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The sum of n consecutive positive integers is 45. What is

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The sum of n consecutive positive integers is 45. What is [#permalink]

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The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even
(2) n < 9
[Reveal] Spoiler: OA

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Last edited by Bunuel on 12 Jun 2013, 03:33, edited 2 times in total.
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enigma123 wrote:
The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9


For me its C. Any thoughts guys?


The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even --> n can be 2: 22+23=45. But it also can be 6 --> x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45 --> x=5. At least two values of n are possible. Not sufficient.

(2) n<9 --> the above example is also valid for this statement, hence not sufficient.

(1)+(2) Still at least two values of n are possible. Not sufficient.

Answer: E.
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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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New post 23 Sep 2012, 22:51
Bunuel

my question is hovv do you knovv vvhere to stop there could have been an 8 too
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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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New post 24 Sep 2012, 04:12
venmic wrote:
Bunuel

my question is hovv do you knovv vvhere to stop there could have been an 8 too


When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

Now, when we consider the two statements together we have that n can be 2 or 6, so we don't have single numerical value of n, which means that the answer is E. We don't need to find whether n can be some other number, since two values are enough to tell that the statements taken together are not sufficient.

Hope it's clear.
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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even
(2) n < 9

First the quick maths.... consecutive numbers added together to make 45..
N + N+1 + N+2 all the way to N +i

The quickest way for me was to disprove both.

With two numbers N + N + 1 = 45 so 2n + 1 = 45, 2n = 44 n = 22 (2 numbers)
With six numbers 6N + 15 = 45, 6N=30 N=5 (6 numbers)

Even if we take both 1 and 2, n could be 2 or 6. Therefore E
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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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New post 06 Oct 2012, 05:52
Bunuel wrote:
venmic wrote:
Bunuel

my question is hovv do you knovv vvhere to stop there could have been an 8 too


When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

Now, when we consider the two statements together we have that n can be 2 or 6, so we don't have single numerical value of n, which means that the answer is E. We don't need to find whether n can be some other number, since two values are enough to tell that the statements taken together are not sufficient.

Hope it's clear.



Hey Bunuel,

How about this approach-

The sum of n consecutive numbers is n (n+1)/2=45
Therefore- n(n+1)=90
...> n^2+n-90=0
....> n-9=0 or n=-10

so both 1 and 2 are sufficient to answer. So the answer must be D right??
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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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rajathpanta wrote:
Bunuel wrote:
venmic wrote:
Bunuel

my question is hovv do you knovv vvhere to stop there could have been an 8 too


When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

Now, when we consider the two statements together we have that n can be 2 or 6, so we don't have single numerical value of n, which means that the answer is E. We don't need to find whether n can be some other number, since two values are enough to tell that the statements taken together are not sufficient.

Hope it's clear.



Hey Bunuel,

How about this approach-

The sum of n consecutive numbers is n (n+1)/2=45
Therefore- n(n+1)=90
...> n^2+n-90=0
....> n-9=0 or n=-10

so both 1 and 2 are sufficient to answer. So the answer must be D right??



The sum of n consecutive numbers is n (n+1)/2=45 NO
The sum of the first n consecutive positive integers 1, 2, 3,..., n is n(n + 1)/2.
Nowhere is stated that we have some number of the first positive integers.
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New post 06 Oct 2012, 09:20
Thanks EVa... got it :)
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The sum of n consecutive positive integers is 45 [#permalink]

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The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

Hi Everyone this problem is from Manhattan Practise test 4. This problem looked very simple to me but I am stuck with it. Despite looking at the explanation which is long and hard to understand, I couldn't understand what so ever. However while I was giving exam I thought that there is a formula for the sum of n consecutive positive integers which is n(n+1)/2 and this problem could be solved easily. But that doesn't apply to this problem. Can someone please help me with this.

OA is E.

Thanks In advance.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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Hi tk1tez7777


The sum of n consecutive integers is 45.

one simple short cut:- as the numbers are consecutive look for the numbers around 45/n

statement 1:- for n = 2 (22 and 23) we will get the sum as 45
for n = 6 (5,6,7,8,9,10) we will get sum as 45

So we cannot decide the value of n. Statement 1 alone is not sufficient.

statement 2:- n>9.

For both n = 2 and n = 6 the sum of the numbers is 45.

So statement 2 alone is not sufficient.

Combining both statements n<9 and n is even.

Again for both n = 2 and n=3 the sum is 45 so combined they are not sufficient.

