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M10-36

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M10-36 [#permalink]

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For a certain sequence \(a_1=9^1\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of \(n\) terms of the sequence, what is the units digit of S?


(1) \(n\) is even

(2) \(n\) is prime
[Reveal] Spoiler: OA

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Re M10-36 [#permalink]

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Official Solution:


We have the sum of \(n\) integers:

(..9) + (..1) + (..9) + (..1) + ...

Statement (1) by itself is sufficient. If \(n\) is even then this sum ends with 0. If it's odd, then this sum ends with 9.

Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd).


Answer: A
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Re: M10-36 [#permalink]

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Bunuel wrote:
Official Solution:


We have the sum of \(n\) integers:

(..9) + (..1) + (..9) + (..1) + ...

Statement (1) by itself is sufficient. If \(n\) is even then this sum ends with 0. If it's odd, then this sum ends with 9.

Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd).


Answer: A


I completely agree with the solution. However, the way this sum is depicted one can assume that n is more than 3, since the powers of 2 and 3 are clearly depicted. So if we assume that n is more than 3 the answer will change.

Maybe the sum should have been depicted as

9^1 + ... + (10^n - 1)^n

in this case one cannot assume that n is more than 3?

Cuz during the test I assumed that N is more than 3 and so I got an answer different from the official one.
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Re: M10-36 [#permalink]

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New post 27 Jun 2015, 14:14
GMAT888

I´m with you, I got the same understanding from the way the question is written
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Re: M10-36 [#permalink]

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New post 17 Jul 2015, 23:43
I guess we should assume that n cannot be 0 either?
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Re: M10-36 [#permalink]

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New post 18 Jul 2015, 00:57
bluesquare wrote:
I guess we should assume that n cannot be 0 either?


Hi,
why can't n be 0?
n is even as per statement 1 and if it is to power 0,which is even, answer will still be 1(anything to pwer of 0 is 1)... suff
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Re: M10-36 [#permalink]

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michaelyb wrote:
GMAT888

I´m with you, I got the same understanding from the way the question is written


Hi,
I think the Q is correct as it is written if you take n to be a positive integer > or <3 will still give you the same answer...
Bunuel , only thing that could have been mentioned is n is a non negative integer... because nonnegative would change the answer
or we take this as a series and therefore n as positive integer but it is not mentioned that it is a sum of a series
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Re: M10-36 [#permalink]

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New post 18 Jul 2015, 16:25
chetan2u wrote:
bluesquare wrote:
I guess we should assume that n cannot be 0 either?


Hi,
why can't n be 0?
n is even as per statement 1 and if it is to power 0,which is even, answer will still be 1(anything to pwer of 0 is 1)... suff


If n can be 0, then in statement 1, n is even means that the term can end in either a 1 or a 9, so statement 1 should be insufficient, no?
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Re: M10-36 [#permalink]

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New post 20 Jul 2015, 04:00
Bunuel wrote:
For a certain sequence \(a_1=9^1\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of \(n\) terms of the sequence, what is the units digit of S?


(1) \(n\) is even

(2) \(n\) is prime


Edited the question. Is it better?
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Re: M10-36 [#permalink]

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please can someone elaborate on the answer and explanation i cannot understand why cannot n be greater than 3.

i tried forming series first as
9,99^2, 999^3, 9999^4...... (10^n - 1)^n


how come series is in form ..9,..1,..9,..1 ??? what is this clearly ??


please help
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Re: M10-36 [#permalink]

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smanujahrc wrote:
For a certain sequence \(a_1=9^1\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of \(n\) terms of the sequence, what is the units digit of S?


please can someone elaborate on the answer and explanation i cannot understand why cannot n be greater than 3.

i tried forming series first as
9,99^2, 999^3, 9999^4...... (10^n - 1)^n


how come series is in form ..9,..1,..9,..1 ??? what is this clearly ??


please help


That the terms of the sequence are given by the formula \(a_n=(10^n - 1)^n\).

\(a_1=9^1=9\)
\(a_2=99^2=...1\)
\(a_3=999^3=...9\)
\(a_4=9999^4=...1\)
...

As you can see odd terms have the units digit of 9 and the even terms have the units digit of 1.

The question asks about the units digit of the sum of \(n\) terms of the sequence. Now, if n is even then this sum ends with 0. If it's odd, then this sum ends with 9.

For example:
If n = 2, then the units digit of the sum of 2 terms is 9^1 + 99^2 = ...0.
If n = 3, then the units digit of the sum of 3 terms is 9^1 + 99^2 + 999^3 = ...9.

Statement (1) by itself is sufficient. n = 2 --> the sum ends with 0.

Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd).


Answer: A.

Hope it's clear.
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Re: M10-36 [#permalink]

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New post 07 Feb 2018, 09:57
Bunuel wrote:
smanujahrc wrote:
For a certain sequence \(a_1=9^1\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of \(n\) terms of the sequence, what is the units digit of S?


please can someone elaborate on the answer and explanation i cannot understand why cannot n be greater than 3.

i tried forming series first as
9,99^2, 999^3, 9999^4...... (10^n - 1)^n


how come series is in form ..9,..1,..9,..1 ??? what is this clearly ??


please help


That the terms of the sequence are given by the formula \(a_n=(10^n - 1)^n\).

\(a_1=9^1=9\)
\(a_2=99^2=...1\)
\(a_3=999^3=...9\)
\(a_4=9999^4=...1\)
...

As you can see odd terms have the units digit of 9 and the even terms have the units digit of 1.

The question asks about the units digit of the sum of \(n\) terms of the sequence. Now, if n is even then this sum ends with 0. If it's odd, then this sum ends with 9.

For example:
If n = 2, then the units digit of the sum of 2 terms is 9^1 + 99^2 = ...0.
If n = 3, then the units digit of the sum of 3 terms is 9^1 + 99^2 + 999^3 = ...9.

Statement (1) by itself is sufficient. n = 2 --> the sum ends with 0.

Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd).


Answer: A.

Hope it's clear.


Brilliant explanation. Thanx a lot. Its is superb.
Re: M10-36   [#permalink] 07 Feb 2018, 09:57
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