Difference between revisions of "2023 AIME II Problems/Problem 5"

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---Solution 1--
-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.
-If this is the case, the sum of the numerator and denominator of 55r is a + <math>\frac{b}{55}</math>. 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 + <math>\frac{b}{11}</math>. This is equal to a+b. Solving this equation gives <math>\frac{a}{b}</math> = r = <math>\frac{5}{22}</math>.
-Case 3: B is divisible by 5.
-In this case, the sum of the numerator and denominator of 55r is 11a + <math>\frac{b}{5}</math>. This is equal to a+b, and solving that equation gives <math>\frac{a}{b}</math> = <math>\frac{2}{25}</math> = r.
-Clearly, B is divisible by 1 doesn't make sense, so these are the only possibilities. Summing these fractions give <math>\frac{169}{550}</math>, so the answer is 169 + 550 = 719.
-~Anonymus

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