Combinatory

Question

Kindly help...thanks

Comments

gcd|(m,n) =1 if both are relatively prime.

so just fin numbers that are relatively prime to 100.

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This problem is equivalent to find a number of generators in the cyclic group.Just u need to find out number of generators of the group of order 100.

φ(100) = φ(2² * 5² )

            =φ(2²) * φ(5²)

            = 2² - 2^2-1   *  5² -5^2-1

           =40

Answer 1

i find factors of  100 = 2 , 2 ,5, 5

we want gcd of two number  (HCF) ,   so 2 ,2 , 5,5 ka factor nai ayga jis number me so hcf =1 ayga  with 100 

i find how many integer between 1 to 100  having factor of 2 like : 2 ,4, 6, ---- 100 = 50

similarly for 5 i got = 20 

5,10,,,,,,,,,100   = 20 terms 
  

hmm ,  there are some repeating numbers too in series

2,4,6,8,10,....................100

5,10,21,........................100

  

like  10 is both factor of 2 and 5

so  they should be subtracted numbers  between 1 to 100 that are factors of 10  is  "10"

50+20 -10 = 60

60 are total numbers between 1 to 100 that  contains 2,2,5,5 as factors  

100 - 60 = 40    such no. are there which donot contain 2,2,5,5 as factors , so their  gcd / hcf with 100 =1 

,no need to take 1 , because gcd(1,100) =1  ,  see gcd (100,100) =100   , which should not be taken in consideration but  in  calculation factors of 2,5,10  { 100 is taken in consideration } and 1 has not taken in consideration  , so it is balanced  

so overall 40 i got

Comments

Sir g its 40, can u tell the approach ?

Answer 2

number of co-prime numbers to 100 can be found using EULER TOTIENT function.

100 =2² *5²

euler totient  = (2² -2) *(5² -5){ Φ(p^k) = p^k – p^k-1. )

                      = 2*20

                       = 40

Comments

 thanks raviyogi bro :)