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The first term of a sequence is 2005. Each succeeding term is the sum

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The first term of a sequence is 2005. Each succeeding term is the sum  [#permalink]

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New post 24 Mar 2019, 23:13
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Difficulty:

  75% (hard)

Question Stats:

39% (01:45) correct 61% (02:22) wrong based on 28 sessions

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The first term of a sequence is 2005. Each succeeding term is the sum  [#permalink]

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New post Updated on: 30 Mar 2019, 00:42
Bunuel wrote:
The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?


(A) 29

(B) 55

(C) 85

(D) 133

(E) 250


first term = 2005
2nd term = 133
third term = 55
4th term = 250
5th term= 133
cyclicity is of 3
2005th term from 1 = 2004 term = 4th term = 250 IMO E
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Originally posted by Archit3110 on 24 Mar 2019, 23:23.
Last edited by Archit3110 on 30 Mar 2019, 00:42, edited 1 time in total.
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Re: The first term of a sequence is 2005. Each succeeding term is the sum  [#permalink]

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New post 28 Mar 2019, 07:02
Archit3110 wrote:
Bunuel wrote:
The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?


(A) 29

(B) 55

(C) 85

(D) 133

(E) 250


first term = 2005
2nd term = 133
third term = 55
4th term = 250
5th term= 133
cyclicity is of 4
2005th term from 1 = 2004 term = 4th term = 250 IMO E




I established the same sequence but then I counted wrong. I always do that, could you try to explain to me how we can establish the 2005th term? I don't understand your reasonening above
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Re: The first term of a sequence is 2005. Each succeeding term is the sum  [#permalink]

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New post 29 Mar 2019, 22:10
Berlin92

first term = 2005

2nd term = 133
3rd term = 55
4th term = 250
5th term= 133
6th term = 55
7th term = 250

Keeping the first term aside
cyclicity is of 3

2005th term = 2004th term from the 2nd term i.e. 133 = 2004/3 = remainder is 0 i.e. 2004 = 3n
Thus the term is the third term in the sequence Ans --> 250
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Re: The first term of a sequence is 2005. Each succeeding term is the sum  [#permalink]

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New post 30 Mar 2019, 00:49
Berlin92 wrote:
Archit3110 wrote:
Bunuel wrote:
The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?


(A) 29

(B) 55

(C) 85

(D) 133

(E) 250


first term = 2005
2nd term = 133
third term = 55
4th term = 250
5th term= 133
cyclicity is of 3
2005th term from 1 = 2004 term = 4th term = 250 IMO E




I established the same sequence but then I counted wrong. I always do that, could you try to explain to me how we can establish the 2005th term? I don't understand your reasonening above


Berlin92
once you have understood the cyclicity of the terms viz 3 in this case , then simply divide the cyclicity term with the desired nth no. i.e 2005 in this case , but since 1st term ie 2005 is not coming up again the 2004 term from 1st is to be determined
so now 2004/3 ; remainder 0 ; ie. 250 will be the value of 2005th term..
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Re: The first term of a sequence is 2005. Each succeeding term is the sum   [#permalink] 30 Mar 2019, 00:49
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