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# one more

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one more [#permalink]  16 May 2007, 14:48
try this
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Re: one more [#permalink]  16 May 2007, 18:07
Sergey_is_cool wrote:
try this

sum first 200 = n x (n+1)/2 = 200 +(200+1)/2 = 20,050
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Re: one more [#permalink]  16 May 2007, 18:31
Himalayan wrote:
Sergey_is_cool wrote:
try this

sum first 200 = n x (n+1)/2 = 200 +(200+1)/2 = 20,050

the formula is correct, but I guess you calculated it wrong 200*100.5=20100
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Re: one more [#permalink]  16 May 2007, 20:14
Sergey_is_cool wrote:
Himalayan wrote:
Sergey_is_cool wrote:
try this

sum first 200 = n x (n+1)/2 = 200 +(200+1)/2 = 20,050

the formula is correct, but I guess you calculated it wrong 200*100.5=20100

i too got by formula but wondering if there is an easier way ? why would they give the first 100 sum ? trying to connect the logic !
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Re: one more [#permalink]  16 May 2007, 20:20
Sergey_is_cool wrote:
try this

S = 5,050 + (100 x 100) + 5,050 = 20,100
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[#permalink]  16 May 2007, 22:22
the only reason I guess they gave us 100 and 5050 so we can check the formula (100+(100+1))/2=5050
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[#permalink]  16 May 2007, 22:26
Sergey_is_cool wrote:
the only reason I guess they gave us 100 and 5050 so we can check the formula (100+(100+1))/2=5050

The reason is in my solution, Sergey!
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[#permalink]  17 May 2007, 07:14
kirakira wrote:
Sergey_is_cool wrote:
the only reason I guess they gave us 100 and 5050 so we can check the formula (100+(100+1))/2=5050

The reason is in my solution, Sergey!

Can you explain your approach to solve this problem because I don't know if you used a specific formula or just logic or just guessed?
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[#permalink]  17 May 2007, 08:22
Sergey_is_cool wrote:
kirakira wrote:
Sergey_is_cool wrote:
the only reason I guess they gave us 100 and 5050 so we can check the formula (100+(100+1))/2=5050

The reason is in my solution, Sergey!

Can you explain your approach to solve this problem because I don't know if you used a specific formula or just logic or just guessed?

Just a simple logic, Sergey!

(1 + 2 + .... + 100) + (101 + 102 + ..... + 200) = (1 + 2 + .... + 100) + (1 + 2 + .... + 100) + (100 + 100 + .....) = 5,505*2 + 100*100
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[#permalink]  29 May 2007, 01:03
kirakira wrote:
Sergey_is_cool wrote:
kirakira wrote:
Sergey_is_cool wrote:
the only reason I guess they gave us 100 and 5050 so we can check the formula (100+(100+1))/2=5050

The reason is in my solution, Sergey!

Can you explain your approach to solve this problem because I don't know if you used a specific formula or just logic or just guessed?

Just a simple logic, Sergey!

(1 + 2 + .... + 100) + (101 + 102 + ..... + 200) = (1 + 2 + .... + 100) + (1 + 2 + .... + 100) + (100 + 100 + .....) = 5,505*2 + 100*100

just to clear things up. OA is E.

n * [(n+1) / 2]
200* (201/2)
100*201
20100

There is a simpler logic behind this formula. if u think about it, the sum of the first 200 integers is 1 through 200. take 1 and 200 and its 201. take 2 and 199 and its 201. so we have 200 numbers and they are paired up into 100 pairs. so its 201/2.

this is an investment banking interview question (although i never was asked it).
[#permalink] 29 May 2007, 01:03
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