Question

Find the number of divisors of 5400 which are divisible by 6 but not by 9.
A. 6
B. 9
C. 12
D. 18

Answer 1

Hint: We first try to find the prime factorisation of 5400. Then we break the given conditions of divisors being divisible by 6 but not by 9. We join them to find a single condition where it has to consist only one 3 and at least one 2. Finally, we find the number of combinations for the rest of the factors.

Complete step-by-step answer:
The divisors of 5400 will be created by taking a certain number of prime factors of the number at a time.
First, we try to find the prime factorisation of 5400.
\[\begin{align}
  & 2\left| \!{\underline {\,
  5400 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2700 \,}} \right. \\
 & 2\left| \!{\underline {\,
  1350 \,}} \right. \\
 & 3\left| \!{\underline {\,
  675 \,}} \right. \\
 & 3\left| \!{\underline {\,
  225 \,}} \right. \\
 & 3\left| \!{\underline {\,
  75 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1 \\
\end{align}\]
So, $5400={{2}^{3}}\times {{3}^{3}}\times {{5}^{2}}$.
We also have been given the condition that the divisors of 5400 have to be divisible by 6 but not by 9.
Now, the divisors to be the multiple of 6, it has to consist of at least one 2 and one 3 as 6 is made of $6=2\times 3$.
The divisors can’t be multiple of 9 which means it can’t have two or more than two 3s as $9=3\times 3$.
Taking both conditions in account we get that it has to consist only one 3 and at least one 2.
So, in $5400={{2}^{3}}\times {{3}^{3}}\times {{5}^{2}}$, the remaining factor parts are ${{2}^{2}}\times {{5}^{2}}$.
Now for every unique factor we have their power number plus one option. This plus one thing is for not choosing any number of those factors.
So, for 2 we have $2+1=3$ options and for 5 we have $2+1=3$ options.
So, the total number of combinations will be $3\times 3=9$.
The divisors will already have one 3 and one 2. They can have other factors from the choices of ${{2}^{2}}\times {{5}^{2}}$.

So, the correct answer is “Option B”.

Note: We need to remember that the conditions cannot be used later to eliminate different options from the total choices as that will be dependent on the choices of factors which can’t be in the divisor. We need to take the options of not taking any factors from ${{2}^{2}}\times {{5}^{2}}$ as in that we get 6 which is itself a choice for a divisor.