How many integers from 0 to 50, inclusive, have a remainder

[mitzwillrockgmat]

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

this is an easy question. when i was doing a test it came up & i picked 16 as the answer as i consider the remainder 1 for every mulitple of 3 (betw 0 & 50) + 1 as the no. giving a remainder of 1. i.e. 3+1 = 4 will give remainder 1 , 6+1 =7 will give remainder 1. hence, there are 16 multiples of 3 between 0 & 50 so i picked b. however, the answer is 17. so my question is should i include 1 in the set as well to make it 17 numbers or is there ANOTHER no. that i'm missing??? but one is not a multiple of 3!

HELP! :0)

[Patrick_GMATFix]

1 divided by 3 has a remainder of 1.

In general whenever integer p is divided by integer q, if p is smaller, then p will also be the remainder.

5/12 has remainder 5.

Best of luck,
-Patrick

[clock60]

the first number is 1
1=3*0+1, and the last 49, 49=3*16+1
follow AP with difference 3
49=1+3(n-1)
n-1=16
n=17

[tpr-becky]

The above are correct - nothing in the question says that the number has to be a multiple of three - just that the number divided by three have a remainder of 1. the number one fits this scenario.

[sethisankalp]

Here is how I thought about it:

1) What is the number of multiples of 3 between 0-50: 17
2) So all other integers (51 - 17 = 34) are either leave a remainder 1 or 2
3) Number of integers that leave a remainder of 1 = 34/2 = 17, number of integers that leave a remainder of 2 = 34/2 = 17

[Scott@TargetTestPrep]

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

The first number that has a remainder of 1 when divided by 3 is 1, and the last number is 49.

Thus, the number of integers from 0 to 50 inclusive that have a remainder of 1 when divided by 3 is:

(49 - 1)/3 + 1 = 17

Answer: C