Aptitude - Numbers - Discussion

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

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

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

Just we need to adjust the factors with the above-given options.

Eg. 3^24*3*4*10.
They separated the 3 from 3^25 to get 3^24.
40 is written as 10.
Multiply this 3 with 10 to get 30 as it is given in the options. This is just checking the factors with the trial method using given options.

Please tell me about sum in brief?

If number is 3 power gives:

Power 1 = 3 (unit place).
Power 2 = 9.
Power 3 = 7.
Power 4 = 1.
Power 5 = 3.

Again power 1, 2, 3, 4, 5....answer so on.

So 24 comes at power 1 place.

So solving question.

=> 3 power 25 + 3 power 26 + 3 power 27 + 3 power 28.

=> 3 power 24 common (3 + 9 + 27 + 81).

=> 3 power 24 common (120).

=> 3 power 24 is 1 as concluded earlier so divide 120/ the given numbers in options and you will see 30 divide absolutely to zero remainder.

How n is taken as 4?

= n(n+1)(2n+1)/6 where n=4.

= 4(4+1)(2*4+1)/6.
= 4(5)(9)/6.
= 180/6 = 30.

Why don't we take 40 tell me please?

First we try to find factor of the number in terms of number given in option,

(3^25 + 3^26 + 3^27 + 3^28) = 325 x (1 + 3 + 32 + 33) = 325 x 40.

= 324 x 3 x 4 x 10.

= (324 x 4 x 30).

Here 30 is a factor of given number.

And every number must be divisible by its factor.

So correct answer is 30.