Aptitude - Numbers - Discussion

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.

Whenever there is a numerator in the form like (number) raised to something. Then Always try to bring denominator in the same form i.e. number (same as that of numerator) raised to something. So that the numbers can be easily cancelled out and our calculation becomes easier.

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.

12 * 12 * 12 * 6 * 6 * 6 * 6/2 * 2 * 2 * 2 * 3 * 3 * 3.

Then 12 * 12 * 12 * 3 = 5184.
It is very simple.

But According to BODMAS rule,

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

Somebody, please explain to me.

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.

@Indrajith.

They converted 432 in terms of multiples 12 & 6 so that it makes easy to cancel.

Here, what is the question?

I can't understand the question itself. Please explain me, anyone?

12*12*12*6*6*6*6/432
3*12*12*6*6*6*6/108
3*12*12*6*6*6/18
3*12*12*6*6/3
12*12*6*6=5184

@Kiran Kumar.

Your method is very helpful to find the right answer quickly. Thanks.