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# The sum of the odd positive integers from 1 to k equals 441.

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Manager
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The sum of the odd positive integers from 1 to k equals 441. [#permalink]  10 Jul 2007, 15:09
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Expln Asw...
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this is an easy problem. K can be an odd or even integer.
If K is odd then the number of odd numbers between 1 and K is (k +1)/2
If K is even then the number of odd numbers between 1 and K is (k)/2

The stem doesnt say whether k is odd or even. So calculate for both scenarios

If k is odd

From Stem
( (k+1)/2*[2 + ((k+1)/2 - 1) * 2] )/2 = 441
From this K = 41. We see the choice there.

Just for fun
If k is even then
k/2[2 + (k/2 -1)*2]/2 = 441 from this k = 42. This choice isnt there.
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Quote:
this is an easy problem. K can be an odd or even integer.
If K is odd then the number of odd numbers between 1 and K is (k +1)/2
If K is even then the number of odd numbers between 1 and K is (k)/2

The stem doesnt say whether k is odd or even. So calculate for both scenarios

If k is odd

From Stem
( (k+1)/2*[2 + ((k+1)/2 - 1) * 2] )/2 = 441
From this K = 41. We see the choice there.

Just for fun
If k is even then
k/2[2 + (k/2 -1)*2]/2 = 441 from this k = 42. This choice isnt there.

Can you please explain the formula for: figuring out how K=41?
for the equation: ( (k+1)/2*[2 + ((k+1)/2 - 1) * 2] )/2 = 441
- what does (k+1)/2 represent?
- what does [2+((k+1)/2-1)*2 represent?
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dahcrap wrote:
this is an easy problem. K can be an odd or even integer.
If K is odd then the number of odd numbers between 1 and K is (k +1)/2
If K is even then the number of odd numbers between 1 and K is (k)/2

The stem doesnt say whether k is odd or even. So calculate for both scenarios

If k is odd

From Stem
( (k+1)/2*[2 + ((k+1)/2 - 1) * 2] )/2 = 441
From this K = 41. We see the choice there.

Just for fun
If k is even then
k/2[2 + (k/2 -1)*2]/2 = 441 from this k = 42. This choice isnt there.

yep, it must be a formula... it is easy not until you know how it works...???
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Re: Sum of first n odd positive integers... [#permalink]  10 Jul 2007, 17:46
boubi wrote:
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Expln Asw...

n = sqrt (441) = 21
so k shouild be (21 x 2) - 1 = 41
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Back test:

Use 37. # of odd integers = 19

441 = 19/2(2(1) + 18(2))? -->361 too low

Use 47. # of odd integers = 24

441 = 24/2(2(1) + 23(2))? --> 576. too high.

must be 41.

Ans B
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boubi wrote:
dahcrap wrote:
this is an easy problem. K can be an odd or even integer.
If K is odd then the number of odd numbers between 1 and K is (k +1)/2
If K is even then the number of odd numbers between 1 and K is (k)/2

The stem doesnt say whether k is odd or even. So calculate for both scenarios

If k is odd

From Stem
( (k+1)/2*[2 + ((k+1)/2 - 1) * 2] )/2 = 441
From this K = 41. We see the choice there.

Just for fun
If k is even then
k/2[2 + (k/2 -1)*2]/2 = 441 from this k = 42. This choice isnt there.

yep, it must be a formula... it is easy not until you know how it works...???

It is not some unknown formula.

It is a simple formula for sum of n numbers

S = n/2[2*a + (n-1)*d ]
n = number of numbers in the sequence
d - difference between contiguous elements in the sequence
a - first number in the equence
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Re: Sum of first n odd positive integers... [#permalink]  11 Jul 2007, 15:11
what crazy math is this??? lol..

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

A) 47
B) 41
C) 37
D) 33
E) 29

Expln Asw...

n = sqrt (441) = 21
so k shouild be (21 x 2) - 1 = 41
Manager
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Excellent stuff guys. I had forgotten this formula.
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Wuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuut???? lol

what happened to sum=avg*#of terms???

thats some crazy stuff
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Re: Sum of first n odd positive integers... [#permalink]  12 Jul 2007, 05:55
Himalayan wrote:
boubi wrote:
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Expln Asw...

n = sqrt (441) = 21
so k shouild be (21 x 2) - 1 = 41

Great!

