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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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02 Aug 2010, 02:06

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

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