【4】The smallest natural number that leaves a remainder of 2 when divided by 3, a remainder of 4 when divided by 5, and a remainder of 4 when divided by 7 is ( ).
【Key Points】Properties of remainders; divisibility of numbers.
【Analysis】(1) Let this number be $a$, then $(a-4)$ can be divided by $5$ and $7$, i.e., $5 \times 7=35$. When $a=39$, it is not possible; when $a=74$, it leaves a remainder of 2 when divided by 3, so the smallest $a$ is 74.
(2) Numbers divisible by 5 end in 0 or 5, so numbers that leave a remainder of 4 when divided by 5 end in 4 or 9. Since we are looking for the smallest natural number, it should be a two-digit number $\square 4$ or $\square 9$.
When it is $\square 4 \div 3=(\quad \ldots \ldots 2$ and 4 divided by 3 leaves a remainder of 1, so dividing by 3 also leaves a remainder of 1. 44 cannot be divided by 7 to leave a remainder of 4 (discard), but 74 can.
When $\square 9 \div 3=(\quad \ldots \ldots 2$, 9 divided by 3 leaves a remainder of 0, so dividing by 3 also leaves a remainder of 2. 59 cannot be divided by 7 to leave a remainder of 4, and $79>74$ (not valid).

【Answer】74