If n = 3^8 - 2^8, which of the following is NOT a factor of

Ans:

we will apply the formula a^2-b^2 here and expand the equation . in the end we get (3-2)(3+2)(3^2+2^2)(3^4+2^4)= 1x5x13x97 , therefore the answer is (C).

\(3^8 - 2^8\)

\(= 81^2 - 16^2\)

= 97 * 65

35 is not a factor; all other options stand fit

Answer = C

Whenever you see anything of the form \(a^2 - b^2\), write it as (a-b)*(a+b)

Coming to the question at hand,
n = \(3^8 - 2^8\) =\((3^4 - 2^4) * (3^4 + 2^4)\)

Again applying the same rule on \((3^4 - 2^4)\) ,
n = \((3^2 - 2^2)*(3^2 + 2^2)*(3^4 + 2^4)\) = \((3 - 2)*(3+2)*(3^2 + 2^2)*(3^4 + 2^4)\)
n = 1*5*(9+4)*(81+16) = 1*5*13*97

On checking the options, we see that 35 cannot be formed by the factors of n, hence the correct answer

Option C

We would never be asked to calculate 3^8 or 2^8, so we must approach this problem not as an arithmetic question but as an algebraic one.

The first thing we must recognize is that we are being tested on the algebraic factoring technique called the "difference of squares." Recall that the general form of the difference of squares is:

x^2 – y^2 = (x + y)(x – y)

Similarly, we can treat 3^8 – 2^8 as a difference of squares, which can be expressed as:

n = (3^4 + 2^4)(3^4 – 2^4)

We can further factor 3^4 – 2^4 as an additional difference of squares, which can be expressed as:

(3^2 + 2^2)(3^2 - 2^2)

This finally gives us:

n = 3^8 - 2^8 = (3^4 + 2^4)(3^2 + 2^2)(3^2 - 2^2)

The numbers are now easy to calculate:

n = (81 + 16)(9 + 4)(9 – 4)

n = (97)(13)(5)

We are being asked which of the answer choices is NOT a factor of n, which we have determined to be equal to the product (97)(13)(5). So we must find the answer choice that does not evenly divide into (97)(13)(5).

Right away we see that 97, 13 and 5 are all factors of (97)(13)(5).

This leaves us with 65 and 35. We should notice that (97)(13)(5) = (97)(65). Thus, 65 also is a factor of n. Only 35 is not.

Answer is C.

Responding to a pm:
Check here for the answer: https://gmatclub.com/forum/if-n-3-8-2-8 ... l#p1154050

What I have given above is the way to eliminate 3 options. That gives you a 50-50 probability of getting the answer (up from 20%). To get to the answer, we need to evaluate n (as done by Bunuel above).

\(3^8\) can be rewritten as \(3^{(4(2))}\) or \((3^4 )^2\). Similarly, \(2^8\) can be rewritten as \(2^{4(2)}\) or \((2^4 )^2\).

Since both are perfect squares, we can apply the property when working with the difference of two squares:

\(a^2-b^2=(a+b)(a-b)\)

Hence,

\(3^8-2^8=(3^4 )^2-(2^4 )^2\)
\(=(3^4+2^4 )(3^4-2^4 )\)
\(=(81+16)(81-16)\)

We have \(3^8-2^8=97 \times 65= 97 \times 5 \times 13\). This shows that \(5\), \(13\), and \(97\) are factors of the total.

In addition, since \(7\) is not a factor of \(3^8-2^8\), \(35\) (which is equal to \(5 \times 7\)) can’t be a factor as well.

The final answer is .

A quick way to solve this it to first recognize that \(3^8 - 2^8\)is a difference of squares, which can be factored.

So, \(n = 3^8 - 2^8\)

\(= (3^4 + 2^4)(3^4 - 2^4)\)

\(= (3^4 + 2^4)(3^2 + 2^2)(3^2 - 2^2)\)

\(= (3^4 + 2^4)(3^2 + 2^2)(3 + 2)(3 - 2)\)

\(= (97)(13)(5)(1)\)

At this point, we can see that 97, 65 (aka 13 x 5), 13 and 5 are all factors of n

Answer: C

Cheers,
Brent

\(n = 3^8 - 2^8\)
\(= (3^4 + 2^4)*(3^4 - 2^4)\)
\(= (3^4 + 2^4)(3^2 + 2^2)(3^2 - 2^2)\)
\(= (3^4 + 2^4)(3^2 + 2^2)(3 + 2)(3 - 2)\)
\(= 97*13*5*1\)
35 is not a factor of n since there is no 7 in n.

Answer (C)

Easiest way is to convert the equation to the formula a^2-b^2

3^8-2^8
=9^4-4^4
=81^2-16^2
=(81-16)(81+16)
=65*97

Therefore, only option C that is 7*5 remains because 7 is neither a factor of 97 nor of 65.



Kudos for some appreciation please, if you like my explanation. I also welcome Critical analysis of my post that will help me reach 700+ level.

Posted from my mobile device

Easiest way is to convert the equation to the formula a^2-b^2

3^8-2^8
=9^4-4^4
=81^2-16^2
=(81-16)(81+16)
=65*97

Therefore, only option C that is 7*5 remains because 7 is neither a factor of 97 nor of 65.



Kudos for some appreciation please, if you like my explanation. I also welcome Critical analysis of my post that will help me reach 700+ level.

Posted from my mobile device

Hi.

97 can be broken down into 7*13, 65 can be broken down into 5*13. When we multiply 65*97, we can get the following expression 35*13^2. Could you please explain why 35 cannot be a divisor?

Thank you.

97 ≠ 7*13
97 is prime

Hi All,

We’re told that N = 3^8 – 2^8. We’re asked which of the follow 5 numbers is NOT a factor of N. The GMAT would NEVER require that you calculate that overall value, so there must be a way to ‘simplify’ that equation. Since it includes Exponents and SUBTRACTION of two values raised to the SAME EVEN power, we should be on the lookout for Classic Quadratics…

You’re probably familiar with X^2 – Y^2 (since that is one of the common “Classic” Quadratics… X^2 – Y^2 = (X + Y)(X – Y)). Similar Quadratics exist for X^4 – Y^4, X^6 – Y^6 and X^8 – Y^8. Quadratic rules apply whether there are variables or numbers involved, so we can replace the X and Y with the “3” and “2”, respectively in the given calculation…

3^8 – 2^8 =
(3^4 + 2^4)(3^4 – 2^4)

We can then ‘factor down’ the 2nd part of that step…

(3^4 + 2^4)(3^4 – 2^4) =
(3^4 + 2^4)(3^2 + 2^2)(3^2 – 2^2)

And then ‘factor down’ the 3rd part of that step…

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

This gives us…
(81+16)(9+4)(5)(1)
(97)(13)(5)(1)

With these results, we can clearly eliminate Answers A, D and E. By multiplying the 13 and 5, we get (13)(5) = 65, so that is ALSO a factor – and we can eliminate Answer B.

Final Answer:

GMAT Assassins aren’t born, they’re made,
Rich

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