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# The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{

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Intern
Status: Yes. It was I who let the dogs out.
Joined: 03 Dec 2012
Posts: 42

Kudos [?]: 64 [1], given: 27

H: B
Concentration: General Management, Leadership
GMAT Date: 08-31-2013
The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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26 Jul 2013, 11:09
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Question Stats:

67% (01:40) correct 33% (01:54) wrong based on 236 sessions

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The infinite sequence $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$, … is such that $$a_{1} = 7$$, $$a_{2} = 8$$, $$a_{3} = 10$$, and $$a_n=a_{n-3} + 7$$ for values of n > 3. What is the remainder when $$a_{n}$$ is divided by 7?

(1) n is a multiple of 3.
(2) n is an even number.

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 94
Question: Page 227
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Jul 2013, 11:35, edited 2 times in total.
Edited the question and moved to DS forum.

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Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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26 Jul 2013, 11:40
The infinite sequence $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$, … is such that $$a_{1} = 7$$, $$a_{2} = 8$$, $$a_{3} = 10$$, and $$a_n=a_{n-3} + 7$$ for values of n > 3. What is the remainder when $$a_{n}$$ is divided by 7?

$$a_{1} = 7$$
$$a_{2} = 8$$
$$a_{3} = 10$$
$$a_{4} = a_1+7=7+7=14$$
$$a_{5} = a_2+7=8+7=15$$
$$a_{6} = a_3+7=10+7=17$$
...
Notice that the remainder upon division the above terms by 7 repeats in blocks of 3: {0, 1, 3} {0, 1, 3}...

(1) n is a multiple of 3 --> every third term has the remainder of 3 ($$a_{3}$$, $$a_{6}$$, $$a_{9}$$, ...). Sufficient.

(2) n is an even number. Not sufficient: consider $$a_2$$ and $$a_4$$.

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Kudos [?]: 139457 [0], given: 12790

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Kudos [?]: 139457 [1], given: 12790

Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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26 Jul 2013, 11:42
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Expert's post
hb wrote:
The infinite sequence $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$, … is such that $$a_{1} = 7$$, $$a_{2} = 8$$, $$a_{3} = 10$$, and $$a_n=a_{n-3} + 7$$ for values of n > 3. What is the remainder when $$a_{n}$$ is divided by 7?

(1) n is a multiple of 3.
(2) n is an even number.

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 94
Question: Page 227
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.

Similar question to practice: a-sequence-a1-64-a2-66-a3-67-an-8-an-3-which-of-the-43871.html
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Kudos [?]: 139457 [1], given: 12790

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Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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03 Jan 2014, 15:25
Hey,

I don't get how the OA can be A. Can anyone please explain?

As per my understanding the OA should be E:

Statement 1 says "n is a multiple of 3."

By applying the formula given in the question stem, we can find that a5=15 and that a7=21. Yet, 15 divided by 7 gives a remainder of 1, while 21 divided by 7 gives a remainder of 0. Hence, IMO statement 1 is insufficient.

Statement 2 says "n is an even number".

Also insufficient: a2=8 gives a remainder of 1, while a4=14 gives a remainder of 0.

Statements 1 and 2 combined say "n is a multiple of 3 and n is an even number".

IMO insufficient. For instance, a9=24 and a14=36. Both are multiples of 3 and are even. However, the former result gives a remainder of 3 whereas the latter one gives a remainder of 1.

Is there something that I'm misunderstanding? Please advise.

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

Math Expert
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Posts: 43322

Kudos [?]: 139457 [0], given: 12790

Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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04 Jan 2014, 05:26
Aurele wrote:
Hey,

I don't get how the OA can be A. Can anyone please explain?

As per my understanding the OA should be E:

Statement 1 says "n is a multiple of 3."

By applying the formula given in the question stem, we can find that a5=15 and that a7=21. Yet, 15 divided by 7 gives a remainder of 1, while 21 divided by 7 gives a remainder of 0. Hence, IMO statement 1 is insufficient.

Statement 2 says "n is an even number".

Also insufficient: a2=8 gives a remainder of 1, while a4=14 gives a remainder of 0.

Statements 1 and 2 combined say "n is a multiple of 3 and n is an even number".

IMO insufficient. For instance, a9=24 and a14=36. Both are multiples of 3 and are even. However, the former result gives a remainder of 3 whereas the latter one gives a remainder of 1.

Is there something that I'm misunderstanding? Please advise.

We need to find the remainder when when $$a_{n}$$ is divided by 7. (1) says n is a multiple of 3. Why are you checking the remainder when $$a_5$$ or $$a_7$$ is divided by 7. Is 5 or 7 a multiple of 3?

Hope it helps.
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Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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08 Jan 2014, 00:45
Thanks a lot for the explanation. Don't know why I confused both.

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Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{ [#permalink]

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11 Oct 2017, 10:08
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Re: The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{   [#permalink] 11 Oct 2017, 10:08
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# The infinite sequence a{1}, a{2}, …, a{n}, … is such that a{

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