Difference between revisions of "2023 AIME II Problems/Problem 5"
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| − | Since 55r is greater than r, we know | + | |
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| + | --Solution 1-- | ||
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| + | Since 55r is greater than r, we know 55r can be simplified. Call r <math>\frac{a}{b}</math>. This means 55r is <math>\frac{55a}{b}</math>. Since 55r can clearly be simplified, let's look at the possibilities, where b is divisible by factors of 55. | ||
Case 1: B is divisible by 55. | Case 1: B is divisible by 55. | ||
Revision as of 16:26, 16 February 2023
--Solution 1--
Since 55r is greater than r, we know 55r can be simplified. Call r 
. This means 55r is 
. Since 55r can clearly be simplified, let's look at the possibilities, where b is divisible by factors of 55.
Case 1: B is divisible by 55.
If this is the case, the sum of the numerator and denominator of 55r is a + 
. This clearly cannot equal a+b, so this case is invalid.
Case 2: B is divisible by 11.
If this is the case, the sum of the numerator and denominator of 55r is 5a + 
. This is equal to a+b. Solving this equation gives = r = 
.
Case 3: B is divisible by 5.
In this case, the sum of the numerator and denominator of 55r is 11a + 
. This is equal to a+b, and solving that equation gives = 
= r.
Clearly, B is divisible by 1 doesn't make sense, so these are the only possibilities. Summing these fractions give 
, so the answer is 169 + 550 = 719.
~Anonymus
