Aptitude - Numbers - Discussion

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.

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.

3^25+3^26+3^27+3^28.
Here if we take a common among these, we get,

3^25*(1+3+3^2+3^3).

Now if we solve the bracket part which can be done easily, we get,
3^25 * 40.

Now factoring in the term 40, and 3^24 * 3=3^25 do not get confused, the equation looks likes,
3^24 * 3 * 4 * 10,
3^24 * 4 * 30.

Here the term 30 will completely divide the equation.

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.

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.

(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 3 x 4 x 10.
Sum = 120 which is divisible by 30.
So, option D is correct.

(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.

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

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

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

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