Given that n has the smallest prime factors as its only prime factors
factors of n= $$2^a\cdot3^b\cdot5^c\left(\ \sin ce\ 1\ is\ not\ a\ prime\ factor\right)$$
How many positive integer divide n completely ?
This means we are looking for for total number of factors which n has and it can be denoted as $$\left(a+1\right)\left(b+1\right)\left(c+1\right)$$
Statement 1
The total number of times the prime factors of n occurs in n is
This means=
a+b+c=5
1+1+3 OR
2+1+2
$$Therefore\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(1+1\right)\left(1+1\right)\left(3+1\right)$$
2*2*4=16
OR
$$Therefore\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(2+1\right)\left(1+1\right)\left(2+1\right)$$
3*2*3=18
The target question cannot be answered with certainty
Statement 1 is NOT SUFFICIENT.
Statement 2
The product of the number of times each prime factors of n occurs in n is if
This means that a*b*c=4
The only possible values of this product is 1,1,4 0r 2,1,2
$$Therefore\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(1+1\right)\left(1+1\right)\left(4+1\right)=2\cdot2\cdot5$$ $$Therefore\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(2+1\right)\left(1+1\right)\left(2+1\right)=3\cdot2\cdot3=18$$
The target question cannot be answered with certainty
statement 2 is NOT SUFFICIENT.
Combining statement 1 and 2 together
a+b+c=5 and a*b*c=4
The only possible value that satisfies the condition in the two conditions in the two statements 1,2,2 or 2,1,2 or 2,2,1 $$Therefore\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(1+1\right)\left(2+1\right)\left(2+1\right)=2\cdot3\cdot3=18$$ $$OR\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(2+1\right)\left(1+1\right)\left(2+1\right)=3\cdot2\cdot3=18$$ $$OR\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(2+1\right)\left(2+1\right)\left(1+1\right)=3\cdot3\cdot2=18$$
A
Total number of factors will always = 18
Both statements together are SUFFICIENT.
$$Answer\ is\ Option\ C$$
