Number properties

[ellexay]

51. 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

Soln: if we arrange this in AP, we get
4+7+10+.......+49

so 4+(n-1)3=49: n=16
C is my pick

I don't know what the OA is, but can someone tell me if C is indeed correct?

Thank you.

[DeepakR]

I think you have to include the number 1 as well.
1,4,7..49 hence a=1 and tn=49 so 49=1+(n-1)3. Therefore n=17 Ans.) D

- Deepak

[ellexay]

Deepak,

I thought the question wanted integers that, when divided by 3, leave R1. That way, 1 wouldn't be included in the answer set.

[billzhao]

I think that 1 should be included in the set because:
1. 1 is an integer
2. 1 is in the range from 0 to 50
3. 1 gives an remainder of 1 when dividing 3.

[sureshbala]

I think that 1 should be included in the set because:
1. 1 is an integer
2. 1 is in the range from 0 to 50
3. 1 gives an remainder of 1 when dividing 3.

Perfect....1 must be included.

[x2suresh]

agree with D

1 should be included.

[raunekk]

wats d OA??

[vittalgmat]

Here is an easy way.

50/3 = 16 integers.
However, this does not include 1, coz 1 divided by 3 has remainder =1.
So add 1.
==> 17 ie D

[ellexay]

Thanks to all! I understand now.

Unfortunately, the source from which I got this problem didn't have the OA for this one =(