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DS: Remainder

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DS: Remainder [#permalink]

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New post 19 Aug 2008, 06:39
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Refer to the question in the attachment.

My query is, in any DS question, if I find answer in variable term, will that be treated a unique answer. In this question, stmt 1 gives the remainder as 1/k and OA seems to be treating this as unique answer.

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Re: DS: Remainder [#permalink]

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New post 19 Aug 2008, 06:40
Oops....missed the attachment....here you go.
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Re: DS: Remainder [#permalink]

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New post 19 Aug 2008, 07:07
scthakur wrote:
Oops....missed the attachment....here you go.



(k+1)^3 = (K^2+2k+1)*(k+1)= k^3 + 3k^2 + 3k + 1

When you divide by k... remainder is always 1.
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Re: DS: Remainder [#permalink]

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New post 19 Aug 2008, 07:18
Or you can go with number picking, seems easier to me. If k=2, we have 27/2, remainder 1. For k=3, we have 64/3, remainder 1.

The algebraic solution gives more certainty, of course.
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Re: DS: Remainder [#permalink]

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New post 19 Aug 2008, 08:13
Nerdboy wrote:
Or you can go with number picking, seems easier to me. If k=2, we have 27/2, remainder 1. For k=3, we have 64/3, remainder 1.

The algebraic solution gives more certainty, of course.



use plugging-in only when there is no way to solve the question algebraically.
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Re: DS: Remainder [#permalink]

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New post 19 Aug 2008, 22:18
Oops.....I realized my mistake....I treated 1/k as remainder.

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Re: DS: Remainder   [#permalink] 19 Aug 2008, 22:18
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