Question

The number of positive divisors of \[{2^5}{3^6}{7^3}\] is
a.14
b.167
c.168
d.210

Answer 1

Hint: We have a number factorized to the powers of their prime factors. We have to find the total number of positive divisors it has. We will check for each prime factor, how many divisors are coming and multiplying them altogether will give us our needed answer.

Complete step-by-step answer:
All the positive divisors will have a prime factorization composed of some subset of the prime factors of the number given.
We can have anything between zero to five 2′s in the divisor's prime factorization, that is 6 ways to take the prime factor 2, one extra for not taking 2 at all (2 raised to 0).
We can have anything between zero to six 3′s in the divisor's prime factorization, that is 7 ways to take the prime factor 3, one extra for not taking 3 at all (3 raised to 0).
We can have anything between zero to three 7′s in the divisor's prime factorization, that is 4 ways to take the prime factor 7, one extra for not taking 7 at all (7 raised to 0).
Using this logic we have now, 6 possibilities for 2, 7 possibilities for 3 and 4 possibilities for 7.
So, to find the total number of divisors we need to multiply them and find our desired result.
Then the total number of divisors of \[{2^5}{3^6}{7^3}\] is, \[6 \times 7 \times 4\]\[ = 168\]
Then we have our answer as, 168, which is option c.

Note: We can find the number of positive divisors of any given number.
Suppose you select 12. It has \[1,2,3,4,6,12\;\]as its divisors; so, the total number of divisors of 12 is 6.
Now the general method: \[x = {p^a}{q^b}\], where p and q are prime numbers. Now, x has \[(a + 1)(b + 1)\]positive divisors.