7^12 - 4^12 is exactly divisible by

(A)33

(B)34

(C)35

(D)36

Explain how?

This question was asked in SSC Exams 2011.

Obviously you don't have to use calculator or anything similar. It's an aptitude test question. You have to answer using skill nad trick.

### 2 Answers | Add Yours

If m is a even number we know that;

`a^m-b^m = (a^(m/2)-b^(m/2))(a^(m/2)+b^(m/2))`

`7^12-4^12`

`= (7^6-4^6)(7^6+4^6)`

`7^6-4^6`

`= (7^3-4^3)(7^3+4^3)`

`7^12-4^12`

`= (7^3-4^3)(7^3+4^3)(7^6+4^6)`

`a^3-b^3 = (a-b)(a^2+ab+b^2)`

`a^3+b^3 = (a+b)(a^2-ab+b^2)`

`7^3-4^3 = (7-4)(7^2+7*4+4^2) = 3*(7^2+7*4+4^2)`

`7^3+4^3 = (7+4)(7^2-7*4+4^2) = 11*(7^2-7*4+4^2)`

`7^12-4^12`

`= (7^3-4^3)(7^3+4^3)(7^6+4^6)`

`= 3*(7^2+7*4+4^2)*11*(7^2-7*4+4^2)(7^6+4^6)`

`= 33*(7^2+7*4+4^2)(7^2-7*4+4^2)(7^6+4^6)`

*So it is divisible by 33.*

**Sources:**

You should notice that the exponent 12 is a multiple of two, hence, you may write the given difference such that:

`7^12 - 4^12 = 7^(6*2) - 4^(6*2)`

Using the exponential identity `a^(b*c) = (a^b)^c` yields:

`7^12 - 4^12 = (7^6)^2 - (4^6)^2`

You should convert the difference of squares into a product such that:

`(7^6)^2 - (4^6)^2 = (7^6 - 4^6)(7^6 + 4^6)`

`7^6 - 4^6 = (7^3)^2 - (4^3)^2`

Converting the difference of squares into a product yields:

`(7^3)^2 - (4^3)^2 = (7^3 - 4^3)(7^3 + 4^3)`

You need to use the following formula, `a^3-b^3 = (a-b)(a^2+ab+b^2),` to convert the difference of cubes into a product such that:

`7^3 - 4^3 = (7-4)(49+28+16)`

`7^3 + 4^3 = (7+4)(49-28+16)`

`(7^6)^2 - (4^6)^2 = (((7-4)(49+28+16))(7^3 + 4^3))(7^6 + 4^6)`

`(7^6)^2 - (4^6)^2 = 3*93*11*37*(7^6 + 4^6)`

`7^12 - 4^12 = 33*37*93*(7^6 + 4^6)`

**Hence, evaluating the given difference of powers, you may notice that it is exactly divisible by 33, thus, you should select the answer A)33.**

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