Question

Given that the divisor of $n={{3}^{p}}\cdot {{5}^{q}}\cdot {{7}^{r}}$ 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

Hint: To solve this kind of question, we will have to take up a number of cases consisting of different values of p, q, r and we will have to develop a relation among these three variables. Once we have developed a relation among three variables we can check the option one by one.

Complete step-by-step answer:
So, now let us start the question with forming the cases:
Case I: - Let the value of q be 1 and the value of p and r are also equal to 1. Thus, $n={{3}^{1}}\times {{5}^{1}}\times {{7}^{1}}=105$. It is of the form $4\lambda +1$ . Let the value of q be 2 and the value of p and r is equal to 1. Thus $n={{3}^{1}}\times {{5}^{2}}\times {{7}^{1}}=525$ . In this case also, the divisors are of the form $4\lambda +1$ . So, we can say that q can contain any value.
Case II: - Now let us check what happens if the value of p becomes 2, q and r remain 1. Thus, $n={{3}^{2}}\times {{5}^{1}}\times {{7}^{1}}=315$ . It is of the form $4\lambda -1$ .Now, let the value of $r=2$ and the values of p and q remain the same. Thus $n={{3}^{2}}\times {{5}^{1}}\times {{7}^{2}}=2205$ . It is of the form $4\lambda +1$ . Thus, from the above cases we can conclude that the q can have any value. Also, when the value of p is odd then the value of r should be odd.
Now, we are going to check options: -
Option (a): - If p is odd then r will also be odd and if p is even then r will also be even. So, $\left( p+r \right)$ will always be even.
Option (b): - From the above option, we can say that $\left( p+r \right)$ is always even, but q can have any value, that is it can be either even or odd. So, the value of $\left( p+q+r \right)$ can be odd or even.
Option (c): - q can have any integral value.
Option (d): - If p is odd and r is even then the divisors will be of the form $\left( 4\lambda -1 \right)$ .
Hence, (a) and (c) are correct.

Note: The alternate method is given by: -
${{3}^{p}}={{\left( 4-1 \right)}^{p}}={}^{p}{{C}_{0}}{{4}^{0}}{{\left( -1 \right)}^{p}}+{}^{p}{{C}_{1}}{{4}^{1}}{{\left( -1 \right)}^{p-1}}+......{}^{p}{{C}_{p}}{{4}^{p}}$
$={{\left( -1 \right)}^{p}}+4\lambda $
${{5}^{p}}={{\left( 4+1 \right)}^{p}}={}^{p}{{C}_{0}}{{4}^{0}}{{\left( 1 \right)}^{p}}+{}^{p}{{C}_{1}}{{4}^{1}}{{\left( 1 \right)}^{p-1}}+......{}^{p}{{C}_{p}}{{4}^{p}}$
$=1+4\lambda $
${{7}^{p}}={{\left( 8-1 \right)}^{p}}={}^{p}{{C}_{0}}{{8}^{0}}{{\left( -1 \right)}^{p}}+{}^{p}{{C}_{1}}{{8}^{1}}{{\left( -1 \right)}^{p-1}}+......{}^{p}{{C}_{p}}{{8}^{p}}$
$={{\left( -1 \right)}^{p}}+4\lambda $
Form above equations also, the option (a) and (c) are satisfying.