francoisph wrote:How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?
A.14 B.15. C.16 D.17 E.18

I am getting D (17)

50/3 = 16; there are 16 numbers between 3 to 50 that are divisible by 3

therefore there must be 16 numbers that are not divisible by 3 and leave a remainder 1

ex: 4, 7, 10 , ....49


apart from the above values, "1" if divided by 3 leaves a remainder 1

therefore, we have a total of 16 +1=17 numbers between 0 to 50 inclusive that leave a remainder 1 when divided by 3.

HTH!

Krazy, I agree with you and I like ur method.

Alternative method:

0 < 3x + 1 < 50 , where x is an interger.

So, 3(0) + 1 = 1, 3(1) + 1 = 4, 3(2) + 1= 7, 3(3) + 1=10, ........ 3(16) + 1=49.

Thus, 16 - 0 + 1 = 17, meaning we have 16 ( from 1 - 16 and 1 when x is zero.

D

imo C is answer

we cant take 1 as it is not divisble by 3 1/3=0.33 can we say that remainder is 1...

hence 16 should be the answer

what is oa

Hi,

1 should be considered and the answer IMO is 17.

Pls post the OA.

Can we get an OA on this please?

I have option C - 16 as the answer.

Answer should be 17. D

pradeepkaushal9518 wrote:imo C is answer

we cant take 1 as it is not divisble by 3 1/3=0.33 can we say that remainder is 1...

hence 16 should be the answer

what is oa

Hi,

every non-negative integer when divided by a positive integer provides a quotient and a remainder.

1 divided by 3 has a quotient of 0 and a remainder of 1.

Another way you can think of remainders is as the numerator in the fraction of a (non-simplified) mixed fraction.

For example, 7/3 can be written as 2 and 1/3. So, the remainder when we divide 7 by 3 is 1.

1/3 can be written as 0 and 1/3. So, the remainder when we divide 1 by 3 is also 1.

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?
A.14 B.15. C.16 D.17 E.18

so basically we are looking for the number of the form 3k+1
where k can take the value of integer .
constraint being it should be between 0 to 50
3k+1
if k=0
then first such integer will be 1
next put k=1
we get intger as 4
so we can say last intger will be for k =16 i.e 3*16+1=49
total number of integer between 0 to 50 will be 1to 16
so 17 possible value