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

(Created page with "Let <math>S</math> be the set of all positive rational numbers <math>r</math> such that when the two numbers <math>r</math> and <math>55r</math> are written as fractions in lo...")
 
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Let <math>S</math> be the set of all positive rational numbers <math>r</math> such that when the two numbers <math>r</math> and <math>55r</math> are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all elements <math>S</math> can be expressed in the form <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
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Since 55r is greater than r, we know 55 r 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.
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Case 1: B is divisible by 55.
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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.
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Case 2: B is divisible by 11.
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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>.
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Case 3: B is divisible by 5.
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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.
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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.
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~Anonymus

Revision as of 16:26, 16 February 2023

Since 55r is greater than r, we know 55 r 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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