how many integers from 0 to 59 , 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 , . . . . . 58 , which are in the form 3 n + 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 e"
divide(59, const_10)