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# M20Q31

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Senior Manager
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18 Nov 2008, 03:46
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The sum of first $$N$$ consecutive odd integers is $$N^2$$ . What is the sum of all odd integers between 13 and 39 inclusive?

(A) 351
(B) 364
(C) 410
(D) 424
(E) 450

Source: GMAT Club Tests - hardest GMAT questions

SOLUTION: m20q31-73009-20.html#p1106480
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20 Nov 2008, 20:55
5
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ventivish wrote:
The sum of first $$N$$ consecutive odd integers is $$N^2$$ . What is the sum of all odd integers between 13 and 39 inclusive?

(C) 2008 GMAT Club - m20#31

* 351
* 364
* 410
* 424
* 450

how about the way that is explained in the question:

The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400
The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36
The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364
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02 Feb 2010, 09:29
5
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A faster way to do this is

Number of odd numbers 39 inclusive = (39 - 13)/2 + 1 = 13 + 1 = 14
Average = (13+39)/2 = 26
26 * 14 = 364
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18 Nov 2008, 12:30
4
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Hey,

First the sum of odd numbers between1 to 39, which represents 20 odd consecutive numbers
is 20*20 = 400, 400 - (1+ 3 + 5 + 7 +9 +11) = 364.

Hope this clear....

Jugolo1
(give kudos)
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20 Nov 2008, 07:44
3
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S=13+15+.......+37+39

39=13+(n-1)2
=>n=14

S=0.5(14)(13+39)=364
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01 Jun 2009, 14:22
2
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The way this question is written makes a person use "their" way (meaning the test maker's way) and in reality, just going through add adding them all up doesn't take more than 60 seconds.

13
15
17
19
21
23
25
27
29
31
33
35
37
39

I approach adding these up as in pairs: 11 & 39, 13 & 37, 15 & 35, 17 & 33, 19 & 31, 21 & 29, 23 & 27. Each of these 6 pairs = 50, so 50 * 6 = 300. Then you have 39 and 25 left, that's 64, so 364.

Not the most "advanced" method, but it certainly won't take over 60 seconds to complete this way.
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06 Jun 2012, 10:45
2
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If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 13 and 39 inclusive?

A 351
B 364
C 410
D 424
E 450

[Reveal] Spoiler:
OA : B

I know its a straight forward question. But somehow I was unable to solve it within 2 minutes.
I took some time to figure out that
[Reveal] Spoiler:
39 = 20th term and 11 = 6 th term
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Re: Sum of Consecutive Integers [#permalink]

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06 Jun 2012, 18:33
2
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manulath wrote:
If the sum of first N consecutive odd integers is N^2, what is the sum of all odd integers between 13 and 39 inclusive?

A 351
B 364
C 410
D 424
E 450

[Reveal] Spoiler:
OA : B

I know its a straight forward question. But somehow I was unable to solve it within 2 minutes.
I took some time to figure out that
[Reveal] Spoiler:
39 = 20th term and 11 = 6 th term

Method of operation:

1. What topics are being tested here? (A) Evenly spaced sets and (B) Statistics

2. What do I know about evenly spaced sets?

(A) average = ( first term + last term ) / 2 = (39+13)/2 = 26
(B) # of terms in the set = [last term - first term)/multiple (which is 2) + 1 = (39-13)/2 + 1 = 14 terms

As such

Sum = 14 * 26 = 364

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17 Feb 2011, 06:15
1
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144144 wrote:
nitya34 wrote:
S=13+15+.......+37+39

39=13+(n-1)2
=>n=14

S=0.5(14)(13+39)=364

can someone please explain me this line -
39=13+(n-1)2

where can i c a summary of ways to know the amount of N in series?

thanks guys

So the problem states that it is every odd number in between 13 and 39 inclusive. To find N integers you just count off how many are included in this series (14). The line in question just sets up an equation to easily solve for N. Subtract 13, then divide by two then add one and you get N=14. To check just count off. By and large it is better to memorize that equation since there are some instances were counting off via a check is too time consuming.

Hope this helps.
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Re: Sum of Consecutive Integers [#permalink]

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06 Jun 2012, 18:24
1
KUDOS
manulath:

Fool proof
2-minute method, list the integers from 13 to 39 and sum them.

