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by ellexay » Tue Feb 10, 2009 5:22 pm
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.
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by DeepakR » Tue Feb 10, 2009 5:32 pm
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
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by ellexay » Tue Feb 10, 2009 5:37 pm
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.
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by billzhao » Tue Feb 10, 2009 7:32 pm
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.
Yiliang
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by sureshbala » Tue Feb 10, 2009 9:26 pm
billzhao wrote: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.
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by x2suresh » Tue Feb 10, 2009 9:32 pm
agree with D

1 should be included.
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by raunekk » Tue Feb 10, 2009 9:44 pm
wats d OA??
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by vittalgmat » Wed Feb 11, 2009 12:57 am
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
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by ellexay » Wed Feb 11, 2009 6:19 am
Thanks to all! I understand now.

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