Aptitude - Numbers - Discussion

First, try to get factors of 432 by factorization (2*2*2*2*3*3*3). Now try to get the same numbers as in numerators to cancel with denominator (i.e 12*36).

By cancelling it you will get the answer.

How 432 split in 12*6 (2)? Please explain.

Do HCF of 432.
Then we get, (2^4)*(3^3)
Which can be written as ( 2 * 3)*(2*3)*(2*2*3) or ( 6^2)*12.

If we know the concept of cyclicity we can find the last digits in each number given.

In 12^3 the end digit will be 8. In 6^4 end digit will be 6 and in 432 it's 2. So it will be 8*6/2 which will be 24. Thus end digit in the final answer should be 4.

Now there are two options with end digits as 4. So the next step is to find the digital sum. In the question the digital sum will be 1 + 2 + 6+ 4 + 3 + 2 = 18 further 1 + 8 is 9. So in whichever option ending with 4 along with digital sum as 9 is the right option. In option 1 the digital sum is 5 + 1 + 8 + 4 which is 18 and further it can be 1+8 that is 9. So the option with end digit 4 and digital sum 9 is option A.

12 * 12 * 12 * 6 * 6 * 6 * 6/27 * 2 * 2 * 2 * 2,
= 5184.

Anyone, please explain me the correct answer in detail.

But According to BODMAS rule,

6^4 ÷ 432 should be solved first, right?

Somebody, please explain to me.

Thanks for the answer @Kiran Kumar.

@Kiran.

Wha is cyclicity?

The correct answer is D.