from 0 to 50 in ....

wilderness · Posted: Wed Jul 23, 2008 9:31 am Post subject: from 0 to 50 in ....

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

The OA is 16 with following explanation. But I think its wrong because its missing 1. So correct answer should be 17. What do you think ?

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

Canman · Posted: Wed Jul 23, 2008 10:48 am Post subject:

Can you explain how you come up with 17?

You can just write out the multiples of 3 and add 1 and count them all up to cross check. 16 as an answer appears to check out ok.

parallel_chase · Posted: Wed Jul 23, 2008 11:35 am Post subject:

The OA should be 17

1 has to be included simply because when 1 is divided by 3 it gives remainder 1.

Canman · Posted: Wed Jul 23, 2008 11:38 am Post subject:

Yeah I just saw the explanation in a previous post. Thanks!

malolakrupa · Posted: Thu Jul 24, 2008 8:14 am Post subject:

But 1 is not exactly divided by 3 ?

Canman · Posted: Thu Jul 24, 2008 8:22 am Post subject:

Right but the problem asks for REMAINDERS of 1 when divided by 3. So if we go back to the general eqn for a number of N=QD+r and rearrange to express as N/D=Q+D/D we find that 1=0*3+r (r=1) or to rearrange in other form of 1 divided by 3 --> 1/3 = 0 + 1/3....you can see you get a remainder of 1

N=number
Q=quotient
D=divisor
r=remainder

Canman · Posted: Thu Jul 24, 2008 8:23 am Post subject:

sorry. N/D=Q+r/D typo above