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Manhattan's challenge!

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SVP
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Manhattan's challenge! [#permalink] New post 13 Oct 2005, 03:09
I came across this one, a very very gripping maths question. You may put a finger on it :wink: .

36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

(A) 14400
(B) 14440
(C) 14460
(D) 14500
(E) 14520

See if anyone have other solutions. I will put my own if no one has the same way of solving.
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 [#permalink] New post 13 Oct 2005, 03:45
It took me 5 min to solve this is equal to 9*36^2+2*36^2+204 which gives 14460
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 [#permalink] New post 13 Oct 2005, 04:07
36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

36^2 + (36+1)^2 + (36+2)^2 + (36+3)^2 + (36+4)^2 + (36+5)^2 + (36+6)^2 + (36+7)^2 + (36+8)^2 =

36^2 + (36^2 + 2(36)(1) + 1) + (36^2 + 2(36)(2) + 4) + (36^2 + 2(36)(3) + 9) + (36^2 + 2(36)(4) + 16) + (36^2 + 2(36)(5) + 25) + (36^2 + 2(36)(6) + 36) + (36^2 + 2(36)(7) + 49) + (36^2 + 2(36)(8) + 64) =

9(36^2) + 2(36)(1+2+3+4+5+6+7+8) + (1+4+9+16+25+36+49+64) = 11664 + 2592 + 204 =14460

(Time Taken: 2 mins, hope it's not too long !)
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 [#permalink] New post 13 Oct 2005, 04:32
I also got it by,

9(36^2) + 2(36)(1+2+3+4+5+6+7+8) + (1+4+9+16+25+36+49+64) = 14460.

but took slightly more than 2 mins.
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 [#permalink] New post 13 Oct 2005, 05:44
My way took about 2.5 minutes.

(40-4)^2+(40-3)^2+.....(40+4)^2
Because (x-y)^2 and (x+y)^2; all -2xy and 2xy products will cancel out,

leaving 40^2*9+2(4^2;3^2;2^2;1^2)

Do the straight math 14400+60= 14460

C.
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 [#permalink] New post 13 Oct 2005, 05:51
Bingo! ...GMATT73's way is the same to mine :wink:
Yeah, if the time is counted as soon as we figure out the shortcut, it gonna be much less. But is it true that everyone can find these shortcuts promptly?! ...hik,anyway, I guess grabbing a calculator and manipulate it this way takes shorter time :-D
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 [#permalink] New post 13 Oct 2005, 16:15
Took me about 2.5 mins....

I used the fact that 1^2 + 2^2 + .....+n^2 = (n(n+1)(2n+1))/6

1^2 + 2^2 + .... + 44^2 = 44*45*89/6 = 22*15*89 = 29370

1^2 + 2^2 + .....+35^2 = 35*36*71/6 = 426*35 = 14910

The answer is 29370 - 14910 = 14460.
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 [#permalink] New post 13 Oct 2005, 19:49
Thank you for reminding us of the formula :!:
The proving the of formula is http://www.cut-the-knot.org/Curriculum/ ... rate.shtml
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 [#permalink] New post 13 Oct 2005, 20:04
laxieqv wrote:
Thank you for reminding us of the formula :!:
The proving the of formula is http://www.cut-the-knot.org/Curriculum/ ... rate.shtml


Well, it's good to know the formula. But if you can't remember it, you can solve it the way we did. I think it's better to remember important formula and rules for the GMAT and if there's still capacity left in your brain, then memorise the formulas! :wink:
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 [#permalink] New post 13 Oct 2005, 20:38
sadsack wrote:
Took me about 2.5 mins....

I used the fact that 1^2 + 2^2 + .....+n^2 = (n(n+1)(2n+1))/6

1^2 + 2^2 + .... + 44^2 = 44*45*89/6 = 22*15*89 = 29370

1^2 + 2^2 + .....+35^2 = 35*36*71/6 = 426*35 = 14910

The answer is 29370 - 14910 = 14460.


Nice one

I didnt know that rule

what happened if it is 1^2+3^2+5^2 ? (if there is an arithmetic difference of 2 betwen each term) ? is there some adaptive formula or not ?
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Re: Manhattan's challenge! [#permalink] New post 13 Oct 2005, 22:13
laxieqv wrote:
I came across this one, a very very gripping maths question. You may put a finger on it :wink: .

36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

(A) 14400
(B) 14440
(C) 14460
(D) 14500
(E) 14520

See if anyone have other solutions. I will put my own if no one has the same way of solving.


Good one. This can be solved as follows..

36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =
(1^2 + 2^ 2 + .... + 44 ^2 ) - (1^2 + 2 ^2 + ... + 35^2)

The formula to compute sum of the squares of first N numbers is n(n+1)(2n+1)/6
Using this we get (1^2 + 2^ 2 + .... + 44 ^2 ) = (44*45*89)/6 = 29370
similarly (1^2 + 2 ^2 + ... + 35^2) = (35*36*71)/6 = 14910
Subtracting both we get 14460 - C

Thanks
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 [#permalink] New post 14 Oct 2005, 06:05
GMATT73's approach is the best because it will take you the least of time. When you see something like this you should be able to see that the middle term (40) is nice and even. You know that (a-1)^2=a^2+2a+1 and (a+1)^2=a^2+2a+1. You know if you start from the middle the second term of (a-1)^2 and (a+1)^2 would cancel out. You accomplish all these within 10 seconds in your head. Then you proceed to write what is left on paper. There are 9 a^2 (which is 9*1600). And 2 of (1+2^2+3^2+4^2). You add them up and get the final answer 14460 within one minute.
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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

  [#permalink] 14 Oct 2005, 06:05
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