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If n is a positive integer, is (12+22+…..+n2)-(1+2+……+n) divisible by

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If n is a positive integer, is (12+22+…..+n2)-(1+2+……+n) divisible by  [#permalink]

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New post 31 Mar 2017, 07:18
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If n is a positive integer, is \((1^2+2^2+…..+n^2) - (1+2+……+n)\) divisible by 8?

1) n is an odd number
2) \(14 < n < 17\)

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Re: If n is a positive integer, is (12+22+…..+n2)-(1+2+……+n) divisible by  [#permalink]

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New post 31 Mar 2017, 08:49
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ziyuen wrote:
If n is a positive integer, is \((1^2+2^2+…..+n^2) - (1+2+……+n)\) divisible by 8?

1) n is an odd number
2) \(14 < n < 17\)


Hi

\((1^2+2^2+…..+n^2) - (1+2+……+n)\) = \(\frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)}{2}\)

simplifying this we'll get \(\frac{n(n + 1)(n - 1)}{3}\)

Three consecutine numbers will be divisible by 3 but we are interested in divisibiliy by \(2^3\), so we'l take into account only numerator.

1) n is an odd number

\(n = 2x + 1\)

\((2x + 1)(2x + 1 + 1)(2x + 1 - 1) = (2x + 1)(x + 1)x*4\)

from the x(x + 1) either x or (x + 1) will be even. Sufficient.

2) \(14 < n < 17\)

\(n=15\) or \(n=16\)

just plug in our values:

\(n=15\) ------> numerator = \(14*15*16\)

\(n=15\) ----> \(numerator = 16*17*15\). In both cases diisible by 8. Sufficient.

Answer D.
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Re: If n is a positive integer, is (12+22+…..+n2)-(1+2+……+n) divisible by  [#permalink]

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New post 01 Apr 2017, 07:08
Is there any way we can avoid arithmetic progression formulas ? Remembering sum of consecutive squares might be tricky. Thanks.

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Re: If n is a positive integer, is (12+22+…..+n2)-(1+2+……+n) divisible by &nbs [#permalink] 01 Apr 2017, 07:08
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