5. Find the number of three-digit numbers ending in zero that give a remainder of 1 when divided by 3, given that the tens or hundreds digit is divisible by 4.
5. According to the condition, three-digit numbers end with zero. Therefore, their general form is: ас 0. Since the number of tens or hundreds is divisible by 4, we get the following two groups of numbers:

$$
\mathrm{a} 00, \mathrm{a} 40, \mathrm{a} 80 \text { and } 4 \mathrm{c} 0,8 \mathrm{c} 0 .
$$

Using the last condition (when divided by 3, the remainder is 1), we determine the required numbers:

$100,400,700$

$340,640,940$

$280,580, \mathbf{8 8 0}$

$400,430,460,490$

$820,850,880$.

Thus, there are fifteen three-digit numbers that satisfy the condition of the problem.

Answer: fifteen.