution 1 (easy to digest) by Power Solve
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution 2 by SpreadTheMathLove
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 Video Solution by Interstigation
- 10 See Also
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?
Solution 1 (BRUTE FORCE)
How many values could be on the first day? Only dollars. The second day, you can either add dollars, or double, so you can have dollars, or . For each of these values, you have values for each. For dollars, you have dollars or , and for dollars, you have dollars or $. Now, you have values for each of these. For dollars, you have dollars or , for dollars, you have dollars or , for dollars, you have dollars or , and for dollars, you have dollars or .
On the final day, there is 11, 11, 16, and 16 repeating, thus 8-4 $ and $ repeat leaving you with different values.
~ cxsmi (minor formatting edits) ~ corrections
Solution 2 (Slightly less BRUTE FORCE)
Continue as in Solution 1 to get , , or dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by or adding ) from here is if or which both aren't true. Hence our answer is .
~ 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.