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Arithmatic Progression Problem - Advise is Needed

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New post 01 Aug 2010, 23:09
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Hello,

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

if we arrange this in AP, we get
4+7+10+.......+49
so 4+(n-1)3=49: n=16

Should we consider 1 in the AP as well? If we do so, then the answer becomes 17.
Please advise.


regards,
Jack
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New post 02 Aug 2010, 00:22
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Yes, you must include 1, since when you divide 1 by 3, the remainder is 1 (and the quotient is zero). That should be obvious if you think of how you first learned division: if you have 1 apple and 3 children, and you need to give each child an equal number of apples, you can't give the children any apples (i.e. the quotient is zero) and you have 1 apple left over (that's the remainder).
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Re: Arithmatic Progression Problem - Advise is Needed [#permalink]

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New post 02 Aug 2010, 01:06
You can turn it into a basic problem.
In fact, integers with a remainder of 1 when divided by 3 have the following shape:
3k+1 with k=0, 1, 2, and so on.
Then:
0<=3k+1<=50
\(\frac{0-1}{3}<=k<=\frac{50-1}{3}\)
\(\frac{-1}{3}<=k<=16+\frac{1}{3}\)
k=0, 1, 2, ..., 15, 16

Result=17!
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New post 02 Aug 2010, 04:30
I made the mistake of not including 1 :(
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Re: Arithmatic Progression Problem - Advise is Needed [#permalink]

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New post 02 Aug 2010, 15:53
I think it's much easier to list them here.


the sequence is the following 3n+1, where n = 0,1,2,3...........

1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49

so, 17.

the trick is not to forget about 1. U cath it, you'll have it.
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Re: Arithmatic Progression Problem - Advise is Needed   [#permalink] 02 Aug 2010, 15:53
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