Difference between revisions of "1990 AIME Problems/Problem 3"
|Line 39:||Line 39:|
Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum
Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum116.
Revision as of 14:21, 29 August 2020
The formula for the interior angle of a regular sided polygon is .
Thus, . Cross multiplying and simplifying, we get . Cross multiply and combine like terms again to yield . Solving for , we get .
This is achievable because the denominator is , making a positive number and .
Like above, use the formula for the interior angles of a regular sided polygon.
This equation tells us divides . If specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is , which does give a solution: . Although, the problem asks for , not . The only conceivable reasoning behind this is that is greater than 1000. This prompts us to look into the second case, where divides . Make . Rewrite the equation using this new information.
Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.
|1990 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|