Question

Total number of divisors of $n=3^5\cdot 5^7\cdot 7^9$ that are of the form $4\lambda +1,1\ge 0$

• A. 240
• B. 30
• C. 120
• D. 15

Answer 1

Answer: (A)

Considering $n=3^5{5}^7{7}^9$ For number of divisors of form $4\lambda +1;1\ge 0$. Let us examine $3,5,7$ and behaviour of its powers.

So even powers of $3$ are of $4\lambda +1$ form and odd powers are $4\lambda -1$ form.
Now each of $5,5^2,5^3,\dots \dots ,5^7$ is of $(4\lambda +1)$ form Considering '$7$'
$7^{\prime }=7;4\lambda -1$ form
$7^2=49;4\lambda +1$ form
$7^3=343;4\lambda -1$ form
$7^4=2401;4\lambda +1$ form
So even powers of $7$ are of $4\lambda +1$ form and odd powers are of $4\lambda -1$ form.
We also have to notice that
$(4\lambda -1)(4\lambda -1)=16{\lambda }^2+1-8\lambda \Rightarrow 4\lambda +1$ form
So, for divisors of $4\lambda +1$ form
We have to consider all possible ways to select powers of $5$ , [even powers of $3$ even power of $7+$ odd powers of $3$ odd powers of $7]$
$=8[3\times 5+3\times 5]=8[30]=240$