CMAT 2014 Solved Question Paper

(c) Since, factors of 7056 = $2^4\times 3^2\times 7^2$
Number of factors/divisors of 7056= (4+1) (2 + 1) (2 + 1)= 45
∴ Product of factors of composite number N = ${\mathit{N}}^{\mathit{n}∕2}$
where, n = Number of factors of N
∴ Product of factors/divisors = ${\left(7056\right)}^{45∕2}$ = ${\left(84\right)}^{45}$