$\triangle$ Let

$$
a=6 q+r
$$

where the remainder $r$ satisfies the inequality $0 \leqslant r \leqslant 5$. We will consider all possible values of $r$.

The case $r=0$ is impossible, otherwise the number $a$ would be divisible by 2 and 3.

The case $r=1$ is also impossible, because then $a$ when divided by 3 would leave a remainder of 1, not 2.

For similar reasons, the cases $r=2, r=3$, and $r=4$ are also ruled out.

The case $r=5$ is possible, for example, when $a=5$ or $a=11$.

Answer: 5.
431. When dividing a natural number $a$ by 2, the remainder is 1, and when dividing by 3, the remainder is 2. What remainder will be obtained when dividing $a$ by 6?