Answer :
Answer:
To solve the problem of finding the maximum carryover when 1000 single-digit numbers (each known to be zero) are added such that the units digit of the sum is 5, we start by understanding the nature of the carryover in addition.
Each digit addition can carry over at most 1 to the next higher digit. Let's consider the units digit of our sum being 5. This suggests that in the addition of all 1000 numbers, the total units place contribution must be 5.
Given that each added number is a single-digit zero, and the sum ends in 5, we need to allocate the carryovers such that the sum of units digits ends in 5. Let's consider the carryovers:
- If all the numbers contributed their maximum possible carryover, which is 1, then the total carryover would be 1000 (since each number can potentially carry over 1).
- However, we need the units digit to be specifically 5, meaning some carryovers must have occurred.
To maximize the carryover while ensuring the units digit is 5, we can explore the following scenario:
- Suppose 999 numbers contribute a carryover of 0, and 1 number contributes a carryover of 5. This configuration satisfies the requirement that the units digit of the sum is 5, as:
\[
999 \times 0 + 1 \times 5 = 5
\]
Thus, in this scenario:
- The maximum carryover is achieved by the single number contributing 5 to the units place.
Therefore, the maximum carryover in this case is \( \boxed{5} \). This solution satisfies the conditions given: all numbers are single-digit zeros, their sum ends in 5, and the maximum possible carryover is accounted for.