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by wilderness » Wed Jul 23, 2008 8:31 am
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
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by Canman » Wed Jul 23, 2008 9:48 am
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.
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by parallel_chase » Wed Jul 23, 2008 10:35 am
The OA should be 17

1 has to be included simply because when 1 is divided by 3 it gives remainder 1.
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by Canman » Wed Jul 23, 2008 10:38 am
Yeah I just saw the explanation in a previous post. Thanks!
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by malolakrupa » Thu Jul 24, 2008 7:14 am
But 1 is not exactly divided by 3 ?
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by Canman » Thu Jul 24, 2008 7:22 am
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
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by Canman » Thu Jul 24, 2008 7:23 am
sorry. N/D=Q+r/D typo above
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