Aptitude - Numbers - Discussion

3^25 x (1 + 3 + 3^2 + 3^3) = 3^25 x 40==>3^25 * (1+3+9+27)==>3^25 X 40.

I didn't understand this. Please anyone help me.

Then Which numbers will completely divide (3^21+ 3^22+ 3^23))?

Thanks for explaining @Mohit Bansal.

(3^25 + 3^26 + 3^27 + 3^28) = 3^25 x (1 + 3 + 32 + 33) = 3^25 x 40.

= 3^24 x 3 x 40
= 3^24 x 120
= 120 is divisible by only option (D) 30.

I cannot understand, please explain in easy method.

Please explain in an easy method.

Please anyone explain in easy method.

Please explain in an easy method.

Guys anyone who didn't understand, please follow this;


3^25 + 3^26 + 3^27 + 3^28) = 3^25 x (1 + 3^1 + 3^2 + 3^3) = 3^25 x 40.

First, we take 3^25 as common to make it easy for finding the factor.

Suppose 2*4 =8, in this 2 is a factor of 8 right?

So like that we took common numbers and made it easy to find factors for this problem.

So now, we got (3^25 *40 = x) we don't know the x value but it is divisible by 40 exactly. Like that we took a 3^1 from 3 ^25 and we modified it like this (3^24*40*3=x) => (3^24*4*10*3=x) => (3^24*30*4=x).

Here x is the number we get when we multiply the 30 with the remaining value so 30 is the factor that divides x.

Why 30?

Because the 40 option isn't there and so we took 3 as common to make it 30 if we take 3^2 then it would be 90 which is also not there in the option so 30 is the only option which satisfies the condition.