A Modular Recursion Sequence?

Let $a_1=x$ and $a_2=y$ . Applying the recursive formula provided, we then have $a_6=n$ , where $n \equiv 3x+5y \pmod{10}$ . We now have two cases:

Case 1: When $y$ is an odd number, we have $n \equiv 3x+5 \pmod{10}$ or equivalently $3x \equiv 1 \pmod{10}$ , since we wish to have $n \equiv 6 \pmod{10}$ . The only possible positive integer value for $x=a_1 \leq 10$ that fits the above equation is $x=a_1=7$

Case 2: When $y$ is an even number, we have $n \equiv 3x \pmod{10}$ or equivalently $3x \equiv 6 \pmod{10}$ , since we wish to have $n \equiv 6 \pmod{10}$ . The only possible positive integer value for $x=a_1 \leq 10$ that fits the above equation is $x=a_1=2$ .

Given that the two possible values for $a_1$ is $2,7$ , their sum is thus $2+7=\fbox{9}$

For the sixth term of the sequence to be six, let us look at the possible fifth and fourth terms which add to give six mod ten. Here is a list:

... 9, 7, 6

... 8, 8, 6

... 7, 9, 6

... 5, 1, 6

... 4, 2, 6

... 3, 3, 6

... 2, 4, 6

... 1, 5, 6

Continuing each sequence to find the first term, we find that the only two first terms are seven and two. Thus the answer is $7+2=\boxed{9}$ .

Noting that all the solutions already given are more elegant than this, here are a few Mathematica commands that will solve it by brute force:

a[n_] := Mod[a[n - 1] + a[n - 2], 10];

a[1] = i; a[2] = j;

Position[Table[a[6], {i, 1, 9}, {j, 1, 9}], 6]

We get only pairs with first elements 2 and 7, so the answer is 9.

After solving in few trial and error method, i found a linear Diophantine equation 3x+5y=4 ; where x is the first term and y is the second term and the solutions of the equation will give the required result.