remainders......

[deadmank2s]

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?
A. 15
B. 16
C. 17
D. 18
E. 19

[shubham_k]

AAAAA When you divide by any number n, there are n possible remainders. For example, when we divide by 3, the 3 possible remainders are 0, 1 and 2. When we divide by 4, the 4 possible remainders are 0, 1, 2 and 3.

So, in this question there are 3 possible remainders.

From 0 to 50 inclusive there are 51 numbers. 51/3 = 17, therefore there are 17 numbers in the set that give each remainder. Thus, choose (C).

If 3 didn't divide evenly into 51, then we'd have to see if we had a "bonus" number in the set; but on this particular question, we see that 51 is evenly divisible by 3 and, if we understand the underlying concepts, get the correct answer in approximately 14 seconds!

[GMATGuruNY]

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?
A. 15
B. 16
C. 17
D. 18
E. 19

An integer that has a remainder of 1 when divided by 3 can be represented as follows:
3k + 1, where k is an integer greater than or equal to 0.

Since the integer must be less than 50:
3k + 1 < 50
3k < 49
k < 16.33.

Thus, 0≤k<16.33, implying that k could any of the 17 integers between 0 and 16, inclusive.
Since there are 17 values for k, there are 17 integers less than 50 that will have a remainder of 1 when divided by 3.
(The smallest is 3(0)+1 = 1; the greatest is 3(16)+1 = 49.)

The correct answer is C.

[Anurag@Gurome]

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?
A. 15
B. 16
C. 17
D. 18
E. 19

1 divided by 3 has a remainder of 1. When an integer a is divided by integer b, and if a is the smaller out of the two, then a will also be the remainder.

Example: 7/15 has remainder 7.

Now, 1 = 3 * 0 + 1
49 = 3 * 16 + 1
This is an AP with common difference = 3. So, 49 = 1 + 3(n - 1) or n = 17

The correct answer is C.