Question

In each of the following 2003 fractions the sum of the numerator and denominator equals 2004: \left[\frac{1}{2003}, \frac{2}{2002}, \frac{3}{2001}, \ldots \ldots . ., \frac{2003}{1}\right]. The number of fractions <1 which are irreducible ( no commom factor between numerator and denominator ) is :

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given In each of the following 2003 fractions the sum of the numerator and denominator equals 2004  \left[\frac{1}{2003}, \frac{2}{2002}, \frac{3}{2001}, \ldots \ldots . ., \frac{2003}{1}\right]. The number of fractions <1 which are irreducible ( no common factor between numerator and denominator ) is :

• Now we have the fractions 1/2004, 1/2003, 1/2001,……..2003 / 1
• So number of fractions less than 1 will be numerator and denominator must be equal.
• Also sum of numerator and denominator equals 2004.
• 1/2003, 2/2002, 3/2001,………..1002 / 1002, 1003 / 1001
• 1/2003, 2/2002, 3/2001……………. ]  1002 / 1002 , 1003 / 1001
• 1/2003, 2/2002, 3/2001………………..1001 / 1003
• Now we need to find that are irreducible but 2/2002 and 3/2001 are reducible
• So 1/2003, 5 / 1999, 7 / 1997, 9 / 1993 …………….1001 / 1003
• Now we need to find the sum and so we have numerator as 1, 5,7,,11,13, 17, 19,23…. so these are the prime numbers and we need to find the number of terms.
• We have two series
• 1,7,13,19……………997 (since less than 1)
• 5,11,17,23…………1001
• So this is in the form of an A.P
• So T n = 997, a = 1, d = 6
• So Tn = a + (n – 1) d
•       997 = 1 + (n – 1) 6
•      997 = 1 + 6n – 6
•     997 = 6n – 5
•   6n = 1002
• Or n = 167
• Now for the next sequence we get
• T x = 1001, a = 5, d = 6
• 1001 = 5 + (x – 1) 6
• 1001 = 5 + 6x – 6
• 1001 = 6x – 1
• 1002 = 6x  
• Or x = 167
• So total value n + x = 167 + 167 = 334
• So the two terms are reducible and have common factors in numerator and denominator.
• Since two terms are reducible, the remaining term will be 334 – 2 = 332
• So 332 fractions are less than 1 and are irreducible.

Reference link will be

https://brainly.in/question/17948134