# Math Help - Number of odd divisors

1. ## Number of odd divisors

Compute the number of odd divisors of 112,000

My book did an example similar to this, but it was very vague. I didn't really understand where it came up with the numbers.

I think this one is giving me a hard time because I am using the trial and error algorithm--if there is such a thing--for getting a prime factorization. Does anyone have any tricks for getting a prime factorization?

Thanks.

2. Well I don't know any short cut method to find prime factorization, but I guess you can look at the number in hand an at least figure out some of the prime factors(this is offcourse not possible always)
has got three tens.....
that is let 112,000 = n * 1000
now in 1000 how many odd prime factors you get?
3 odd prime factors, namely 5.... since 10 = 5 * 2.....

now in the residue, 112, this is obviously divisible by 2, well if you check for divisibility test of 3, it fails, obviously this is not divisible by 5(since 5/0 is not the last digit), try 7.... the next prime..... 7 * 16 = 112.....
now 16 is all having even prime factor namely 2....

therefore, the number of odd prime factor = three 5s + one 7 = 4 odd prime factor which is 1 less than number of total prime factors....

3. Originally Posted by natarajchakraborty
the number of odd prime factor = three 5s + one 7 = 4 odd prime factor which is 1 less than number of total prime factors....
Just to modify that answer slightly: you have found the prime factorisation of 112000 as $2^7\times5^3\times7$. So you can get odd divisors of the number by taking any product of the 5s and the 7. There will be seven such divisors, namely

$5,\ 5^2,\ 5^3,\ 7,\ 7\times5,\ 7\times5^2,\ 7\times5^3$.

4. Opalg... Ahha you are right... I misread the questions as number of odd prime factors, and so I said that its 1 less than the total! my mistake!

5. Originally Posted by Opalg
Just to modify that answer slightly: you have found the prime factorisation of 112000 as $2^7\times5^3\times7$. So you can get odd divisors of the number by taking any product of the 5s and the 7. There will be seven such divisors, namely

$5,\ 5^2,\ 5^3,\ 7,\ 7\times5,\ 7\times5^2,\ 7\times5^3$.
Great. So I'm assuming it's the same procedure for even divisors?

How would you works that for distinct divisors? Just use some sort of subtraction?

EDIT: Never mind, I figured it out.