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The sum of the odd positive integers from 1 to k equals to

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Senior Manager
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The sum of the odd positive integers from 1 to k equals to [#permalink]

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New post 15 Aug 2008, 09:21
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The sum of the odd positive integers from 1 to k equals to 441. What is the value of k?

47
41
37
33
29

Kudos [?]: 68 [0], given: 0

Director
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Joined: 12 Jul 2008
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Kudos [?]: 162 [0], given: 0

Schools: Wharton
Re: PS: Sum to 441 [#permalink]

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New post 15 Aug 2008, 12:23
judokan wrote:
The sum of the odd positive integers from 1 to k equals to 441. What is the value of k?

47
41
37
33
29


Fastest way, I think, is just to do the addition. There is a trick for adding evenly spaced numbers. Just do the lowest number first. Then you can do quick addition to get the other ones.

If you want to do it mathematically:

Number of odd digits between 1 and k = (k+1)/2

If (k+1)/2 is odd, then

Sum of digits = ((k-1)/4)*(k+1) + (k+1)/2

If (k+1)/2 is even, then

Sum of digits = ((k+1)/4)*(k+1)

A: 576
B: 441
C: 380
D: 289
E: 225

Kudos [?]: 162 [0], given: 0

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Re: PS: Sum to 441 [#permalink]

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New post 15 Aug 2008, 13:02
You can also know that

Sum of odd positive integers from 1 to \(k = (\frac{k+1}{2})^2\)

so here, we have 47+1= 48 / 2 = 24 * 24 > 441

41+1=42/2 = 21 * 21 = 441 - bingo

for EVEN NUMBERS

Sum of even positive integers from 2 to k where k is even \(= (\frac{k}{2})*(\frac{k}{2}+1)\)
Example:

The sum of 2 to k = 30

\((\frac{k}{2})*(\frac{k}{2}+1)=30\)

\((\frac{k^2}{4}+\frac{k}{2})=30\)

\((\frac{k^2}{4}+\frac{2k}{4})=30\)

\(\frac{k^2 + 2k}{4}=30\)

\(k^2 + 2k=120\)

\(k^2 + 2k - 120 = 0\)

\((k + 12)(k - 10) = 0\)

k = -12 or k = 10

k must be positive, so k = 10

judokan wrote:
The sum of the odd positive integers from 1 to k equals to 441. What is the value of k?

47
41
37
33
29

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Kudos [?]: 615 [0], given: 32

Re: PS: Sum to 441   [#permalink] 15 Aug 2008, 13:02
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