The sum of all three-digit numbers that leave the remainder 2 when divided by 5 is:

The numbers are 102, 107, …, 997 — an arithmetic progression with $a=102,d=5$ and last term 997. From $997=102+(n-1)5$ we get $n=180$, so the sum is $(180/2)(102+997)=90\times 1099=98910$. (98550 is the sum of the three-digit multiples of 5 instead; 197820 forgets the factor $n/2$ and uses $n$; 49455 halves the correct answer a second time.)