Question

Given that the divisors of $n=3^{p}c{\dot{5}}^q\cdot 7^r$ are of the form $4\lambda +1,\lambda \ge 0$. Then

• A. $p+r$ is always even
• B. $p+q+r$ is always odd
• C. $q$ can be any integer
• D. If $p$ is odd then $r$ is even

Answer 1

Answer: (A), (C)

The correct options are
A $p+r$ is always even
C $q$ can be any integer

$3^p=(4-1{)}^p=4{\lambda }_1+(-1{)}^p$
$5^q=(4+1{)}^q=4{\lambda }_2+1$
$7^r=(8-1{)}^r=4{\lambda }_3+(-1{)}^r$
Hence, any positive integer power of 5 will be in the form of $4{\lambda }_2+1$. Even power of 3 and 7 will be in the form of $4\lambda +1$ and odd power of 3 and 7 will be form of $4\lambda -1$. Hence, both p and r must be even. Thus p+r is always even. Also, p+q+r can be odd or even.