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

Question

a 0  = 0, a 1  = 1. a n  is the remainder when a n-1  + a n-2  is divided by  3  for  n ≥ 2 . What is the value of a 101  + a 102  + a 103  + a 104  + a 105  + a 106  + a 107  + a 108 ?
 
A. 0 
B. 3 
C. 5 
D. 7 
E. 9

prabsahi · Answer

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!!

KanishkM · Answer

The trick was to find a pattern.
Keyword = Remainder when divided by 3
a n  is the remainder when a n-1  + a n-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

prabsahi · Answer

I see .I made a mistake

Thanks

KanishkM · Answer

prabsahi , your approach was right.

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

prabsahi · Answer

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

Thanks

Archit3110 · Answer

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

prabsahi · Answer

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 !!

MathRevolution · Answer

=>

a 0  = 0
a 1  = 1
a 2  = a 1  + a 0  = 1 + 0 = 1
a 3  = a 2  + a 1  = 1 + 1 = 2
a 4  = 0 since a 3  + a 2  = 2 + 1 = 3 = 3(1)+0, and the remainder when a 3  + a 2  is divided by 3 is zero.
a 5  = a 4  + a 3  = 0 + 2 = 2
a 6  = a 5  + a 4  = 2 + 0 = 2
a 7  = 1 since a 6  + a 5  = 2 + 2 = 4= 3(1)+1, and the remainder when a 6  + a 5  is divided by 3 is 1.
a 8  = 0 since a 7  + a 6  = 1 + 2 = 3= 3(1)+0, the remainder is 0, when a 7  + a 6  is divided by 3.
a 9  = a 8  + a 7  = 0 + 1 = 1
Thus, the sequence is periodic, with period 8.
a 101  + a 102  + a 103  + a 104  + a 105  + a 106  + a 107  + a 108 
= a 5  + a 6  + a 7  + a 0  + a 1  + a 2  + a 3  + a 4  
 = 2 + 2 + 1 + 0 + 1 + 1 + 2 + 0 = 9 
Therefore, the answer is E. 
Answer: E

SWAPNILP · Answer

I dont understand what quotient to take here. Please explain.

bumpbot · Answer

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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

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