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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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Intern
Joined: 25 Nov 2012
Posts: 3

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Updated on: 08 Sep 2017, 22:53
3
2
00:00

Difficulty:

35% (medium)

Question Stats:

69% (01:30) correct 31% (02:08) wrong based on 113 sessions

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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

Originally posted by yaj on 08 Aug 2013, 13:11.
Last edited by Bunuel on 08 Sep 2017, 22:53, edited 2 times in total.
Edited the question, renamed the topic.
Director
Joined: 14 Dec 2012
Posts: 742
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6

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08 Aug 2013, 13:16
8
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

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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##### General Discussion
Intern
Joined: 25 Nov 2012
Posts: 3

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08 Aug 2013, 13:28
That was easy! Thanks
Senior Manager
Joined: 10 Jul 2013
Posts: 312

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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

OA:B

Thanks so much!!

...

make a common of 9.
so 9 * 285 = 2565
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Asif vai.....
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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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Joined: 04 Dec 2015
Posts: 750
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)

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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;

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

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

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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