Better method
Let's use the equation given, we are looking for:

13 + 15 + ... + 39 = (1+3+ ... + 39) - (1+3+ ... + 11) = sum of odd numbers from 1 to 39 less sum of odd numbers from 1 to 11

Sum of odd numbers from 1 to 39 = 20^2
- there are twenty odd numbers: (39-1)/2+1 = 20
- use the equation given, 20^2

Sum of odd numbers from 1 to 11 = 6^2
- there are six odd numbers (11-1)/2+1 = 6

20^2 - 6^2 = 364
B)

Robust method
Question is for an arithmetic series, starting number = 13, constant difference = 2, number of items = 14
sum = 14*(13+(14-1)/2*2) = 14*13 = 364
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07 Jun 2012, 22:03
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it's the best that you write the whole matrix out, like -

1 3 5 7 9
11 13 15 17 19
21 .. .. .. ..
31 .. .. .. ..
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12 Mar 2013, 07:56
1
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Bunuel wrote:
The sum of first $$N$$ consecutive odd integers is $$N^2$$ . What is the sum of all odd integers between 13 and 39, inclusive?

A. 351
B. 364
C. 410
D. 424
E. 450

The sum of all odd integers between 13 and 39, inclusive equals to the sum of all integers from 1 to 39, inclusive minus the sum of all integers from 1 to 11, inclusive.

Since there are 20 odd integers from 1 to 39, inclusive then the sum of all integers from 1 to 39, inclusive is $$20^2$$;
Since there are 6 odd integers from 1 to 11, inclusive then the sum of all integers from 1 to 11, inclusive is $$6^2$$;

So, the required sum is $$20^2-6^2=364$$.

"The sum of all odd integers between 13 and 39, inclusive equals to the sum of all integers from 1 to 39, inclusive minus the sum of all integers from 1 to 11, inclusive."

Should be:

"The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive."

The explanation in the problem M20-31 from GMAT Club Tests is most likely copied from Bunuel's explanation and presents the same problem.
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23 Nov 2008, 11:34
GMAT TIGER wrote:
ventivish wrote:
The sum of first $$N$$ consecutive odd integers is $$N^2$$ . What is the sum of all odd integers between 13 and 39 inclusive?

(C) 2008 GMAT Club - m20#31

* 351
* 364
* 410
* 424
* 450

how about the way that is explained in the question:

The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400
The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36
The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364

Hi thank you,
this is the OA
However I used the same methodology, but I included 13 as in the sum of consecutive odd integers from 1 to 13 and subtracted that from 400, thereby getting 351 as my answer. Why do we only calculate the sum from 1 to 11?
I know I am missing something, sorry if this is a stupid question!
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04 Dec 2008, 17:30
ventivish wrote:
GMAT TIGER wrote:
ventivish wrote:
The sum of first $$N$$ consecutive odd integers is $$N^2$$ . What is the sum of all odd integers between 13 and 39 inclusive?

(C) 2008 GMAT Club - m20#31

* 351
* 364
* 410
* 424
* 450

how about the way that is explained in the question:

The sum of consecutive odd integers from 1 to 39 = [(39+1)/2]^2 = 400
The sum of consecutive odd integers from 1 to 11 = [(11+1)/2]^2 = 36
The sum of consecutive odd integers from 13 to 39 = 400 - 36 = 364

Hi thank you,
this is the OA
However I used the same methodology, but I included 13 as in the sum of consecutive odd integers from 1 to 13 and subtracted that from 400, thereby getting 351 as my answer. Why do we only calculate the sum from 1 to 11?
I know I am missing something, sorry if this is a stupid question!

"13 and 39 inclusive" so if you subtract 13 also then you are only finding the sum of all odd numbers from 15-39
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02 Feb 2010, 21:19
th sum of first N odd integers = $$N^2$$

so the sum of 1 to 39 inclusive
N=(39-1)/2+1=20
so $$N^2=400$$

so the sum of 1 to 11 inclusive
N=(11-1)/2+1 =6

so $$N^2=36$$

so the sum of 13 to 39 inclusive = 400-36 =364
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07 Feb 2011, 06:51

20^2-6^2
400-36
364
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07 Feb 2011, 12:23
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16 Feb 2011, 19:20
(13+39) = 52

52/2 = 26

26x14 = 364

I would estimate this question has to be on the far easier side.
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16 Feb 2011, 22:39
nitya34 wrote:
S=13+15+.......+37+39

39=13+(n-1)2
=>n=14

S=0.5(14)(13+39)=364

can someone please explain me this line -
39=13+(n-1)2

where can i c a summary of ways to know the amount of N in series?

thanks guys
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09 Feb 2012, 05:22
Total no of terms = n=((39-13)/2)+1 = 14
so sum = 14/2(39+13) = 364 so OA is B
Re: M20Q31   [#permalink] 09 Feb 2012, 05:22

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# M20Q31

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