2023 AIME II Problems/Problem 5

--Solution 1--

Since 55r is greater than r, we know 55r can be simplified. Call r = $\frac{a}{b}$ . This means 55r is $\frac{55a}{b}$ . 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 + $\frac{b}{55}$ . 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 + $\frac{b}{11}$ . This is equal to a+b. Solving this equation gives $\frac{a}{b}$ = r = $\frac{5}{22}$ .

Case 3: B is divisible by 5.

In this case, the sum of the numerator and denominator of 55r is 11a + $\frac{b}{5}$ . This is equal to a+b, and solving that equation gives $\frac{a}{b}$ = $\frac{2}{25}$ = r.

Clearly, B is divisible by 1 doesn't make sense, so these are the only possibilities. Summing these fractions give $\frac{169}{550}$ , so the answer is 169 + 550 = 719.

~Anonymus

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2023 AIME II Problems/Problem 5