Question

The number of divisors of 1080000 of the form 6m + 3 is:
(a) 20
(b) 105
(c) 90
(d) 15

Answer 1

Hint: First of all find the prime factorization of 1080000. Then equate those factors to 6m + 3. Now, take 3 as common form 6m + 3 and then we will get some relation between the prime factorization of 1080000 and 6m + 3. We know that, the number of divisors of x in ${{x}^{a}}{{y}^{b}}$ where x and y are the prime numbers is equal to $\left( a+1 \right)\left( b+1 \right)$. Similarly, find the number of the divisors of the form 6m + 3 of 1080000.

Complete step-by-step solution:
We are asked to find the number of divisors of 1080000 so we are going to write its prime factorization.
The prime factorization of 1080000 is equal to:
${{2}^{6}}\times {{3}^{3}}\times {{5}^{4}}$
Now, we want that number of divisors is in the form of 6m + 3. This means that the prime factorization of 1080000 is equal to 6m + 3 because we want those factors of 1080000 which are in the form of 6m + 3.
Equating 6m + 3 to the prime factorization of 1080000 we get,
${{2}^{6}}\times {{3}^{3}}\times {{5}^{4}}=6m+3$
Taking 3 as common from 6m + 3 in the above equation we get,
${{2}^{6}}\times {{3}^{3}}\times {{5}^{4}}=3\left( 2m+1 \right)$
One of the 3 will be cancelled out from both the sides and we get,
${{2}^{6}}\times {{3}^{2}}\times {{5}^{4}}=\left( 2m+1 \right)$
Now, to find the number of divisors in ${{2}^{6}}\times {{3}^{2}}\times {{5}^{4}}$ which are in the form of 2m + 1 is calculated as follows:
$\begin{align}
  & \left( 6+1 \right)\left( 2+1 \right)\left( 4+1 \right) \\
 & =7\left( 3 \right)\left( 5 \right) \\
 & =105 \\
\end{align}$
From the above, we got the number of divisors of the form 6m + 3 of 1080000 is equal to 105.
Hence, the correct option is (b).

Note: The point to be noted in using the formula of finding the number of divisors is that the numbers of whose powers you add with 1 individually and then multiplying them with each other should be prime numbers.
For e.g. in the above problem, we have calculated the number of divisors of ${{2}^{6}}\times {{3}^{2}}\times {{5}^{4}}$ of the form 2m + 1 as follows:
$\left( 6+1 \right)\left( 2+1 \right)\left( 4+1 \right)$
In the above expression, we took the power of 2 which is 6 in the first bracket, power of 3 which is 2 in the second bracket. Similarly, we have used 4 in the third bracket which is the power of 5. So, you can see that all the powers that we used are of the prime numbers (2, 3, 5).