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 $x+y+z+w\le a+c+d$ is

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

Answer 1

Answer: (D)

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

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