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Q.
Total number of divisors of $n =3^{5} \cdot 5^{7} \cdot 7^{9}$ that are of the form $4 \lambda+1,1 \geq 0$
Permutations and Combinations
Solution:
Considering $n =3^{5} 5^{7} 7^{9}$ For number of divisors of form $4 \lambda+1 ; 1 \geq 0$. 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 $5,5^{2}, 5^{3}, \ldots \ldots, 5^{7}$ 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 $7]$
$=8[3 \times 5+3 \times 5]=8[30]=240$
