5. (3 points) There is a basket of apples. If divided among Class A, each person gets 3 and there are 10 left; if divided among Class B, each person gets 4 and there are 11 left; if divided among Class C, each person gets 5 and there are 12 left. Then the basket of apples has at least $\qquad$ apples.

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【Analysis】Since $10 \div 3=3 \cdots 1, 11 \div 4=2 \cdots 3, 12 \div 5=2 \cdots 2$, the number of apples leaves a remainder of 1 when divided by 3, is 1 less when divided by 4, and leaves a remainder of 2 when divided by 5. The smallest number that satisfies leaving a remainder of 1 when divided by 3 and is 1 less when divided by 4 is 7, and 7 divided by 5 leaves a remainder of 2. Since the number of apples is greater than 12, and the least common multiple of 3, 4, and 5 is 60, the minimum number of apples in the basket is: $60+7=67$; solve accordingly.

【Solution】Solution: $10 \div 3=3 \cdots 1$,
$$
\begin{array}{l}
11 \div 4=2 \cdots 3, \\
12 \div 5=2 \cdots 2,
\end{array}
$$

The smallest number that satisfies leaving a remainder of 1 when divided by 3 and is 1 less when divided by 4 is 7, and 7 divided by 5 leaves a remainder of 2. Since the number of apples is greater than 12,

So the minimum number of apples: $3 \times 4 \times 5+7=60+7=67$ (apples);
Answer: The basket has at least 67 apples.
Therefore, the answer is: 67.