if a number when divided by 3 and has reminder of 1 then the number is written as 3K+1, so we can consider all the multiples of 3 which are less than 50(because if 50 is considered then 50+1 is 51, which is 51), so numbers will be 0, 3,6...48(which are multiple of 3 less than 50) and last multiple is 48, that is 3*16, so total count is 17(including 0). Hence Answer is C

Remind that the number of terms in an arithmetic sequence is $\frac{l-f}{d} + 1$, where $l, f$ and $d$ are its last term, it first term and its common difference, respectively.

The numbers between $0$ and $50$, inclusive with the remainder $1$, when divided by $3$ are $1, 4, 7, ... , 49$, which is an arithmetic sequence.

Thus, the number of terms is $\frac{49-1}{3}+1 = \frac{48}{3}+1 = 16+1 = 17$.

Therefore, the right answer is C.

Although there are different (and clever) ways to answer this question, I think the fastest (and most accurate) approach is to simply list the values in your head as you count on your fingers (or use a tally sheet)

We get: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49

Answer: C


Cheers,
Brent

If you want to solve problems like these under 1 minute then always approach it by dividing the number by the given integer

how to know the pattern
3*1 + 1= 4 (rem = 1when div by 3)
3*2 + 1 = 7 (rem = 1when div by 3)
.....

50/3= 16.667
so we know 3*16 = 48 (hence 16 numbers are to be there which will give rem = 1) because 48 + 1 is < 50
finally add 1 because it'll give the same rem so
Answer is C ->17

@greenlightprep

Sir, but why do we include 1 in the list?

how can 1 divided by 3 yield a remainder of 1?

We need to find How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3y

Theory: Dividend = Divisor*Quotient + Remainder

Number, n -> Dividend
3 -> Divisor
q -> Quotient (Assume)
1 -> Remainder

=> n = 3*q + 1 = 3q + 1

( Watch this video to learn about Basics of Remainders )

Between 0 and 50
If we put q = 0, we will get the starting number as
n = 3*0 + 1 = 1

If we put q = 16, we will get the ending number as
n = 3*16 + 1 = 48 + 1 = 49

(Hint: here itself we can get the number of possible numbers as 16 + 1 = 17)

So, we have the numbers as 1 , 4, 7, ...., 49

This is an Arithmetic Sequence with

First term, a = 1
Common difference, d = 3
Last term, $T_n$ = 49

=> Number of terms, n = ($T_n$ - a) / d + 1 = $\frac{49 - 1}{3}$ + 1 = $\frac{48}{3}$ + 1 = 17

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Sequence problems