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

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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.
 
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.
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Case 1: B is divisible by 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.
 
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.
 
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>.
 
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.
 
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.
 
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.
 
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
 
~Anonymus

Revision as of 16:28, 16 February 2023


--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