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

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Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1322
Location: Malaysia
If n is a positive integer, is (12+22+…..+n2)-(1+2+……+n) divisible by  [#permalink]

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31 Mar 2017, 06:18
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00:00

Difficulty:

95% (hard)

Question Stats:

41% (03:02) correct 59% (02:49) wrong based on 38 sessions

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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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Joined: 13 Oct 2016
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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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31 Mar 2017, 07:49
2
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.

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Joined: 18 Feb 2017
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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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01 Apr 2017, 06:08
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, 06:08
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