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Can someone clarify the question stem? I was able to do that math easily and get to the fact that the x=3^17*13*2 , but I thought the question asked how many different values n there could possibly be, which I did not interpret to mean how many prime factors does X have?

I chose 4 because i figured these were all the different possible combination of n to make x/n an integer...
N=13*2
N=13*3
N=3*3
N=2*3
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healthjunkie
Can someone clarify the question stem? I was able to do that math easily and get to the fact that the x=3^17*13*2 , but I thought the question asked how many different values n there could possibly be, which I did not interpret to mean how many prime factors does X have?

I chose 4 because i figured these were all the different possible combination of n to make x/n an integer...
N=13*2
N=13*3
N=3*3
N=2*3

Hi healthjunkie,

The question tells us that n is a positive integer that has exactly two factors. We know that a number that has exactly two factors is a prime number. The factors of a prime number are 1 and the number itself. For example consider the prime number 5 which has two factors 1 & 5.

Further the question tells us that x/n is an integer i.e. x should have n as its factor. Since n is prime, to know if n is a factor of x we need to know the prime factors of x which you have rightly calculated to be 2, 3, and 13. As there are 3 prime factors of x, n can take only one of these values. Hence there are 3 possible values which n can take i.e 2 or 3 or 13.

Hope its clear!

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Harsh
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x = 9^10 – 3^17 and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

so I'm lost on how Manhattan Prep decided that we were looking for factors on N in this problem, rather than values of N.

so let me know if my exponents math is wrong or there is a rule preventing what im about to do.

9^10=(3^10)*2= 3^20 now we have a same base of 3, and the exponents can no be subtracted 20-17 to get 3 or 3^3 =X or X=81

now we know N has to be a prime number because is only has two factors 1 & itself, and X/N must be an integer or whole number.

well if X is 81 the only prime numbers that divide into 81 that create an integer are prime numbers 27 & 3. so why isn't the answer two values for N can be used?

the book says that the factors are 2, 3, and 13? Where in this question does it ever ask for factors of anything.

I believe the answer should be two values of N
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dajr1984
x = 9^10 – 3^17 and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

so I'm lost on how Manhattan Prep decided that we were looking for factors on N in this problem, rather than values of N.

so let me know if my exponents math is wrong or there is a rule preventing what im about to do.

9^10=(3^10)*2= 3^20 now we have a same base of 3, and the exponents can no be subtracted 20-17 to get 3 or 3^3 =X or X=81

now we know N has to be a prime number because is only has two factors 1 & itself, and X/N must be an integer or whole number.

well if X is 81 the only prime numbers that divide into 81 that create an integer are prime numbers 27 & 3. so why isn't the answer two values for N can be used?

the book says that the factors are 2, 3, and 13? Where in this question does it ever ask for factors of anything.

I believe the answer should be two values of N

This is where you are wrong.

\(X = 3^{20} - 3^{17}\) is not \(3^3\)

When bases are added or subtracted, the exponents cannot be added or subtracted. All you can do is take something common.
Only when bases are multiplied or divided, then you add or subtract the exponents.

\(3^5 / 3^2 = 3^{5 - 2}\)
But \(3^5 - 3^2\) is not \(3^{5-2}\)
\(3^5 - 3^2 = 3^2(3^3 - 1) = 9 * (27 -1) = 9 * 26\)

Similarly, here \(X = 3^{20} - 3^{17} = 3^{17} * (3^3 - 1) = 3^{17} * 26\)
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Bunuel
x = 9^10 – 3^17 and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five

Kudos for a correct solution.


Paraphrasing question:
1. \(\frac{x}{n}\) is an integer=> n is a factor of x
2. n is a positive integer that has exactly 2 factors => n is a prime

3. Combining statement 1 and 2, we get: n is a prime factor of x i.e. n=prime factors of x

4. how manydifferent values for n are possible? => How many prime factors of x are possible? (because n=prime factors of x.)

So, we are to find the prime factors of x.

Given,
x=\(9^{10} – 3^{17}\)
=\((3^2)^{10}-3^{17}\)
=\(3^{20}-3^{17}\)
=\(3^{17}[3^3-1]\)
=\(3^{17}[27-1]\)
=\(3^{17}. 26\)
=\(3^{17}. 13. 2\)

Hence, # of prime factors of x possible=3
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Bunuel
x = 9^10 – 3^17 and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five

Kudos for a correct solution.

9^10 - 3^17
3^20 - 3^17
3^17 [ (3^3)- 1)
3^17 (26)
3^17 * 2 * 13
Hence Option C
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Bunuel
\(x = 9^{10}– 3^{17}\) and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five

Since n has exactly two factors, n must be prime. Thus, we should prime factorize x:

x = (3^2)^10 - 3^17 = 3^20 - 3^17 = 3^17(3^3 - 1) = 3^17(26) = 3^17(2)(13)

Thus, n could be 3, 2, or 13.

Answer: C
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Case of:
-breaking the expression down to prime bases
- taking common the Exponential Term with the lowest valued Exponent
-and calculating how many unique prime factors make up the prime factorization of X

X = (9)^10 - (3)^17

(3)^20 - (3)^17

(3)^17 * ( 3^3 - 1)

thus we know 3 is a prime factor. We just need to evaluate the 2nd part.

(3)^3 - 1

27 - 1 = 26 = (13)(2)

3 unique prime factors can be N——-

3 - 2 - 13

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