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# S98-18

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Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

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16 Sep 2014, 01:52
Expert's post
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Difficulty:

95% (hard)

Question Stats:

29% (02:05) correct 71% (02:51) wrong based on 90 sessions

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An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?

A. $$\frac{17}{300}$$
B. $$\frac{1}{15}$$
C. $$\frac{2}{25}$$
D. $$\frac{1}{10}$$
E. $$\frac{3}{25}$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128901 [0], given: 12183

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

### Show Tags

16 Sep 2014, 01:52
Expert's post
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This post was
BOOKMARKED
Official Solution:

An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?

A. $$\frac{17}{300}$$
B. $$\frac{1}{15}$$
C. $$\frac{2}{25}$$
D. $$\frac{1}{10}$$
E. $$\frac{3}{25}$$

First, make sure that you grasp the question. To find the desired probability, we need to count the integers between 1 and 300 that fit the given constraint. (We will then divide this count by 300, to determine the final answer.)

The constraint is worded in a confusing way, so we should attempt to reword it. If necessary, put in sample numbers to make the conditions make sense. We need an "integer raised to an exponent that is an integer greater than 1." In other words, we need an integer raised to the 2nd, 3rd, 4th, etc. power. In even other words, we need perfect squares, perfect cubes, etc. that are between 1 and 300, inclusive.

Now we need to count these perfect squares, etc. in an efficient way. Let's start with the squares. What's the biggest perfect square less than 300? Test numbers if necessary. $$16^2 = 256$$ and $$17^2 = 289$$, but $$18^2 = 324$$. Thus, we have 12 through 172, for 17 perfect squares.

Now, let's count the cubes. Leave out $$1^3$$, since we've already counted 1 as $$1^2$$. $$5^3 = 125$$ and $$6^3 = 216$$, but $$7^3 = 343$$. Thus, we seem to have 5 more integer powers ($$2^3$$ through $$6^3$$, inclusive) - but careful! $$4^3 = 64$$, which we've already counted as a square ($$8^2$$), so we only have 4 more integer powers. This gives us a cumulative total of 21.

What about the fourth powers - $$2^4$$, $$3^4$$, etc.? We've already counted any of these that matter, because they are also perfect squares: $$2^4 = 42$$, $$3^4 = 9^2$$, etc. We can leave out any higher even powers for the same reason ($$2^6 = 8^2$$, $$2^8 = 16^2$$, etc.).

However, we must consider additional odd powers, continuing to leave out 1 raised to any power. $$2^5 = 32$$, $$3^5 = 243$$, but $$4^5 =$$ something greater than 300, as we can see by considering that $$4^5 = 4^4 \times 4 = 16^2 \times 4 = 256 \times 4$$. So we have two more integer powers ($$2^5$$ and $$3^5$$), for a cumulative total of 23.

Seventh powers: $$2^7 = 128$$, but $$3^7 =$$ something much greater than 300 (since $$3^7 = 3^5 \times 3^2 = 243 \times 9$$). Be sure to stop calculating when you see that the result is outside the bounds of the problem.

We have 1 more power, for a cumulative total of 24.

Ninth powers: $$2^9 = 128 \times 4 =$$ greater than 300. So we can stop here.

Finally, we compute $$\frac{24}{300} = \frac{2}{25}$$.

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Kudos [?]: 128901 [0], given: 12183

Intern
Joined: 17 Jan 2015
Posts: 13

Kudos [?]: 4 [0], given: 0

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23 Mar 2015, 01:57
Any other method.....it is too long..?

Kudos [?]: 4 [0], given: 0

Manager
Joined: 11 Nov 2011
Posts: 74

Kudos [?]: 11 [0], given: 96

Location: United States
Concentration: Finance, Human Resources
GPA: 3.33
WE: Consulting (Non-Profit and Government)

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21 Jul 2015, 08:14
any other simpler/faster method?

Kudos [?]: 11 [0], given: 96

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

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21 Jul 2015, 08:20
evdo wrote:
any other simpler/faster method?

Check here: an-integer-between-1-and-300-inclusive-is-chosen-at-random-87439.html
_________________

Kudos [?]: 128901 [0], given: 12183

Manager
Joined: 20 Apr 2014
Posts: 116

Kudos [?]: [0], given: 23

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26 Jul 2016, 20:24
is it a real gmat question ?
it takes twoo much time.

Kudos [?]: [0], given: 23

Intern
Joined: 23 May 2016
Posts: 8

Kudos [?]: [0], given: 5

Location: India
GPA: 3.04
WE: Analyst (Retail)

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26 Dec 2016, 08:13
is there any shortcut method? Seems the only method to get this sorted out is working through all options. is it really he case with real GMAT questions?

Kudos [?]: [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

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26 Dec 2016, 10:48
vnitnagpur wrote:
is there any shortcut method? Seems the only method to get this sorted out is working through all options. is it really he case with real GMAT questions?

Check here: an-integer-between-1-and-300-inclusive-is-chosen-at-random-87439.html
_________________

Kudos [?]: 128901 [0], given: 12183

Intern
Joined: 20 Aug 2017
Posts: 2

Kudos [?]: 0 [0], given: 0

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15 Sep 2017, 02:45
can anyone please explain me this quant ?

Kudos [?]: 0 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

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15 Sep 2017, 02:48
Ajstyles wrote:
can anyone please explain me this quant ?

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Kudos [?]: 128901 [0], given: 12183

Re: S98-18   [#permalink] 15 Sep 2017, 02:48
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# S98-18

Moderator: Bunuel

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