Number Systems -Sequence and Series

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Number Systems -Sequence and Series

by sukhman » Thu Oct 17, 2013 10:17 am
A certain club has exactly 5 new members at the end of its first week. Every subsequent week,each of the previous week's new members (and only these members) brings exactly x new members into the club.If y is the number of new members brought into the club during the twelfth week, which of the following could be y?
(A) 5^1/12 (B) 3^11 . 5 ^11 (C) 3^12 . 5^12 (D) 3^11 . 5^12 (E)60^12
Source: — Data Sufficiency |

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by Mike@Magoosh » Thu Oct 17, 2013 2:21 pm
sukhman wrote:A certain club has exactly 5 new members at the end of its first week. Every subsequent week,each of the previous week's new members (and only these members) brings exactly x new members into the club.If y is the number of new members brought into the club during the twelfth week, which of the following could be y?
(A) 5^1/12 (B) 3^11 . 5 ^11 (C) 3^12 . 5^12 (D) 3^11 . 5^12 (E)60^12
Dear sukman,
Why this is posted in DS is a little peculiar. This is a PS question. I am happy to help. :-)

I think the easiest way to proceed would be to start listing the numerical results of what happens and see if a pattern emerges.

End of week #1 = 5 new members
Each one of those 5 members brings x new members, so
End of week #2 = 5x new members
Now, each one of those 5x members brings x new members, so
End of week #3 = 5*x^2 new members
End of week #4 = 5*x^3 new members
End of week #5 = 5*x^4 new members
There's our pattern. Extending this to the twelfth week
End of week #12 = 5*x^11 new members
We know y = 5*x^11 for some x.

Choice (A) can't be written in this form at all, because it's a root. In choice (E), 60^12 = (5^12)*(12^12): nothing is to the power of 11, which is a problem. This can be the answer.
Notice that if x is a multiple of 5, say, x = 5p, then x^11 = (5^11)*(p^11), and then 5*x^11 = (5^12)*(p^11), so the answer could be of the form --- 5 to the 121th and some other factor to the 11th. Only choice (D) has this form, so this must be the answer.

BTW, (3^11)*(5^12) = 4.3248779 x 10^13 (!!), which is considerably more than all the people alive now and all the people who have ever been alive. If that many new members entered a gym, the gym wold have to be building with an area larger than the area of North America. On any word problem on the real GMAT, any in all better GMAT prep sources, the numerical answers are actually reasonable in the situation described. In this problem, the story in the word problem is a complete farce, and the problem is just asking a question about abstract number properties. This falls quite short of the high standards that the GMAT keeps.

Does all this make sense?
Mike :-)
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https://gmat.magoosh.com/

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by sukhman » Fri Oct 18, 2013 12:54 am
I did not understand what I got was 3^11*5^11=5*x^11 is this equation right ?

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by PrashantShinagare » Fri Oct 18, 2013 2:27 am
3^11*5^11=5*x^11 ==> 15^11

But we require it in the form 5*x^11.
Hence D

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by PrashantShinagare » Fri Oct 18, 2013 2:29 am
3^11*5^11=5*x^11 ==> 15^11

But we require it in the form 5*x^11.
Hence D

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by Scott@TargetTestPrep » Mon Dec 18, 2017 7:42 am
sukhman wrote:A certain club has exactly 5 new members at the end of its first week. Every subsequent week,each of the previous week's new members (and only these members) brings exactly x new members into the club.If y is the number of new members brought into the club during the twelfth week, which of the following could be y?

(A) 5^1/12
(B) 3^11 . 5 ^11
(C) 3^12 . 5^12
(D) 3^11 . 5^12
(E)60^12

The club has the following members:

At the end of week 1: 5 new members

At the end of week 2: 5x new members

At the end of week 3: 5x^2 new members

Thus, at the end of week 11, the club has 5x^10 new members, and at the end of week 12, the club has 5x^11 new members.

We see that y = 5x^11, i.e., there are 5x^11 new members brought into the club during the 12th week. Now we need to see which of the given choices can be 5x^11 for some integer value of x.

It can't be choice A, since 5^1/12 is not even an integer. Let's check the remaining choices by setting 5x^11 equal to the number given in the choice.

(B) 3^11 * 5^11

5x^11 = 3^11 * 5^11

x^11 = 3^11 * 5^10

x = ^11√(3^11 * 5^10) = 3 * ^11√(5^10)

We see that x is not an integer, so choice B is not correct.

(C) 3^12 * 5^12

5x^11 = 3^12 * 5^12

x^11 = 3^12 * 5^11

x = ^11√(3^12 * 5^11) = 5 * ^11√(3^12)

We see that x is not an integer, so choice C is not correct.

(D) 3^11 * 5^12

5x^11 = 3^11 * 5^12

x^11 = 3^11 * 5^11

x = ^11√(3^11 * 5^11) = 3 * 5 = 15

We see that x is an integer, so choice D is the correct answer.

Answer: D

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