Divisors of $75600$ of the type of $4n+2$

Question

Find the total no. of divisors of $75600$ which of the type of $4n+2$ where $n\in \mathbb{N}$ and $75600=2^4 \cdot 3^3 \cdot 5^2 \cdot 7^1$

Now I think divisors of type $(4n+2)$ should be of type $2^1\cdot 3^a\cdot 5^b\cdot 7^c$. Am I correct?

Yes, you are right. Note that one or more of $a,b,c$ can be $0$. — André Nicolas, Feb 07 '16 at 18:10
@juantheron Don't know. But why not $24 (432)$? Which case are you eliminating? — Ananya, Feb 07 '16 at 18:19
Bcz you have mention that $n\in \mathbb{N}$. So I eliminate the case when all are $0$ — juantheron, Feb 07 '16 at 18:24
Do you include $0 \in \mathbb{N}$? The answer depends on the definition of $\mathbb{N}$ that you are using. — N. F. Taussig, Feb 07 '16 at 19:29
@juantheron: I would tend to guess that the author of this question intended to count all non-negative integer divisors, including $2$ (generated with $n=0$). Check out the answers to Is $0$ a natural number?. — barak manos, Feb 07 '16 at 19:31

score 0 · Answer 1 · answered Feb 07 '16 at 18:22

I think you're correct and that's the complete solution. (4n+2)=2(2n+1).

So, the prime factorization of every divisor can have only $2^1$ and not more. Apart from that, the combination of any power or 3,5,7 would be an odd number, and any odd number can be expressed as (2n+1).

Possible powers of 3 can be from 0 to 3 = 4. Possible powers of 5 can be from 0 to 2 = 3. Possible powers of 7 can be from 0 to 1 = 2.

So, the total possibilities = (4).(3).(2) = 24.

Thus, there are 24 such divisors.

Hope it helps :)

score 0 · Answer 2 · answered Feb 07 '16 at 18:47

First of all, you are right.

Therefore, you need to count the number of divisors of $3^3\cdot5^2\cdot7^1$:

In each divisor, the factor $3$ can appear between $0$ and $3$ times, i.e., $4$ different combinations
In each divisor, the factor $5$ can appear between $0$ and $2$ times, i.e., $3$ different combinations
In each divisor, the factor $7$ can appear between $0$ and $1$ times, i.e., $2$ different combinations

Hence there are $4\cdot3\cdot2=24$ divisors:

$3^0\cdot5^0\cdot7^0$
$3^1\cdot5^1\cdot7^1$
$3^2\cdot5^2\cdot7^0$
$3^3\cdot5^0\cdot7^1$
$3^0\cdot5^1\cdot7^0$
$3^1\cdot5^2\cdot7^1$
$3^2\cdot5^0\cdot7^0$
$3^3\cdot5^1\cdot7^1$
$3^0\cdot5^2\cdot7^0$
$3^1\cdot5^0\cdot7^1$
$3^2\cdot5^1\cdot7^0$
$3^3\cdot5^2\cdot7^1$
$3^0\cdot5^0\cdot7^0$
$3^1\cdot5^1\cdot7^1$
$3^2\cdot5^2\cdot7^0$
$3^3\cdot5^0\cdot7^1$
$3^0\cdot5^1\cdot7^0$
$3^1\cdot5^2\cdot7^1$
$3^2\cdot5^0\cdot7^0$
$3^3\cdot5^1\cdot7^1$
$3^0\cdot5^2\cdot7^0$
$3^1\cdot5^0\cdot7^1$
$3^2\cdot5^1\cdot7^0$
$3^3\cdot5^2\cdot7^1$

Each of those divisors should be multiplied by $2^1$ of course.

Divisors of $75600$ of the type of $4n+2$

2 Answers2