If $n$ is $a$ positive integer, which one of the following numbers must have $a$ remainder of 3 when divided by any of the numbers 4, 5 and 6?

Step 1 (Set up Modular Congruences): Let the number be $N$. We are given that $N$ leaves a remainder of 3 when divided by 4, 5, and 6:

1. $N\equiv 3(\mathrm{mod}4)⟹N-3\equiv 0(\mathrm{mod}4)$

2. $N\equiv 3(\mathrm{mod}5)⟹N-3\equiv 0(\mathrm{mod}5)$

3. $N\equiv 3(\mathrm{mod}6)⟹N-3\equiv 0(\mathrm{mod}6)$

Step 2 (Find Common Multiple): This implies that $N-3$ is a common multiple of $4,5,$ and $6$. The smallest positive common multiple is the Least Common Multiple (LCM):

Therefore:

Step 3 (Conclusion): Any such number must be of the form 60n+3 (or 63 for $n=1$).