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

Options

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

Answer

A, B, C

Explanation

`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.