Home
Class 12
MATHS
Given that the divisors of n=3^(p)*5^(q)...

Given that the divisors of `n=3^(p)*5^(q)*7^(r)` are of of the form `4lamda+1,lamdage0`. Then,

A

p+r is always even

B

p+q+r is even orr odd

C

q can be any integer

D

if p is even, then r is odd

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`because 3^(p)=(4-1)^(p)=4lamda_(1)+(-1)^(p)`
`5^(q)=(4+1)^(q)=4lamda_(2)+1`
and `7^(r)=(8-1)^(r)=8lamda_(3)+(-1)^(r)`
hence, both p annd r must be odd or both must be even. Thus, p+r is always even. Also `p+q+r` can be odd or even.
Promotional Banner

Similar Questions

Explore conceptually related problems

Sum of divisors of 2^(5).3^(7).5^(3).7^(2) is :

Total number of divisors of n = 3^5. 5^7. 7^9 that are in the form of 4lambda + 1; lamda >=0 is equal to

Total number of divisors of N=2^(5)*3^(4)*5^(10)*7^(6) that are of the form 4n+2,n ge 1 , is equal to

The sum of the divisors of 2^(5)xx3^(4)xx5^(2) , is

The number of proper divisors of 2^(p)*6^(q)*21^(r),AA p,q,r in N , is

Number of odd proper divisors of 3^(p).6^(m).21^(n) is :

The number of odd proper divisors of 3^(p)*6^(q)*15^(r),AA p,q,r, in N , is

The simplified form of (p^^q)vv(p^^r)=

Number of divisors of the form 4n+2(n ge 0) of integer 240 is :

Statement-1: The number of divisors of 10! Is 280. Statement-2: 10!= 2^(p)*3^(q)*5^(r)*7^(s) , where p,q,r,s in N.