How many integers from 0 to 53, inclusive, have a remainder of 1 when divided by 3 ?

My ans is also C.17.
Explanation:
1 also gives 1 remainder when divided by 3, another number is 4, then 7 and so on.
Hence we have an arithmetic progression: 1, 4, 7, 10,..... 52, which are in the form 3n+1.
Now we have to find out number of terms.
tn=a+(n-1)d, where tn is the nth term of an AP, a is the first term and d is the common difference.
so,52 = 1+(n-1)3
or, (n-1)3 = 51
or, n-1 = 17
or, n = 18
D