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m10 #36 isn't ambiguous? [#permalink]
 15 Nov 2009, 01:52
What is the last digit of (9)^1 + (99)^2 + (999)^3 + ... + (10^n - 1)^n ? 1. n is even 2. n is prime The question stem shows n>3 (because first 3 terms are explicitly stated). So I chose D, all prime > 2 will be odd making the output=9. OA: OE: Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd). Thoughts?
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Re: m10 #36 isn't ambiguous? [#permalink]
15 Nov 2009, 02:33
gmattokyo wrote: What is the last digit of (9)^1 + (99)^2 + (999)^3 + ... + (10^n - 1)^n ? 1. n is even 2. n is prime The question stem shows n>3 (because first 3 terms are explicitly stated). So I chose D, all prime > 2 will be odd making the output=9. OA: OE: Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd). Thoughts? If the question stem doesnot say n>3 explicitly, you cannot assume that n>3 just because the first 3 terms are stated the way as above in the question. Then 1. n is even is suff as n is even only. 2. n is prime is not suff as n could be odd or even. That only A is suff.
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Re: m10 #36 isn't ambiguous? [#permalink]
15 Nov 2009, 06:18
GMAT TIGER wrote: gmattokyo wrote: What is the last digit of (9)^1 + (99)^2 + (999)^3 + ... + (10^n - 1)^n ? 1. n is even 2. n is prime The question stem shows n>3 (because first 3 terms are explicitly stated). So I chose D, all prime > 2 will be odd making the output=9. OA: OE: Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd). Thoughts? If the question stem doesnot say n>3 explicitly, you cannot assume that n>3 just because the first 3 terms are stated the way as above in the question. Then 1. n is even is suff as n is even only. 2. n is prime is not suff as n could be odd or even. That only A is suff. Thanks. Have to be careful about these interpretations. Next test (m11 16), another question which had me baffled (similar interpretation thing): Which of the following represents the range for all X which satisfy |1 - X| \lt 1 ? * (-1, 1) * (-1, 2) * (0, 1) * (0, 2) * (1, 2) I went the same way as the solution goes and reached 0<X<2. Now looking at the options, (0,2) looks as if including 0 and 2 which is not correct... but then there is no other option 
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Re: m10 #36 isn't ambiguous? [#permalink]
15 Nov 2009, 11:33
gmattokyo wrote: Next test (m11 16), another question which had me baffled (similar interpretation thing): Which of the following represents the range for all X which satisfy |1 - X| \lt 1 ? * (-1, 1) * (-1, 2) * (0, 1) * (0, 2) * (1, 2) I went the same way as the solution goes and reached 0<X<2. Now looking at the options, (0,2) looks as if including 0 and 2 which is not correct... but then there is no other option Thats the value: 0< X < 2. You are correct. The answer choices look strange however none other than (0, 2) are near to the possible range. If range means excluding, (0, 2) is fine but does range defined that way? Probably not.
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Re: m10 #36 isn't ambiguous? [#permalink]
30 Sep 2011, 08:09
gmattokyo wrote: What is the last digit of (9)^1 + (99)^2 + (999)^3 + ... + (10^n - 1)^n ? 1. n is even 2. n is prime The question stem shows n>3 (because first 3 terms are explicitly stated). So I chose D, all prime > 2 will be odd making the output=9. OA: OE: Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd). Thoughts? I also interpreted the question stem the same way and chose D. Very surprised that GMATTIGER's position is the correct one. Any expert would like to comment on this issue?
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Re: m10 #36 isn't ambiguous? [#permalink]
30 Sep 2011, 10:27
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cano wrote: I also interpreted the question stem the same way and chose D.
Very surprised that GMATTIGER's position is the correct one.
Any expert would like to comment on this issue? As per request, I am giving my position on this question: Actually, no. There is no ambiguity here. The OE is correct. 9^1 + (99)^2 + (999)^3 + ... + (10^n - 1)^n only implies that n is 1 or any other positive integer. It is implied that the first term of the series is 9^1. If there is a second term, it will be (99)^2 and so on... Whenever we have such a series, we always try to put n = 1 to figure out the sum in the simplest case. Then we put n = 2 to figure out the sum in that case. This gives us some idea of where we are going. It only shows you what the series looks like in the general case.
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Re: m10 #36 isn't ambiguous? [#permalink]
01 Oct 2011, 10:38
Thanks Karishma for confirming this!
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Re: m10 #36 isn't ambiguous? [#permalink]
04 Oct 2011, 04:27
my pick is D.
S1: all 9 powers to even results 1 as the unit digit, and odd power with 9. So, unit digit of the sum is "0". Sufficient
S2: Only prime with even is 2. As the question already has powers more than 2, then "n" has to odd. So, the unit digit of sum has to be "9". Sufficient
But, both the answers contradict. Still i will go with D, before seeing others explanations.
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Re: m10 #36 isn't ambiguous? [#permalink]
04 Oct 2011, 04:32
Again got it wrong...  Thanks karishma for the clarification.
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Re: m10 #36 isn't ambiguous?
[#permalink]
04 Oct 2011, 04:32
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