Difference between revisions of "2023 AIME II Problems/Problem 5"
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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 the elements of <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> | 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 the elements of <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> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Denote <math>r = \frac{a}{b}</math>, where <math>\left( a, b \right) = 1</math>. | ||
+ | We have <math>55 r = \frac{55a}{b}</math>. | ||
+ | Suppose <math>\left( 55, b \right) = 1</math>, then the sum of the numerator and the denominator of <math>55r</math> is <math>55a + b</math>. | ||
+ | This cannot be equal to the sum of the numerator and the denominator of <math>r</math>, <math>a + b</math>. | ||
+ | Therefore, <math>\left( 55, b \right) \neq 1</math>. | ||
+ | |||
+ | Case 1: <math>b</math> can be written as <math>5c</math> with <math>\left( c, 11 \right) = 1</math>. | ||
+ | |||
+ | Thus, <math>55r = \frac{11a}{c}</math>. | ||
+ | |||
+ | Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | a + 5c = 11a + c . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Hence, <math>2c = 5 a</math>. | ||
+ | |||
+ | Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>. | ||
+ | Thus, <math>a = 2</math> and <math>c = 5</math>. | ||
+ | Therefore, <math>r = \frac{a}{5c} = \frac{2}{25}</math>. | ||
+ | |||
+ | Case 2: <math>b</math> can be written as <math>11d</math> with <math>\left( d, 5 \right) = 1</math>. | ||
+ | |||
+ | Thus, <math>55r = \frac{5a}{c}</math>. | ||
+ | |||
+ | Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | a + 11c = 5a + c . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Hence, <math>2a = 5 c</math>. | ||
+ | |||
+ | Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>. | ||
+ | Thus, <math>a = 5</math> and <math>c = 2</math>. | ||
+ | Therefore, <math>r = \frac{a}{11c} = \frac{5}{22}</math>. | ||
+ | |||
+ | Case 3: <math>b</math> can be written as <math>55 e</math>. | ||
+ | |||
+ | Thus, <math>55r = \frac{a}{c}</math>. | ||
+ | |||
+ | Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | a + 55c = a + c . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Hence, <math>c = 0</math>. This is infeasible. | ||
+ | Thus, there is no solution in this case. | ||
+ | |||
+ | Putting all cases together, <math>S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}</math>. | ||
+ | Therefore, the sum of all numbers in <math>S</math> is | ||
+ | <cmath> | ||
+ | \[ | ||
+ | \frac{2}{25} + \frac{5}{121} = \frac{169}{550} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is <math>169 + 550 = \boxed{\textbf{(719) }}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 17:46, 16 February 2023
Let be the set of all positive rational numbers
such that when the two numbers
and
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 the elements of
can be expressed in the form
where
and
are relatively prime positive integers. Find
Solution
Denote , where
.
We have
.
Suppose
, then the sum of the numerator and the denominator of
is
.
This cannot be equal to the sum of the numerator and the denominator of
,
.
Therefore,
.
Case 1: can be written as
with
.
Thus, .
Because the sum of the numerator and the denominator of and
are the same,
Hence, .
Because ,
.
Thus,
and
.
Therefore,
.
Case 2: can be written as
with
.
Thus, .
Because the sum of the numerator and the denominator of and
are the same,
Hence, .
Because ,
.
Thus,
and
.
Therefore,
.
Case 3: can be written as
.
Thus, .
Because the sum of the numerator and the denominator of and
are the same,
Hence, . This is infeasible.
Thus, there is no solution in this case.
Putting all cases together, .
Therefore, the sum of all numbers in
is
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)