3. a) Determine the remainder of the division of a natural number by 42, knowing that when divided by 6 it gives a remainder of 5 and when divided by 7 it gives a remainder of 3.

b) Show that the number $21^{33} \cdot 33^{77} \cdot 77^{21}$ is a perfect square.
3. a) Using the division theorem with remainder, we deduce that $n=6a+5$, and $n=7b+3$. Then $7n=42a+35$, and $6n=42b+18$. By subtracting, we obtain that $n=42c+17$, with $17<42$, so the sought remainder is 17.

$$
\text { b) } 21^{33} \cdot 33^{77} \cdot 77^{21}=3^{110} \cdot 7^{54} \cdot 11^{98}=\left(3^{55} \cdot 7^{27} \cdot 11^{49}\right)^{2}
$$