MHB What Is the Smallest n Such That Sequence Sum S_n Exceeds 10000?

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The discussion revolves around finding the smallest integer n such that the sum S_n of the sequence 4, 7, 1, 8, 9, 7, 6, ... exceeds 10,000. The sequence is defined such that for n > 2, each term is the units digit of the sum of the two preceding terms. Participants share their approaches and calculations to determine the value of n. Several members successfully provide correct solutions, highlighting their problem-solving strategies. The focus remains on the mathematical exploration of the sequence and its cumulative sum.
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Here is this week's POTW:

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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. What is the smallest value of $n$ for which $S_n>10000$?

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Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. kaliprasad
3. lfdahl

Solution from castor28:
The sequence is defined by the recurrence relation $x_n=(x_{n-1}+x_{n-2})\bmod{10}$.
Looking at the first five terms, we observe that the period of the sequence is $3\pmod{2}$ and $4\pmod{5}$; the period is therefore $12$.

We list the first $12$ terms, and the cumulative sums:
$$
\begin{array}{c|r|r|r|r|r|r|r|r|r|r|r|r|}
n&1&2&3&4&5&6&7&8&9&10&11&12\\
\hline
a_n&4&7&1&8&9&7&6&3&9&2&1&3\\
\hline
S_n&4&11&12&20&29&36&42&45&54&56&57&60
\end{array}
$$
As each period contributes $60$ to the total, we need $166$ full periods, giving a total of $9960$. To reach $10001$, we must increase that total by at least $41$. Looking at the table, we see that we must include $7$ additional terms, giving a total of $10002$. The total number of terms is therefore $12\times166+7=\bf1999$.
 
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