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18 Mar 2019, 04:11
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B
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Difficulty:
95%
(hard)
Question Stats:
44% (02:58) correct
56%
(03:01)
wrong
based on 59
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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
(A) 1992
(B) 1999
(C) 2001
(D) 2002
(E) 2004
18 Mar 2019, 05:06
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Bunuel
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
19 Mar 2019, 19:09
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Bunuel
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.
Answer: B
