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Question

Find the number of divisors of number 239000?

Answer 1

First, obtain the prime factorization.

1000 x 239 =
10 x 100 x 239 =
2 x 5 x 10 x 10 x 239
2 x 5 x 2 x 5 x 2 x 5 x 239

Let's order this from least to greatest.

2 x 2 x 2 x 5 x 5 x 5 x 239

Now, let's write this as an expression of powers:

(2^3) (5^3) (239)

To find the number of divisors, all you have to do is take each power, add 1, and multiply the numbers out. In this case, we have a power of 3, a power of 3, and a power of 1. These numbers get multiplied out:

(3 + 1) (3 + 1) (1 + 1)
You take the powers, add one to each of them, then multiply.

The number of divisors should then be
(4)(4)(2) = 32

There's actually a function out there specifically for the number of divisors of a number n, denoted σ(n), where

σ(p^n) = n + 1 (for p a prime number).
More generally, if n = p^a * q^b
σ(n) = σ( [p^a][q^b]) = σ(p^a)σ(q^b) = (a + 1)(b + 1)
{In other words, the function σ works such that the σ of a product is the product of the σ's}

In the above case,

σ(239000) = σ[(2^3) (5^3) (239^1)] =
σ(2^3)σ(5^3)σ(239^1) = (3 + 1)(3 + 1)(1 + 1) = (4)(4)(2) = 32

Answer 2

Since 239 is prime, a divisor of 239000 is of the form p or 239 p where p is a divisor of 1000. that is 1 2 4 o 8 times 1 5 25 or 125. So you have 2 times 4 times 4 divisors, that is 32.

Answer 3

5,25,125,250,500,4,20,100,4780,9560,1912,926,463,1,2,10,1000
then the number of divisors are (17) numbers

Answer 4

239000=2^3*29875=2^3*5^3*239