This is from OG 13 #117 in Problem Solving:
If N = 3^8  2^8, which of the following is NOT a factor of n?
A) 97
B) 65
C) 35
D) 13
E) 5
OA: C
The book posts a pretty obscure way to solve, would like to hear others opinions on this question.
Arithmetic  Properties of Numbers: If n = ...
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 neelgandham
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If N = 3^8  2^8, which of the following is NOT a factor of n?
3^8  2^8 = (3^42^4)(3^4+2^4)  a^2 b^2 = (a+b)*(ab)
3^8  2^8 = (3^22^2)(3^2+2^2)(3^4+2^4)  a^2 b^2 = (a+b)*(ab)
3^8  2^8 = (32)(3+2)(3^2+2^2)(3^4+2^4)
3^8  2^8 = (1)(5)(13)(97), so 97,65(13*5),13,5 are factors of 3^8  2^8. The only number left is 35, which isn't a factor of 3^8  2^8
3^8  2^8 = (3^42^4)(3^4+2^4)  a^2 b^2 = (a+b)*(ab)
3^8  2^8 = (3^22^2)(3^2+2^2)(3^4+2^4)  a^2 b^2 = (a+b)*(ab)
3^8  2^8 = (32)(3+2)(3^2+2^2)(3^4+2^4)
3^8  2^8 = (1)(5)(13)(97), so 97,65(13*5),13,5 are factors of 3^8  2^8. The only number left is 35, which isn't a factor of 3^8  2^8
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3^8  2^8=(3^42^4)(3^4+2^4)=
=(8116)(81+16)=65*97 (a and b are out)
65*97=5*13*97 (d and E are out)
answ is C
=(8116)(81+16)=65*97 (a and b are out)
65*97=5*13*97 (d and E are out)
answ is C
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Solution:wied81 wrote:This is from OG 13 #117 in Problem Solving:
If N = 3^8  2^8, which of the following is NOT a factor of n?
A) 97
B) 65
C) 35
D) 13
E) 5
OA: C
The book posts a pretty obscure way to solve, would like to hear others opinions on this question.
It is very unlikely that a problem would require us to calculate 3^8 or 2^8, so we should approach this problem not as an arithmetic question but as an algebraic one.
The first thing we should 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).
We immediately see that 97, 13 and 5 are all factors of (97)(13)(5).
This leaves us with 65 and 35. Notice that (97)(13)(5) = (97)(65). Thus, 65 also is a factor of n. Only 35 is not.
Answer: C
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I want to add something about the concept of algebraic manipulations without variables:
The GMAT loves to test our algebra skills (like factoring), HOWEVER many students associate algebra with variables only, so they don't see that we can apply algebraic principles to numbers as well (which should make sense, since those variables are, indeed, representing numbers).
So, for example, many students are fine with the following:
 Factoring 6x + 3 to get 3(2x + 1)
 Factoring 6xâ�µ + 2xÂ³ + 8xÂ² to get 2xÂ²(3xÂ³ + x + 4)
 Factoring xÂ² + 5x + 6 to get (x + 2)(x + 3)
 Factoring xÂ²  yÂ² to get (x + y)(x  y)
On the GMAT, we need to recognize that we can also factor expressions that have no variables.
So, for example, a GMAT question might ask us to evaluate 54Â²  53Â²
If we recognize that this is a difference of squares in the form xÂ²  yÂ², we can factor it to get:
54Â²  53Â² = (54 + 53)(54  53) = (107)(1) = 107
Likewise, the expression 2Â¹â�°â�°  2â�¹â�¶ is no different from xÂ¹â�°â�°  xâ�¹â�¶
xÂ¹â�°â�°  xâ�¹â�¶ = xâ�¹â�¶(xâ�´  1) in the exact same way that 2Â¹â�°â�°  2â�¹â�¶ = 2â�¹â�¶(2â�´  1)
Cheers,
Brent
The GMAT loves to test our algebra skills (like factoring), HOWEVER many students associate algebra with variables only, so they don't see that we can apply algebraic principles to numbers as well (which should make sense, since those variables are, indeed, representing numbers).
So, for example, many students are fine with the following:
 Factoring 6x + 3 to get 3(2x + 1)
 Factoring 6xâ�µ + 2xÂ³ + 8xÂ² to get 2xÂ²(3xÂ³ + x + 4)
 Factoring xÂ² + 5x + 6 to get (x + 2)(x + 3)
 Factoring xÂ²  yÂ² to get (x + y)(x  y)
On the GMAT, we need to recognize that we can also factor expressions that have no variables.
So, for example, a GMAT question might ask us to evaluate 54Â²  53Â²
If we recognize that this is a difference of squares in the form xÂ²  yÂ², we can factor it to get:
54Â²  53Â² = (54 + 53)(54  53) = (107)(1) = 107
Likewise, the expression 2Â¹â�°â�°  2â�¹â�¶ is no different from xÂ¹â�°â�°  xâ�¹â�¶
xÂ¹â�°â�°  xâ�¹â�¶ = xâ�¹â�¶(xâ�´  1) in the exact same way that 2Â¹â�°â�°  2â�¹â�¶ = 2â�¹â�¶(2â�´  1)
Cheers,
Brent

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Here's an approach that doesn't involve difference of squares.
If a number divides by 65, it must divide by 13 and by 5. So D and E *CANNOT* be the answers: if the number is not divisible by 5, it is also not divisible by 65. (Same logic for 13 and 65.) Now that we know the answer divides by 5 and 13, it must also divide by 65, so B CANNOT be the answer.
Now we'll consider A and C. Since we know our number divides by 5, it will divide by 35 if and only if our number also divides by 7. (This is because 35 = 5 * 7).
At this point, we recall that 3â�¸  2â�¸ will divide by 7 if 3â�¸ and 2â�¸ have the same remainder when divided by 7. 2â�¸ = 256, which has a remainder of 4 when divided by 7. 3â�¸ = 6561, which has a remainder of 2.
So our number doesn't divide by 7, and hence can't divide by 35. We're done, and we don't need to bother testing A.
If a number divides by 65, it must divide by 13 and by 5. So D and E *CANNOT* be the answers: if the number is not divisible by 5, it is also not divisible by 65. (Same logic for 13 and 65.) Now that we know the answer divides by 5 and 13, it must also divide by 65, so B CANNOT be the answer.
Now we'll consider A and C. Since we know our number divides by 5, it will divide by 35 if and only if our number also divides by 7. (This is because 35 = 5 * 7).
At this point, we recall that 3â�¸  2â�¸ will divide by 7 if 3â�¸ and 2â�¸ have the same remainder when divided by 7. 2â�¸ = 256, which has a remainder of 4 when divided by 7. 3â�¸ = 6561, which has a remainder of 2.
So our number doesn't divide by 7, and hence can't divide by 35. We're done, and we don't need to bother testing A.

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Factoring Formula  xÂ²  yÂ² to get (x + y)(x  y) is the key to solve this question.
35 is the answer.
35 is the answer.
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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: C
GMAT Assassins aren’t born, they’re made,
Rich
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: C
GMAT Assassins aren’t born, they’re made,
Rich