2024 AMC 8 Problems/Problem 8: Difference between revisions
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==Solution 1== (BRUTE FORCE) | ==Solution 1== (BRUTE FORCE) | ||
How many values could be on the first day? Only <math>2. The second day, you can either add 3, or double, so you can have </math>5, or <math>4. For each of these values, you have 2 values for each. For </math>5, you have <math>10 or </math>8, and for <math>4, you have </math>8 or <math>7. Now, you have 2 values for each of these. For </math>10, you have <math>13 or </math>20, for <math>8, you have </math>16 or 11, for 8, you have 16 or 11, and for 7, you have 14 or 10. | How many values could be on the first day? Only <math>2. The second day, you can either add 3, or double, so you can have </math>5, or <math>4. For each of these values, you have 2 values for each. For </math>5, you have <math>10 or </math>8, and for <math>4, you have </math>8 or <math>7. Now, you have 2 values for each of these. For </math>10, you have <math>13 or </math>20, for <math>8, you have </math>16 or <math>11, for </math>8, you have <math>16 or </math>11, and for <math>7, you have </math>14 or <math>10. | ||
</math>11,$16 repeat leaving you with 8-2 = 6 different values | |||
Revision as of 15:36, 25 January 2024
Problem
==Solution 1== (BRUTE FORCE)
How many values could be on the first day? Only 
5, or 
5, you have 
8, and for 
8 or 
10, you have 
20, for 
16 or 
8, you have 
11, and for 
14 or 
11,$16 repeat leaving you with 8-2 = 6 different values
