n consecutive positive integers

sum of is 45? are you missing somthing?

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)

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

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.

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

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

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

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.

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.

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

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.

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.

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?

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