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# m20 #31 doubt on question phrasing?

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Senior Manager
Joined: 18 Aug 2009
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m20 #31 doubt on question phrasing? [#permalink]

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19 Nov 2009, 06:19
If the sum of $$N$$ consecutive odd integers is $$N^2$$ , what is the sum of all odd integers between 13 and 39 inclusive?

* 351
* 364
* 410
* 424
* 450

I calculated this as (the stem says N consecutive odd integers):
$$(39-13)/2 + 1=14$$
$$14^2=196$$, which is not in answer list.

OE says calculate the range 1-39 and subtract the range 1-11. $$20^2 - 6^2$$
I think this applies if the question stem says "sum of 1st N consecutive odd integers?"
Else please comment on which part is causing the misunderstanding...
Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
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Re: m20 #31 doubt on question phrasing? [#permalink]

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19 Nov 2009, 08:01
gmattokyo wrote:
If the sum of $$N$$ consecutive odd integers is $$N^2$$ , what is the sum of all odd integers between 13 and 39 inclusive?

* 351
* 364
* 410
* 424
* 450

I calculated this as (the stem says N consecutive odd integers):
$$(39-13)/2 + 1=14$$
$$14^2=196$$, which is not in answer list.

OE says calculate the range 1-39 and subtract the range 1-11. $$20^2 - 6^2$$
I think this applies if the question stem says "sum of 1st N consecutive odd integers?"
Else please comment on which part is causing the misunderstanding...

Yes. It has to be N consecutive odd integers starting from 1.

In fact, this is a property and not just specific for this particular question.

The property is that for n consecutive odd integers starting from 1 (or -1 when all values are negative),
(a) sum = $$n^2$$ for all positive values.
(b) sum = $$-n^2$$ for all negative values.

Another interesting property is that for n consecutive even integers starting from 0 (0 included) ,
(a) sum = $$n^2 - n$$ for all positive values.
(b) sum = $$-(n^2 - n)$$ for all negative values.

Note: In either case, the property is not valid when the set contains both positive and negative values.

Ps. We can even combine both of them to find the sum of N consecutive integers starting from 0.
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compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

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Senior Manager
Joined: 18 Aug 2009
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Re: m20 #31 doubt on question phrasing? [#permalink]

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20 Nov 2009, 09:25
Thanks for checking it. hope dzyubam has a look and confirms
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Re: m20 #31 doubt on question phrasing? [#permalink]

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20 Nov 2009, 15:45
I see. So would adding "first" clear the doubt?

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?

* 351
* 364
* 410
* 424
* 450
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Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
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Re: m20 #31 doubt on question phrasing? [#permalink]

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20 Nov 2009, 16:16
dzyubam wrote:
I see. So would adding "first" clear the doubt?

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?

* 351
* 364
* 410
* 424
* 450

You'd still have to specify that the numbers are positive.

Phrase it as 'sum of first $$N$$ consecutive natural numbers' instead. That should clear up the confusion.
Besides, it will also provide an added challenge for test takers to recognize the usage of 'natural numbers'.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Senior Manager
Joined: 18 Aug 2009
Posts: 303
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Kudos [?]: 200 [0], given: 9

Re: m20 #31 doubt on question phrasing? [#permalink]

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20 Nov 2009, 17:21
Thanks for checking this.
Yes, I think "first N consecutive odd integers" will be better (e.g. 5,7,9... can be considered as N consecutive integers, and the question doesn't mean that).
And as sriharimurthy said, "first N positive odd consecutive integers" would make it crystal clear what the Q is looking for, though sounds bit "wordy"
Not so sure about natural number though, didn't see "natural", "whole" used so much in gmat?
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Re: m20 #31 doubt on question phrasing? [#permalink]

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21 Nov 2009, 01:35
1
KUDOS
Another way to solve this is the following:

1st: Find the average of the numbers from 13-39. (13+39)/2=26
2nd: Count the consecutive odd numbers between 13-39, so 39-13=26 and then divide by 2 since its every other number, so 13+1=14 (N) to include the inclusive part.
3rd. Multiply the average by the N, so 26*14 = 364.

I think this is correct.
Re: m20 #31 doubt on question phrasing?   [#permalink] 21 Nov 2009, 01:35
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# m20 #31 doubt on question phrasing?

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