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a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f

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a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 02:42
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Question Stats:

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[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9

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a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post Updated on: 08 Feb 2019, 04:51
MathRevolution wrote:
[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9


This is cyclicity question :

The series is as below
A1- 1
A2- 1
A3- 2
A4- 0
A5- 2
A-6 2
A7- 1
A8- 0

So together the sum of first segments gives 9 ..Now picking up any random segments will lead to same set of digits which will sum up to 9

So for even A161+ A162+ ...A168 =sum of digits will be 9 .
or A3 to A11 :)



Hope it helps!!

Originally posted by prabsahi on 08 Feb 2019, 03:52.
Last edited by prabsahi on 08 Feb 2019, 04:51, edited 2 times in total.
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 04:18
1
MathRevolution wrote:
[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9


The trick was to find a pattern.
Keyword = Remainder when divided by 3
an is the remainder when an-1 + an-2 is divided by \(3\)

a_0 = 0
a_1 = 1
a_2 = 1 (1+0)/3
a_3 = 2 (1+1)/3
a_4 = 0 (2+1)/3
a_5 = 2 (0+2)/3
a_6 = 2 (0+2)/3
a_7 = 1 (2+2)/3
a_8 = 0(2+1)/3

If you go forward you will notice the same series.

So now the series starts from 101 ...... 108
1+1+2+0+2+2+1+0
9

E
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 04:32
KanishkM wrote:
MathRevolution wrote:
[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9


The trick was to find a pattern.
Keyword = Remainder when divided by 3
an is the remainder when an-1 + an-2 is divided by \(3\)

a_0 = 0
a_1 = 1
a_2 = 1 (1+0)/3
a_3 = 2 (1+1)/3
a_4 = 0 (2+1)/3
a_5 = 2 (0+2)/3
a_6 = 2 (0+2)/3
a_7 = 1 (2+2)/3
a_8 = 0(2+1)/3

If you go forward you will notice the same series.

So now the series starts from 101 ...... 108
1+1+2+0+2+2+1+0
9

E



I see .I made a mistake :)

Thanks
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 04:38
prabsahi wrote:

I see .I made a mistake :)

Thanks


prabsahi, your approach was right. :thumbup:

But i still feel that is not the easiest way to solve it. :)
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 04:44
KanishkM wrote:
prabsahi wrote:

I see .I made a mistake :)

Thanks


prabsahi, your approach was right. :thumbup:

But i still feel that is not the easiest way to solve it. :)


yes..You are right KanishkM..Will update it :)

Thanks
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 05:11
MathRevolution wrote:
[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9



the pattern here followed is of divisibility of a no with 3 ; which is always for consective no is series of 0,1,2,0,1,2
0,1,2 is our pattern
so the 99th term would be 2 and 101st would be 1
so sum = 1+2+0+1+2+0+1+2; 9
IMO E
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a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 08 Feb 2019, 05:14
Archit3110 wrote:
MathRevolution wrote:
[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9



the pattern here followed is of divisibility of a no with 3 ; which is always for consective no is series of 0,1,2,0,1,2
0,1,2 is our pattern
so the 99th term would be 2 and 101st would be 1
so sum = 1+2+0+1+2+0+1+2; 9
IMO E



Hi Archit,

I guess you made a similar mistake like I did.

Please check the pattern mentioned in my previous post or KanishkM's.I have corrected it.

Its 0 1 1 2 0 2 2 1 0

Hope it helps !!
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 10 Feb 2019, 18:14
=>

a0 = 0
a1 = 1
a2 = a1 + a0 = 1 + 0 = 1
a3 = a2 + a1 = 1 + 1 = 2
a4 = 0 since a3 + a2 = 2 + 1 = 3 = 3(1)+0, and the remainder when a3 + a2 is divided by 3 is zero.
a5 = a4 + a3 = 0 + 2 = 2
a6 = a5 + a4 = 2 + 0 = 2
a7 = 1 since a6 + a5 = 2 + 2 = 4= 3(1)+1, and the remainder when a6 + a5 is divided by 3 is 1.
a8 = 0 since a7 + a6 = 1 + 2 = 3= 3(1)+0, the remainder is 0, when a7 + a6 is divided by 3.
a9 = a8 + a7 = 0 + 1 = 1
Thus, the sequence is periodic, with period 8.
a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108
= a5 + a6 + a7 + a0 + a1 + a2 + a3 + a4
\(= 2 + 2 + 1 + 0 + 1 + 1 + 2 + 0 = 9\)
Therefore, the answer is E.
Answer: E
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f  [#permalink]

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New post 13 Feb 2019, 07:51
KanishkM wrote:
MathRevolution wrote:
[GMAT math practice question]

a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by \(3\) for \(n ≥ 2\). What is the value of a101 + a102 + a103 + a104 + a105 + a106 + a107 + a108?

A. 0
B. 3
C. 5
D. 7
E. 9


The trick was to find a pattern.
Keyword = Remainder when divided by 3
an is the remainder when an-1 + an-2 is divided by \(3\)

a_0 = 0
a_1 = 1
a_2 = 1 (1+0)/3
a_3 = 2 (1+1)/3
a_4 = 0 (2+1)/3
a_5 = 2 (0+2)/3
a_6 = 2 (0+2)/3
a_7 = 1 (2+2)/3
a_8 = 0(2+1)/3

If you go forward you will notice the same series.

So now the series starts from 101 ...... 108
1+1+2+0+2+2+1+0
9

E

I dont understand what quotient to take here. Please explain.
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Re: a0 = 0, a1 = 1. an is the remainder when an-1 + an-2 is divided by 3 f   [#permalink] 13 Feb 2019, 07:51
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