Difference between revisions of "2023 AIME II Problems/Problem 5"
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− | + | ==Problem== | |
+ | 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 | + | 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> | ||
+ | \[ | ||
+ | \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is <math>169 + 550 = \boxed{\textbf{(719) }}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by The Power of Logic== | ||
+ | https://youtu.be/qUJtReB_9sU | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2023|num-b=4|num-a=6|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:13, 18 June 2023
Problem
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)
Video Solution by The Power of Logic
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.