GATE CSE
First time here? Checkout the FAQ!
x
+9 votes
503 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 (19.3k points)  
edited ago by | 503 views

4 Answers

+10 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 (286k points)  
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
+6 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 (20.7k points)  
+5 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 (6.1k points)  

Please provide derivation of ∅(p2). 

–3 votes
1 is the answer
answered by (465 points)  
reshown by
how?


Top Users Jun 2017
  1. Bikram

    3912 Points

  2. Arnab Bhadra

    1550 Points

  3. Hemant Parihar

    1502 Points

  4. Niraj Singh 2

    1501 Points

  5. Debashish Deka

    1480 Points

  6. junaid ahmad

    1432 Points

  7. pawan kumarln

    1366 Points

  8. Arjun

    1246 Points

  9. Rupendra Choudhary

    1242 Points

  10. rahul sharma 5

    1240 Points

Monthly Topper: Rs. 500 gift card
Top Users 2017 Jun 26 - Jul 02
  1. pawan kumarln

    498 Points

  2. akankshadewangan24

    404 Points

  3. Arjun

    286 Points

  4. Debashish Deka

    234 Points

  5. Abhisek Das

    230 Points


23,435 questions
30,153 answers
67,633 comments
28,503 users