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# Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2,

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Math Expert
Joined: 02 Sep 2009
Posts: 59712
Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2,  [#permalink]

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18 Mar 2019, 04:11
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75% (hard)

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48% (03:29) correct 52% (03:45) wrong based on 25 sessions

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Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2, the n-th term of the sequence is the units digit of the sum of the two previous terms. Let $$S_n$$ denote the sum of the first n terms of this sequence. The smallest value of n for which $$S_n>10,000$$ is:

(A) 1992
(B) 1999
(C) 2001
(D) 2002
(E) 2004

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Math Expert
Joined: 02 Aug 2009
Posts: 8309
Re: Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2,  [#permalink]

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18 Mar 2019, 05:06
3
Bunuel wrote:
Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2, the n-th term of the sequence is the units digit of the sum of the two previous terms. Let $$S_n$$ denote the sum of the first n terms of this sequence. The smallest value of n for which $$S_n>10,000$$ is:

(A) 1992
(B) 1999
(C) 2001
(D) 2002
(E) 2004

You cannot calculate till the sum becomes 10,000, so there has to be some repetition.
Let us check the sequence till it repeats..
WHEN will it happen ? - When two digits come one after another again because thereafter the units digit will remain the same.
so 4, 7, 1, 8, 9, 7, 6, .. will become 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7.... 4 and 7 have got repeated
So how many in a sequence before it repeats 4, 7...1, 3, so 12 numbers.
What is the total of these 12 numbers 4+7+...+1+3 = 60

Now, the total has to be greater than 10000?
Let us divide 10000 by 60, we get 10,000=60*166+40, so we have 166 sets and then we require digits totaling more than 40..
Now, we have to add starting from 4, 7, till it becomes >40...
4+7+1+8+9+7+6=42, so 7 numbers more..

Total = $$12*166+7=1992+7=1999$$

B
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Re: Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2,  [#permalink]

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19 Mar 2019, 19:09
1
Bunuel wrote:
Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2, the n-th term of the sequence is the units digit of the sum of the two previous terms. Let $$S_n$$ denote the sum of the first n terms of this sequence. The smallest value of n for which $$S_n>10,000$$ is:

(A) 1992
(B) 1999
(C) 2001
(D) 2002
(E) 2004

The best way to solve this problem is to write out the terms, and hopefully we can discover a pattern (for example, there is a sequence of numbers that repeats).

4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, …

We can see that the sequence {4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3} repeats. The sequence has 12 numbers, and their sum is 60. Since 10,000/60 = 166 R 40. We need to have at least 12 x 166 = 1992 numbers and a few more that sum up to be more than 40.

Since 4 + 7 + 1 + 8 + 9 + 7 + 6 = 42 > 40, we need 7 more numbers. Therefore, the total numbers we need such that S_n > 10,000 is 1992 + 7 = 1999.

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Re: Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, ... For n > 2,   [#permalink] 19 Mar 2019, 19:09
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