How many integers from 0 to 59, inclusive, have a remainder of 1 when divided by 3 ?  Choose one:
A) 15
B) 16
C) 17
D) 18
E) 20
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,..... 58, 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, 58 = 1+(n-1)3   or, (n-1)3 = 57   or, n-1 = 19   or, n = 20   20
<result>E</result>
How many integers from 0 to 59, inclusive, have a remainder of 1 when divided by 3 ?