I see it is still the same formula, indeed very simple

(average of odds terms from 1 to K)*(Nb of terms in the sequence)= SUM

( (1+k)/2 ) * ( (k+1)/2 ) = 441 = 21 * 21

interesting parallelism, with the square root, your formula assumes that avg of odds terms from one to k is always equal to the nb of terms in the sequence...isn't it?

hence

( (1+k)/2 )=21
1+k=42
k=41

hope i am right.
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The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Sum = (No. of terms in seq) * (Average)

Average = (first term + last term)/2
Average = (k + 1)/2

Num. of terms = [(last-term - first-term)/2] + 1
Num. of terms = (k - 1)/2 + 1
Num. of terms = (k + 1)/2

Sum = (k + 1)/2 * (k + 1)/2
441 = (k + 1)^2/4
(k + 1)^2 = 49 * 9 * 4
K + 1 = 7 * 3 * 2
k = 41
B
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asaf wrote:
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Sum = (No. of terms in seq) * (Average)

Average = (first term + last term)/2
Average = (k + 1)/2

Num. of terms = [(last-term - first-term)/2] + 1
Num. of terms = (k - 1)/2 + 1
Num. of terms = (k + 1)/2

Sum = (k + 1)/2 * (k + 1)/2
441 = (k + 1)^2/4
(k + 1)^2 = 49 * 9 * 4
K + 1 = 7 * 3 * 2
k = 41
B

You are assuming K is odd here. What if K is even Remember if K was 42, the sum is still 441
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dahcrap wrote:
asaf wrote:
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Sum = (No. of terms in seq) * (Average)

Average = (first term + last term)/2
Average = (k + 1)/2

Num. of terms = [(last-term - first-term)/2] + 1
Num. of terms = (k - 1)/2 + 1
Num. of terms = (k + 1)/2

Sum = (k + 1)/2 * (k + 1)/2
441 = (k + 1)^2/4
(k + 1)^2 = 49 * 9 * 4
K + 1 = 7 * 3 * 2
k = 41
B

You are assuming K is odd here. What if K is even Remember if K was 42, the sum is still 441

yes that 's it! either k=42 or k=41 the sum will be the same 441, because we are asked for the sum of only odd integers in [1;k]

but if we are assuming such a thing, it is to play on the obvious shortcut.
knowing that 21*21=(k+1)/2*(k+1)/2
21=(k+1)/2
k=41

Assuming k is odd and you have all the answers that is either k=41 since it will be the last odd integer in the evenly spaced sequence or k=42 since it could be the next even integer but still not accounted in the sum of odd integers.
No reason to bother yourself dude...
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By the wat OA is B, good job guys
i learned a lot of stuff on this one!
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Simple, use following Formulas

tn=a + (n-1)d
sn=n(2a+ (n-1)d)/2

where a=first term (1)
d=common diff (2)
tn=nth term...k
sn=sum..441

plug in and get the answer!

FP
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Re: Sum of first n odd positive integers... [#permalink]  13 Jul 2007, 07:54
boubi wrote:
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Expln Asw...

If I use the AP formula I get K=21 (obviously wrong).

Sum = (n/2)(2a+(n-1)d) and Sum =441, a=1, d=2, n=k
441 = k/2(2+2k-2)
441=k^2 => k=21

Why is applying this formula doesn't get me the right result?
Director
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Folks,

I too used formula for arithmetic progression.
Sum of n terms = n/2(2a+(n-1)d)

where n = number of terms
a = first term
d= common difference

I hace come across many problems like this which can be easily solved by applying these formulas. This is a very important formula which we should remember. Also we should remember formuls for geometric progression
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asaf wrote:
The sum of the odd positive integers from 1 to k equals 441.
What is the value of k?

A) 47
B) 41
C) 37
D) 33
E) 29

Sum = (No. of terms in seq) * (Average)

Average = (first term + last term)/2
Average = (k + 1)/2

Num. of terms = [(last-term - first-term)/2] + 1
Num. of terms = (k - 1)/2 + 1
Num. of terms = (k + 1)/2

Sum = (k + 1)/2 * (k + 1)/2
441 = (k + 1)^2/4
(k + 1)^2 = 49 * 9 * 4
K + 1 = 7 * 3 * 2
k = 41
B

?? # of odd terms = (last term - first term)/2 + 1??

how many odd terms between 10 and 5 inclusive? 3 terms (5 7 9)

how many odd terms between 10 and 5 inclusive (10-5)/2 +1 = 3.5 ??

im just learning how to use AP, I never used it like this however.

Since AP isn't part of the official guide - how would someone figure this out
without the AP formula? do they really expect someone to just derive this off the top of their head?
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Guys, I usually remember this another formula as well:

Sum of AP = n/2 (A1+An), where A1 is first term and An is nth term.

In our question,

Sum of AP = 441
n = 21 (using standard formula for sum of AP)
A1 = 1
An = ?

Just plug in the numbers and you will get An = 41.
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