7.1. [5-6.5 (а) - 20 points, б) - 20 points)] Vovochka adds numbers in a column in the following way: he does not remember the tens, and under each pair of digits in the same place value, he writes their sum, even if it is a two-digit number. For example, for the sum $248+208$, he would get the value 4416.

a) In how many cases will Vovochka get the correct answer when adding all possible pairs of three-digit numbers? (If Vovochka has already added some two different numbers in a different order previously, he does not notice it.)

б) Find the smallest possible difference between the correct answer and Vovochka's answer for all other pairs of three-digit numbers.
Answer: a) 244620 b) 1800.

Solution. a) Vovochka will get the correct answer if and only if the sum of the last and second-to-last digits also results in a digit, i.e., there is no carry-over. Let's count how many pairs of digits have a sum that does not exceed 9. If one digit is 0, the other can be any of 10 digits (any digit); if one digit is 1, the other can be 9 digits (from 0 to 8), and so on; if one digit is 9, the other can only be 1 digit (0). In total, there are $1+2+\ldots+10=55$ variants. The first digit of each of the three-digit numbers being added can be any digit from 1 to 9. Thus, we get $9 \cdot 9 \cdot 55 \cdot 55 = 245025$ variants. However, identical pairs of numbers for which Vovochka will get the correct answer are counted twice. There are $9 \cdot 5 \cdot 5 = 405$ such pairs. Therefore, Vovochka will get the correct answer in $245025 - 405 = 244620$ cases.

b) Let $\overline{abc}$ and $\overline{xyz}$ be the three-digit numbers Vovochka is adding ($a, x \geqslant 1$, since the numbers are three-digit), their sum is $100(a+x) + 10(b+y) + (c+z)$. If Vovochka got the wrong answer, then the sum of the last or second-to-last digits (or both) was a two-digit number, i.e., there are three cases.

1) If $b+y \geqslant 10, c+z < 10$, then Vovochka will get the answer $1000(a+x) + 10(b+y) + (c+z)$, and the difference between the correct answer and Vovochka's answer is $900(a+x)$.
2) If $b+y < 10, c+z \geqslant 10$, then Vovochka's answer is $1000(a+x) + 100(b+y) + (c+z)$, which differs from the correct answer by $900(a+x) + 90(b+y)$.
3) If $b+y \geqslant 10, c+z \geqslant 10$, then Vovochka will get the answer $10000(a+x) + 1000(b+y) + (c+z)$, in this case, the difference is $9900(a+x) + 990(b+y)$.

Thus, in each case, the difference between the answers will be no less than $900(a+x) \geqslant 1800$. The difference of 1800 can be obtained, for example, by adding the numbers 105 and 105: Vovochka will get the answer 2010 when the correct answer is 210.