Question

Consider, $M =2^{ a } \cdot 3^3 \cdot 7^{ c } \cdot 11^{ d }$, where $a , c , d \in N$ such that number of odd divisors of $M$ is 140 and product of all the divisors of $M$ is $( M )^{280}$.
The number of non-negative integral solutions of the inequality $x+y+z+w \leq a+c+d$ is

• A. ${ }^9 C _4$
• B. ${ }^{12} C _4$
• C. ${ }^{13} C _4$
• D. $ { }^{17} C _4$

Answer 1

Answer: (D)

$\text { Number of odd divisors }=140 $
$ 4 \times(c+1)(d+1)=140 \Rightarrow (c+1)(d+1)=35$
$\text { Product of all the divisors }=(M)^{n / 2} \Rightarrow (n=\text { total number of divisors of } M)$
$\therefore n=560 \Rightarrow a=3$
$(a+1) \cdot 4(c+d)(d+1)=560 $
$\Rightarrow a+1=4 \Rightarrow a=3$

$x + y + z + w \leq 13 \Rightarrow x + y + z + w + t =13 $
$\therefore \text { Number of non-negative integral solutions }={ }^{13+5-1} C _{5-1}={ }^{17} C _4$