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