What is the largest 2-digit number that satisfies 5x=6 (mod 4) is
a) 96
b) 97
c) 98
d) 99

Hint:
The value of any number mod 4 is the remainder when divided by 4. From number theory we know that every number $\text{N}$ can be written as $\text{N}=4\text{m}+\text{r}$, where $m \in N$ and $\text{r} \in \{0, 1, 2, 3 \}$

Complete step by step solution:
I am assuming that we are looking for the value x. We want the largest '2 digit number' value for x (so x is an integer) such that \[5x{\text{ }} \div {\text{ }}4\] has a remainder of 2.
That is, \[5x{\text{ }} = {\text{ }}4y{\text{ }} + {\text{ }}2\], where y is an integer.
A value of \[x{\text{ }} = {\text{ }}2\] satisfies the equation (y is also 2).
If we increase x by any positive multiple of 4, say 4z, that would also satisfy the equation, because the remainder would be equal to
Remainder \[\left\{ {\left( {4\left( {y + z} \right){\text{ }} + {\text{ }}2} \right){\text{ }} \div {\text{ }}4} \right\}{\text{ }} = {\text{ }}2\].
We want the largest 2 digit value of x. That would be
\[98{\text{ }} = {\text{ }}2{\text{ }} + {\text{ }}4 \times 24\]
\[\Rightarrow 5x = 490\]
This is \[4 \times 144{\text{ }} + {\text{ }}2\].

The answer is 98.

Note:
This question played a game with students. We know remainders modulo 4 can only be 0, 1, 2 or 3. So, students may confuse how is it possible to get remainder 6.
If we interpret the question to require a number x where 5x = 2 modulo 4, then we can reason like this.
2, 6 and 10 are the first numbers = 2 modulo 4. Of these, 10 is the first multiple of 5. The next number that is a multiple of 5 and = 2 mod 4 is 10 + 20 = 30, then the next one's are 50, 70 and 90 (rising in steps of 20). So the largest 2 digit number that satisfies the criteria is 90.