5. Do there exist two numbers such that the first is 2016 times greater than the second, and the sum of its digits is less than the sum of the digits of the second by 2016 times?
Answer. Yes, they do exist. Solution. Consider, for example, a number $A$ of the form $11...15$, where the number of ones is greater than 10. Multiplying $A$ by 2016, we get:

$$
11 \ldots 15 \times 2016=10080+20160+201600+2016000+\ldots+201600 \ldots 0=2240 \ldots .007840
$$

(here, many nines form due to $2+0+1+6=9$, but they all disappear due to the running carry over ten). The sum of the digits of the product is 27 regardless of the number of ones in the representation of the number $A$. For the sum of the digits of the number $A$ to be 2016 times greater, there should be $27 \times 2016-5$ ones in its representation.