remainder

There is not one UNIQUE solution but an infinite number of them. Here is the general form of the solution: x =2 + 8c, where c is any positive integer. So, the numbers would start with:

2, 10, 18, 26, 34......etc.

The 2nd part of your question depends on what value you take.

So, [2 + 9] mod 8=3. And so the remainder will be 3, no matter what value you choose.

Previous guest is wrong. Answer is 25/8

Solution

Let x /8 = Q  + 2    so  x = 8(Q +2)   where Q is the quotient and 2 is the remainder

then   x  + 9  = 8(Q +2)  +9   so

(x  +9) /8 = {  8(Q +2) +9 }  /8

              = Q +2 +9/8

which is the quotient plus a remainder of 25/8 or 3 plus a quarter.

Guest #2 is wrong:

"x /8 = Q  + 2    so  x = 8(Q +2)   where Q is the quotient and 2 is the remainder"       is INCORRECT

                                x = 8(Q)  + 2   is correct

anyway , if you start with a reaminder of  2   and add 9  its like a remainder of 11....pull another '8' out of there for a remainder of '3'

\(x\pmod 8 \equiv 2\\ x+8\pmod 8 \equiv 2\\ x+9\pmod 8 \equiv 3\)

Guest #2 is actually wrong. Note that x is an positive integer and 8 is also a positive integer. The remainder won't be a fraction.

This could be solved using modulus maths like Guest #1 and I did.

Positive integer x leaves a remainder of 2 when divided by 8.
What is the remainder when x+9 is divided by 8

\(\begin{array}{|rcll|} \hline x & \equiv & 2 \pmod{8} \\ x+{\color{red} 9 } & \equiv & 2+{\color{red} 9 } \pmod{8} \\ x+9 & \equiv & 11 \pmod{8} \\ x+9 & \equiv & 11-8 \pmod{8} \\ x+9 & \equiv & 3 \pmod{8} \\ \hline \end{array}\)