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If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=?

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If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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New post 08 Aug 2013, 13:11
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Question Stats:

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If \(1^2+2^2+3^2+...+9^2=285\), \(3^2+6^2+9^2+12^2+15^2+...+27^2=?\)

(A) 855
(B) 2,565
(C) 7,695
(D) 23,085
(E) 69,255

Source: Unknown
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Sep 2017, 22:53, edited 2 times in total.
Edited the question, renamed the topic.

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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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New post 08 Aug 2013, 13:16
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yaj wrote:
Hello!

Anyway to solve this one ?

If 1^2+2^2+3^2+....+9^2=285, 3^2+6^2+9^2+12^2+15^2+¡-27^2=?
(A) 855
(B) 2,565
(C) 7,695
(D) 23,085
(E) 69,255

Source: Unknown

[Reveal] Spoiler:
OA:B

Thanks so much!!

\(3^2+6^2+9^2+12^2+15^2+....+27^2\)
=\(3^2(1^2+2^2+3^2+....+9^2)\)===>JUST TAKING OUT 3^2 COMMON FROM EACH TERM
=\(9*285\)
=\(2565\)
HENCE B

hope it helps
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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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New post 08 Aug 2013, 13:28
That was easy! Thanks :)

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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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New post 08 Aug 2013, 22:32
yaj wrote:
Hello!

Anyway to solve this one ?

If 1^2+2^2+3^2+....+9^2=285, 3^2+6^2+9^2+12^2+15^2+¡-27^2=?
(A) 855
(B) 2,565
(C) 7,695
(D) 23,085
(E) 69,255

Source: Unknown

[Reveal] Spoiler:
OA:B


Thanks so much!!

...

make a common of 9.
so 9 * 285 = 2565
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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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New post 29 Jul 2017, 23:52
Since \(1^2+2^2+3^2+....+9^2\)=285,

\(3^2*1^2 + 3^2*2^2 + 3^2*3^2 + 3^2*4^2 + 3^2*5^2 + .... + 3^2*9^2\)
= \(3^2(1^2+2^2+3^2+4^2+5^2+....+9^2)\)
= \(3^2*285\)
= \(9*285 = 2565\)(Option B)
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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=? [#permalink]

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New post 16 Aug 2017, 23:12
yaj wrote:
Hello!

Anyway to solve this one ?

If 1^2+2^2+3^2+....+9^2=285, 3^2+6^2+9^2+12^2+15^2+¡-27^2=?
(A) 855
(B) 2,565
(C) 7,695
(D) 23,085
(E) 69,255

Source: Unknown

Thanks so much!!

\(1^2+2^2+3^2+....+9^2=285\) ------ (i)

Multiplying \(3^2\) to (i), we get;

\((3^2)(1)+(3^2)(2^2)+(3^2)(3^2)+(3^2)(4^2)+(3^2)(5^2)+...+(3^2)(9^2)= (3^2)(285) = (9)(285)\)

\((3^2)+(6^2)+(9^2)+(12^2)+(15^2)+...+(27^2)= 2565\)

Answer (B)...

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Re: If 1^2+2^2+3^2+...+9^2=285, 3^2+6^2+9^2+12^2+15^2+...+27^2=?   [#permalink] 16 Aug 2017, 23:12
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