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# If, for all integers n ≥ 1, x_n is the probability of selecting at ran

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Math Expert
Joined: 02 Sep 2009
Posts: 50001
If, for all integers n ≥ 1, x_n is the probability of selecting at ran  [#permalink]

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14 Sep 2015, 05:21
1
15
00:00

Difficulty:

95% (hard)

Question Stats:

37% (02:53) correct 63% (02:31) wrong based on 156 sessions

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If, for all integers n ≥ 1, $$x_n$$ is the probability of selecting at random an integer with an odd number of unique positive factors from among the positive integers less than or equal to n^2, then the sum of the first 5 values of $$x_n$$ is

(A) less than 1
(B) between 1 and 2
(C) between 2 and 3
(D) between 3 and 4
(E) greater than 4

Kudos for a correct solution.

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If, for all integers n ≥ 1, x_n is the probability of selecting at ran  [#permalink]

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14 Sep 2015, 05:50
2
7
Bunuel wrote:
If, for all integers n ≥ 1, $$x_n$$ is the probability of selecting at random an integer with an odd number of unique positive factors from among the positive integers less than or equal to n^2, then the sum of the first 5 values of $$x_n$$ is

(A) less than 1
(B) between 1 and 2
(C) between 2 and 3
(D) between 3 and 4
(E) greater than 4

Kudos for a correct solution.

CONCEPT: Only Perfect squares i.e. 1, 4, 9, 16, etc. have Odd Number of factors

Also from 1 to n^2 there will always be n perfect square
Hence probability of picking a Perfect square from 1 to n^2 = n/n^2 = 1/n

$$x_1 = 1/1$$
$$x_2 = 2/4 = 1/2$$ (Probability of selected Number to have odd number of factors from 1 to 2^2)
$$x_3 = 3/9 = 1/3$$ (Probability of selected Number to have odd number of factors from 1 to 3^2)
$$x_4 = 4/16 = 1/4$$ (Probability of selected Number to have odd number of factors from 1 to 4^2)
$$x_5 = 5/25 = 1/5$$ (Probability of selected Number to have odd number of factors from 1 to 5^2)

Sum of all $$x_n$$ = (1) + (1/2) + (1/3) + (1/4) + (1/5) = (60+30+20+15+12)/60 = 137/60 = 2.28

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##### General Discussion
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Joined: 17 Mar 2014
Posts: 5
Re: If, for all integers n ≥ 1, x_n is the probability of selecting at ran  [#permalink]

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18 Sep 2015, 18:17
1
x1 : 1/1
x2 : 1,4 of 1,2,3,4 ----2/4
x3 : 1,4,9 of 1,2...9 ----3/9
x4 : 1,4,9,16----------4/16
x5 : 1,4,9,16,25--------5/25

Total 1+2/4+3/9+4/16+5/25 = 2.3

C is the right answer
Math Expert
Joined: 02 Sep 2009
Posts: 50001
If, for all integers n ≥ 1, x_n is the probability of selecting at ran  [#permalink]

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20 Sep 2015, 21:04
Bunuel wrote:
If, for all integers n ≥ 1, $$x_n$$ is the probability of selecting at random an integer with an odd number of unique positive factors from among the positive integers less than or equal to n^2, then the sum of the first 5 values of $$x_n$$ is

(A) less than 1
(B) between 1 and 2
(C) between 2 and 3
(D) between 3 and 4
(E) greater than 4

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

According to the question, $$x_n$$ is “the probability of selecting at random an integer with an odd number of unique positive factors from among the positive integers less than or equal to n^2.” That definition contains a lot of words. Let’s simplify by focusing on just one value: $$x_1$$.

Plug in 1 in place of n. Now, because we are examining a complicated probability scenario, we should ask ourselves two questions.

(a) What is the pool we are picking from?
We are picking from among the positive integers less than or equal to 1^2—in other words, we are picking from this group: {1}.

(b) What would success be?
Success would be picking an integer “with an odd number of unique positive factors.” Well, 1 has just one “unique positive factor” (namely, 1 itself), so 1 is a successful pick.

Thus, the probability of picking 1 out of the set {1} is 100%, or 1. Now we know that $$x_1 = 1$$. (By the way, since none of the remaining probabilities can be negative, we can rule out (A) at this point.)

What about $$x_2$$? Let’s start making a table and looking for the pattern. What will help at this point is to decipher one bit of code: integers with “an odd number of unique positive factors” are perfect squares. Simply put, perfect squares always have a single unpaired factor – the square root. For instance, 1 has just 1 factor. 4 has 3 factors (1, 2, and 4), as does 9. 16 has 5 factors (1, 2, 4, 8, and 16, where 1 and 16 form a factor pair, 2 and 8 form a factor pair, and 4 is the odd man out).

So we are really looking for the probability of picking a perfect square out of the positive integers less than or equal to n^2.

So we are really just adding up 1 + 1/2 + 1/3 + 1/4 + 1/5. We don’t need an exact number—we just need to know what integers that sum falls between. At this point, we might be strategic and grab the middle fractions. 1/2 + 1/3 + 1/4 is just a little more than 1 (since 1/2 + 1/4 + 1/4 would be exactly 1). Adding 1/5 would not get us very far toward 2; this sum would be between 1 and 2. (If we compute 1/2 + 1/3 + 1/4 + 1/5 exactly, we get 30/60 + 20/60 + 15/60 + 12/60 = 77/60, which is definitely between 1 and 2.)

Finally, don’t forget to add in $$x_1 = 1$$. Thus, the sum we want (1 + 1/2 + 1/3 + 1/4 + 1/5) is between 2 and 3.

This problem combines several different topics—divisibility, probability, and fractions. You’ll encounter such “hybrid” problems at the harder end of the GMAT. Not every hard problem is a hybrid, but hybrids tend to be hard, because they force you to solve more than one unrelated problem quickly and correctly.

The correct answer is C.

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Re: If, for all integers n ≥ 1, x_n is the probability of selecting at ran  [#permalink]

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18 Apr 2018, 09:09
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# If, for all integers n ≥ 1, x_n is the probability of selecting at ran

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