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# x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that

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Math Expert
Joined: 02 Sep 2009
Posts: 47037
x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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08 May 2015, 05:43
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55% (hard)

Question Stats:

60% (01:05) correct 40% (01:12) wrong based on 401 sessions

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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?

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

Kudos for a correct solution.

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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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08 May 2015, 11:15
4
The expression can be rewritten as (3^2)^10 - 3^17.

Factoring the above we get: 3^17*[(3^3)-1)] = 3^17(26) = 3^17*13*2

The question asks us to find the number of prime factors (number with exactly two factors) of the expression.

From it's factored form we can find that only 2, 3 and 13 will satisfy the above conditions.

So 3?
Math Expert
Joined: 02 Sep 2009
Posts: 47037
Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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11 May 2015, 06:30
1
1
Bunuel wrote:
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.

MANHATTAN GMAT OFFICIAL SOLUTION:

Now let's look at answers you might give to the Polya questions:

What exactly is the problem asking for?
The number of possible values for n.
This means that n might have multiple possible values. In fact, it probably can take on more than one value.
I may not need these actual values. I just need to count them.

What are the quantities I care about?
I'm given x and n as variables. These are the quantities I care about.

What do I know?
x = 9^10 – 3^17. That is, x = a specific large integer, expressed in terms of powers of 9 and 3.

x/n is an integer. That is, x is divisible by n, or n is a factor of x.

n is a positive integer that has exactly two factors. This should make me think of prime numbers. Primes have exactly two factors. So I can rephrase the information: n is a positive prime number.

What don't I know?

Here's something I don't know: I don't know the value of x as a series of digits. Using a calculator or Excel, I could find out that x equals 3,357,644,238. But I don't know this number at the outset. Moreover, because this calculation is far too cumbersome, I must not need to find this number.

What is this problem testing me on?
From the foregoing, I can infer that this problem is testing us on divisibility and primes. We also need to manipulate exponents, since we see them in the expression for x.

You can ask these questions in whatever order is most helpful for the problem. For instance, you might not look at what the problem is asking for until you've understood the given information.

Now you should think about how you will solve the problem. Ask yourself a few more Polya questions to build your Plan:

Is a good approach already obvious?
From your answers above, you may already see a way to reach the answer. Don't try to work out all the details in your head ahead of time. If you can envision the rough outlines of the right path, then go ahead and get started.

If not, what in the problem can help me figure out a good approach?
If you are stuck, look for particular clues to tell you what to do next. Revisit your answers to the basic questions. What do those answers mean? Can you rephrase or
reword them? Can you combine two pieces of information in any way, or can you rephrase the question, given everything you know?

Can I remember a similar problem?
Try relating the problem to other problems you've faced. This can help you categorize the problem or recall a solution process.

For the Try-It problem, we have already rephrased some of the given information. We should go further now, combining information and simplifying the question.

Given: n is a prime number AND n is a factor of x
Combined: n is a prime factor of x
Question: How many different values for n are possible?
Combined: How many different values for n, a prime factor of x, are possible?
Rephrased: How many different prime factors does x have?

Thus, we need to find the prime factorization of x. Notice that n is not even in the question any more. The variable n just gave us a way to ask this underlying question.

Now, we look at the other given fact: x = 9^10 – 3^17. Earlier, we did not try to do very much with this piece of information. We just recognized that we were given a particular value for x, one involving powers. It can be helpful initially to put certain complicated facts to the side. Try to understand the kind of information you are given, but do not necessarily try to interpret it all of it right away.

At this stage, however, we know that we need the prime factors of x. So now we know what we need to do with 9^10 – 3^17. We have to factor this expression into primes.

This goal should motivate us to replace 9 (which is not prime) with 32 (which displays the prime factors of 9). We can now rewrite the equation for x:

x = (3^2)^10 – 3^17 = 3^20 – 3^17

Now, we still have not expressed x as a product of prime numbers. We need to pull out a common factor from both terms. The largest common factor is 317, so we write
x = 3^20 – 3^17 = 3^17(3^3 – 1) = 3^17(27 – 1) = 3^17(26) = 3^17(13)(2)

Now we have what we need: the prime factorization of x. We can see that x has three different prime factors: 2, 3, and 13.

The correct answer is C.
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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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12 May 2015, 06:34
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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Joined: 04 Jan 2015
Posts: 1755
Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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12 May 2015, 11:06
1
healthjunkie wrote:
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!

Regards
Harsh
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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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17 Dec 2015, 10:01
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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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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18 Dec 2015, 21:05
1
dajr1984 wrote:
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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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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29 Aug 2016, 01:20
Bunuel wrote:
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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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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29 Aug 2016, 04:43
Bunuel wrote:
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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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that [#permalink]

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01 Oct 2017, 12:16
1
Bunuel wrote:
$$x = 9^{10}– 3^{17}$$ 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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Re: x = 9^10– 3^17 and x/n is an integer. If n is a positive integer that   [#permalink] 01 Oct 2017, 12:16
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