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a0 = 0, a1 = 1. an is the remainder when an1 + an2 is divided by 3 f
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08 Feb 2019, 02:42
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[GMAT math practice question] a 0 = 0, a 1 = 1. a n is the remainder when a n1 + a n2 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
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a0 = 0, a1 = 1. an is the remainder when an1 + an2 is divided by 3 f
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Updated on: 08 Feb 2019, 04:51
MathRevolution wrote: [GMAT math practice question]
a0 = 0, a1 = 1. an is the remainder when an1 + an2 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 A6 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 an1 + an2 is divided by 3 f
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08 Feb 2019, 04:18
MathRevolution wrote: [GMAT math practice question]
a0 = 0, a1 = 1. an is the remainder when an1 + an2 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 a n is the remainder when a n1 + a n2 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 an1 + an2 is divided by 3 f
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08 Feb 2019, 04:32
KanishkM wrote: MathRevolution wrote: [GMAT math practice question]
a0 = 0, a1 = 1. an is the remainder when an1 + an2 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 a n is the remainder when a n1 + a n2 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 an1 + an2 is divided by 3 f
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08 Feb 2019, 04:38
prabsahi wrote: I see .I made a mistake Thanks prabsahi, your approach was right. 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 an1 + an2 is divided by 3 f
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08 Feb 2019, 04:44
KanishkM wrote: prabsahi wrote: I see .I made a mistake Thanks prabsahi, your approach was right. 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 an1 + an2 is divided by 3 f
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08 Feb 2019, 05:11
MathRevolution wrote: [GMAT math practice question]
a0 = 0, a1 = 1. an is the remainder when an1 + an2 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 an1 + an2 is divided by 3 f
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08 Feb 2019, 05:14
Archit3110 wrote: MathRevolution wrote: [GMAT math practice question]
a0 = 0, a1 = 1. an is the remainder when an1 + an2 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 0Hope it helps !!



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Re: a0 = 0, a1 = 1. an is the remainder when an1 + an2 is divided by 3 f
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10 Feb 2019, 18:14
=> 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
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Re: a0 = 0, a1 = 1. an is the remainder when an1 + an2 is divided by 3 f
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13 Feb 2019, 07:51
KanishkM wrote: MathRevolution wrote: [GMAT math practice question]
a0 = 0, a1 = 1. an is the remainder when an1 + an2 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 a n is the remainder when a n1 + a n2 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.




Re: a0 = 0, a1 = 1. an is the remainder when an1 + an2 is divided by 3 f
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