The ten numbers 1, 11, 111, 1111, … up to the string of ten 1s are added together. What remainder does the total leave on division by 9?

Every number leaves the same remainder as its digit sum on division by 9, and the kth term here is a string of k ones whose digit sum is simply k. So the whole total leaves the same remainder as 1 + 2 + 3 + … + 10 = 55, and $55=9\times 6+1$. The remainder is 1. Working it out the long way agrees: the sum is 1234567900, whose digits add to 37, then to 10, then to 1. The trap 0 assumes such a tidy-looking pile of 1s must land on a multiple of 9, and 5 stops at the 55 without dividing it by 9.