Sequential Fractions

{ 1 2003 , 2 2002 , 3 2001 , 4 2000 , , 2003 1 } \large \left \{\dfrac{1}{2003}, \dfrac{2}{2002}, \dfrac{3}{2001}, \dfrac{4}{2000}, \ldots, \dfrac{2003}{1} \right \}

For each of the 2003 fractions above, the sum of the numerator and denominator equal 2004. Find the number of fractions less than 1 which are irreducible.


The answer is 332.

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2 solutions

Relevant wiki: Euler's Totient Function

We note that the general form of the fraction is n 2004 n = 1 2004 n 1 \dfrac n{2004-n} = \dfrac 1{\frac {2004}n - 1} , where n = 1 , 2 , 3 , . . . 2003 n =1,2,3,...2003 . We also note that the fraction is irreducible when 2004 n \dfrac {2004}n is irreducible or n n and 2004 2004 are coprime integers. The number of 2003 fractions which is irreducible is the number of coprime integers of 2004 less than 2004, which is Euler's totient function of 2004:

ϕ ( 2004 ) = 2004 ( 1 1 2 ) ( 1 1 3 ) ( 1 1 167 ) where 2, 3 and 167 are prime factors of 2004. = 664 \begin{aligned} \phi (2004) & = 2004\left(1-\frac 1{\color{#3D99F6}{2}} \right)\left(1- \frac 1{\color{#3D99F6}{3}} \right) \left(1- \frac 1{\color{#3D99F6}{167}} \right) & \small \color{#3D99F6}{\text{where 2, 3 and 167 are prime factors of 2004.}} \\ & = 664 \end{aligned} .

Since: { 1 2003 , 2 2002 , 1001 1003 , 1002 1002 , 1003 1001 , 2002 2 , 2003 1 } = { 1 2003 , 2 2002 , 1001 1003 , 1 , 1 1001 1003 , 1 2 2002 , 1 1 2003 } \left \{\frac{1}{2003}, \frac{2}{2002}, \cdots \frac{1001}{1003}, \color{#3D99F6}{\frac{1002}{1002}}, \frac{1003}{1001}, \cdots \frac{2002}{2}, \frac{2003}{1} \right \} = \left \{\frac{1}{2003}, \frac{2}{2002}, \cdots \frac{1001}{1003}, \color{#3D99F6}{1}, \frac 1{\frac{1001}{1003}}, \cdots \frac 1{\frac 2{2002}}, \frac 1{\frac 1{2003}} \right \} , we note that half of the irreducible fractions are less than 1 and the other half greater than 1. Therefore, the number of irreducible less than 1 is 664 2 = 332 \dfrac {664}2 = \boxed{332} .

Qin Haichen
Aug 7, 2016

Hi Firstly, This question is asking for fractions lesser than one. Hence we only consider 1 2003 \frac{1}{2003} to 1001 1003 \frac{1001}{1003} Now, for the fraction n 2004 n \frac{n}{2004-n} to be reducible, n must be a factor or 2004. The factors of 2004 which are smaller than 1002 are 2,3.4.6,12,167,334,501,668(not considering 1). So what we are trying to find now is the number of fractions from 1 2003 \frac{1}{2003} to 1001 1003 \frac{1001}{1003} which have their numerators being a factor of 2004. Hence we need to find the amount of numbers which are a multiple of 2,3,167 from 1-1001(The other factors are taken out as they are just multiple of those prime factors). Set A:Multiple of 2s Set B:Multiple of 3s Set C:Multiple of 167s AUBUC=A+B+C-AnB-BnC-AnC+AnBnC =500+333+5-166-1-2+0(not applicable) =669 This are the number of fractions which are REDUCIBLE. Hence the number of non-reducible fractions are 1001-669 =332 U stands for union while n stands for intersects(i am bad at latex)

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