Sum of $ n + $ 2 digits

Question

If none of the parcels of an addition has more than n digits, and in the sum there are n + 2 digits, how many are at least the parcels?

Example for n = 3

888 + 888 + 888 + ............ 888 (11 installments) ---> 11,888 = 9 768 ---> has 3 + 1 digits.

999 + 999 + 999 + ............ 999 (11 installments) ---> 11,999 = 10 989 ---> has 3 + 2 digits

If none of the plots of a aExample for n = 3

How do I explain this using variables? Or the logic behind it

Answer 1

To show that the least number of parcels equals 11 you need to show two things.

1. There is one case where adding 11 n-digit numbers will result in a (n+2)-digit number.

2. Adding 10 n-digit numbers will always result in a (n+1) digit number.

The second is indeed true (try to explain why). You have effectively shown the first part. Your example works not only for 3, but generally for n digits (again try to explain why).

PS I am simply expanding on the comment from user saulspatz