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n consecutive positive integers

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n consecutive positive integers

by logitech » Sun Dec 07, 2008 4:38 pm
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

OA E
Last edited by logitech on Sun Dec 07, 2008 4:49 pm, edited 1 time in total.
LGTCH
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by hwiya320 » Sun Dec 07, 2008 4:45 pm
sum of is 45? are you missing somthing?
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by cramya » Sun Dec 07, 2008 5:18 pm
Did u post it in the wrong section? :-)

Stmt I

n is even

n = 2

22 23

n=6

5 6 7 8 9 10

INSUFF


Stmt II

n<9

Same 2 cases above

Combined

same 2 cases above

INSUFF

Choose E)
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by hwiya320 » Sun Dec 07, 2008 5:22 pm
I guess you answered it before I got to finish typing it..

No, I didn't post it in the wrong section. You forgot the "n" in your first post I thought?

anyways, i had the same logic behind what you wrote.

when n=even, then 22,23

when n<9, 45, 22,23, or 14,15,16..

too many choices

together, well, 22,23, or 5,6,7,8,9,10..

insuf
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by cramya » Sun Dec 07, 2008 5:40 pm
No, I didn't post it in the wrong section
Oops, I meant Logitech since this was a ds question on the ps section of the forum.
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by logitech » Sun Dec 07, 2008 5:59 pm
cramya wrote:
No, I didn't post it in the wrong section
Oops, I meant Logitech since this was a ds question on the ps section of the forum.
This was a tricky DS question and I made it even more trickier by posting it in wrong section :oops:
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by cramya » Sun Dec 07, 2008 6:03 pm
This was a tricky DS question and I made it even more trickier by posting it in wrong section
We can only take so much tricks! The 2nd one(more trickier by posting it in wrong section) was a classic :-)
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by ravikirancheni » Thu Dec 11, 2008 5:50 am
well whatever it might be.....
The answer would always be E
The reason being it was said that ----> n(n+1)/2 = 45 ==> n = 9.
we need no supporting statements a,b so the ans is E
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by adilka » Thu Dec 11, 2008 1:12 pm
Im confused as to how you actually come up with all the possible combinations. It took me a few mins to manually pick the numbers and then determine that neither (1) nor (2) are sufficient.
THere's gotta be a quicker way/formula.

PS: I also didnt follow this last formula from ravikirancheni
ravikirancheni wrote: The reason being it was said that ----> n(n+1)/2 = 45 ==> n = 9.
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by 720dreaming » Thu Dec 11, 2008 1:39 pm
Took me a few min too, had 22, 23 right away, but others came more slowly. Also looking for a formula...

As for the n(n+1)/2 formula that just ads up all the numbers up to a certain number starting with 1. So I am also not quite sure how that is being used.
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by weena82 » Thu Dec 11, 2008 10:23 pm
I've got one point to ask; if the stem said n consecutive integers, can we conclude that it start with 1? If not we can not use the formular (n(n+1))/2
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by cramya » Sat Dec 13, 2008 12:02 am
I've got one point to ask; if the stem said n consecutive integers, can we conclude that it start with 1?
The stem does say n consecutive positive intgers but we cannot assume it starts at 1( unless explicitly stated)

So we cannot use n(n+1)/2 formula here.
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by earth@work » Sat Dec 13, 2008 11:19 am
sum = (a+1) + (a+2) +.....(a+n) for any positive integer a
= na+n(n+1)/2 = 45
now to find the value of n we need to know 'a' which is given in neither of statements...so both insuff
ans E .... do let me know if u see some error here, as this is the best i cud think of without plugging nos.
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by adilka » Sat Dec 13, 2008 12:31 pm
earth@work wrote:sum = (a+1) + (a+2) +.....(a+n) for any positive integer a
= na+n(n+1)/2 = 45
now to find the value of n we need to know 'a' which is given in neither of statements...so both insuff
ans E .... do let me know if u see some error here, as this is the best i cud think of without plugging nos.
There is a slight error in your calculations. It should be na+n(n-1)/2=45
i.e. (n-1) not (n+1) the reason for that is that the first number in sequence, which you assumed to be "a" has an adder of 0, hence you have (n-1) numbers that start with 1.

I also tried to use this logic, but it led me to nowhwere. I argue that even though we do end up with an equation with 2 variables - a and n that are unknown, there are some restrictions given on these variables, hence we cannot definitively say that the answer is E based on this formula alone (since restrictions can potentially eliminate multiple potential solutions).
Restrictions are:

1. Both a and n are integers
2. n is even
3. n <9

Anyone has a solution? Stuart? Ron?
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by earth@work » Sat Dec 13, 2008 2:53 pm
adilka wrote:
earth@work wrote:sum = (a+1) + (a+2) +.....(a+n) for any positive integer a
= na+n(n+1)/2 = 45
now to find the value of n we need to know 'a' which is given in neither of statements...so both insuff
ans E .... do let me know if u see some error here, as this is the best i cud think of without plugging nos.
There is a slight error in your calculations. It should be na+n(n-1)/2=45
i.e. (n-1) not (n+1) the reason for that is that the first number in sequence, which you assumed to be "a" has an adder of 0, hence you have (n-1) numbers that start with 1.
Hi adilka,
my first number of the sequence is (a+1) and not 'a', that is the reason why my sum is na+n(n+1)/2
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