GATE CSE
First time here? Checkout the FAQ!
x
+10 votes
704 views

Let n = p2q, where p and q are distinct prime numbers. How many numbers m satisfy 1 ≤ m ≤ n and gcd (m, n) = 1? Note that gcd (m, n) is the greatest common divisor of m and n.

  1. p(q - 1)
  2. pq
  3. (p2- 1) (q - 1)
  4. p(p - 1) (q - 1)
asked in Set Theory & Algebra by Veteran (21.2k points) 466 748 812
edited by | 704 views

4 Answers

+13 votes
Best answer

n = p2q, where p and q are prime. 

So, number of multiple of p in n = pq 

Number of multiples of q in n = p2

Number of multiples of pq in n = p

Since prime factorisation of n consists of only p and q, gcd(m, n) will be a multiple of these or 1. So, number of possible m such that gcd(m, n) is 1 will be n - number of multiples of either p or q.

= n - p2-pq+p

= p2q-p2-pq+p

= p(pq-p-q+1)

=p(p-1)(q-1)

answered by Veteran (319k points) 577 1442 2960
selected by
Sir means the numbers we get after subtracting from "n" are the ones which do not contain p or q or pq in them,, and so their gcd with n will result to 1?
After subtraction we get numbers which are not multiple of p or q.
@arjun sir

can you please explain it a bit more, actually not getting it.
Since prime factorisation of n consists of only p and q, gcd(m, n) will be a multiple of these or 1. So, number of possible m such that gcd(m, n) is 1 will be n - number of multiples of either p or q
Got it. Thanx
Maybe you want to tell that you have used Inclusion - Exclusion Theorom
yes, for gcd being 1 , m will not divisible by any multiple of p or q

for pq it is double subtraction. So, added it
+7 votes

Euler's totient function $\phi (n)$ is being asked here :

Euler's totient function $\phi (n)$ = Number of positive integers which are $\leq n$ and relatively prime or co-prime to n . (ie. co-prime means if  $\gcd (a,b)=1$ )

It is given by $\phi (n)= n\times( \frac{(P_1-1)(P_2-1)...(P_k-1)}{P_1 P_2..P_k} )$

$\text{where } P_1 P_2..P_k \ \ \text{distinct prime divisors of }n$

We have $n=p^{2}q$.

Therefore,
$\begin{align*} \phi(n)&= n(\frac{(p-1)(q-1)}{p q}) \\ &= p^{2}q(\frac{(p-1)(q-1)}{p q}) \\ &= p (p-1)(q-1) \end{align*}$


 

answered by Veteran (22.8k points) 46 215 355
+6 votes

using Eulers function we can find out the no of relatively prime factors.

If we find out gcd of n with any of these prime factor,it will be always 1.

Eulers function is ∅(pn) is pn-1(p-1) 

given that n=p2q(p,q prime)

∅(p2q)=∅(p2)*∅(q)

          =p(p-1)(q-1)

so D is the answer

answered by Boss (6.5k points) 3 14 34

Please provide derivation of ∅(p2). 

–4 votes
1 is the answer
answered by (463 points) 2 6 11
reshown by
how?


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Top Users Oct 2017
  1. Arjun

    23240 Points

  2. Bikram

    17038 Points

  3. Habibkhan

    7096 Points

  4. srestha

    6008 Points

  5. Debashish Deka

    5430 Points

  6. jothee

    4928 Points

  7. Sachin Mittal 1

    4762 Points

  8. joshi_nitish

    4278 Points

  9. sushmita

    3954 Points

  10. Rishi yadav

    3744 Points


Recent Badges

Notable Question tajar
Notable Question Imarati Gupta
Notable Question set2018
Popular Question jothee
Notable Question set2018
Notable Question Pavan Kumar Munnam
Notable Question iarnav
Popular Question makhdoom ghaya
Popular Question Satyam
Popular Question radha gogia
27,254 questions
35,075 answers
83,756 comments
33,185 users