SO answer is E
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 19 May 2013, 00:34
tk1tez7777 wrote:
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9


Let the first term be a, which is a positive integer. Thus, given that\(\frac{n}{2}[2a+(n-1)]\) = 45

From F.S 1, we know that n=even, thus 2a+(n-1) = even+odd=odd.

Thus,\(\frac{n}{2}*odd\) =\(\frac{10}{2}*9\). It could also be =\(\frac{6}{2}*15\)
Insufficient.

From F.S 2, we know that n<9. Thus, \(\frac{90}{n}\) must be an integer.We have n=1 or 2 or 3 or 5 or 6.Insufficient.
Taking both together , we have n = 2 or 6. Insufficient.
E.
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Re: The sum of n consecutive positive integers is 45 [#permalink]

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New post 19 May 2013, 03:18
tk1tez7777 wrote:
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

Hi Everyone this problem is from Manhattan Practise test 4. This problem looked very simple to me but I am stuck with it. Despite looking at the explanation which is long and hard to understand, I couldn't understand what so ever. However while I was giving exam I thought that there is a formula for the sum of n consecutive positive integers which is n(n+1)/2 and this problem could be solved easily. But that doesn't apply to this problem. Can someone please help me with this.

OA is E.

Thanks In advance.


Merging similar topics. Please refer to the solutions above.
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Hello Friends, Bunuel, @VertiasPrepKarishma, Vyshak, Skywalker18
I understand the solution given by students above. However, what i don't understand is how to arrive at that solution. What i mean to ask is that what made you think about 6? Did you guys do trial and error or something else?
Can someone please elaborate on that...

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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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Hello Friends, Bunuel, @VertiasPrepKarishma, Vyshak, Skywalker18
I understand the solution given by students above. However, what i don't understand is how to arrive at that solution. What i mean to ask is that what made you think about 6? Did you guys do trial and error or something else?
Can someone please elaborate on that...

Thanks


First we have to know that the max. possible number of positive consecutive integers can be 9.
According to Statement 1, n is even. When the number of consecutive integers is even, the mean will be 'x.5'.
45/2 = 22.5 --> Possible
45/6 = 7.5 --> Possible
So only 2 and 6 are possible.

Hope it helps.
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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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New post 18 Oct 2016, 07:08
Vyshak wrote:
manishtank1988 wrote:
Hello Friends, Bunuel, @VertiasPrepKarishma, Vyshak, Skywalker18
I understand the solution given by students above. However, what i don't understand is how to arrive at that solution. What i mean to ask is that what made you think about 6? Did you guys do trial and error or something else?
Can someone please elaborate on that...

Thanks


First we have to know that the max. possible number of positive consecutive integers can be 9.
According to Statement 1, n is even. When the number of consecutive integers is even, the mean will be 'x.5'.
45/2 = 22.5 --> Possible
45/6 = 7.5 --> Possible
So only 2 and 6 are possible.

Hope it helps.




Absolutely, thanks alot :yes
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The sum of n consecutive positive integers is 45. What is [#permalink]

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New post 10 Dec 2016, 03:02
This is an excellent Question.
Here are my 2 cents on this Questions.

Given data --> sum of N consecutive integers is 45.
We need to get the value of N.

Statement 1-->
N is even
N=2 => 22+23 => HIT...!!
N=4=> N can never be 4.
Reasoning -> E+O+E+O => Even and O+E+O+E=> Even
Hence the sum of 4 consecutive will always be even.
Hence n≠4
N=6 => 7+7+7+7+7+7=42
Lets arrange to get 45 on right and consecutive integers on left -->8+6+7+7+7+7=42 => 6+7+8+9+5+7 =42 Hence 5+6+7+8+9+10=> 45

Hence it a HIT.

Hence not sufficient.


Statement 2 -->
N<9

Clearly insufficient.

Combining the two Statements-->
we get N can definitely be 2 or 6.
Hence insufficient

Hence E.



Additionally n can never be 8 as the sum of 8 consecutive integers will always be even
Hence n can never be 8.


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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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New post 11 Dec 2016, 11:11
Found a Great Solution to this problem.

Follow the link -> statistics-made-easy-all-in-one-topic-203966.html#p1563176


Excellent Post.
So much to learn from that post.


Regards
Stone Cold.

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Re: The sum of n consecutive positive integers is 45. What is [#permalink]

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I want to add my bit here . Immaterial of what the question asks for i found an approach that makes understanding this question easier .

For Any number to be a product of consecutive terms it has to be an arithmetic progression and hence the formula of sum Sn=n/2(2a+(n-1)*1) is applicable

If the number is 45 then it follows that 90=n(2a+ (n-1))

The factors of 90 in its basic form can be written as 90=3^2*5*2

Now to find the series that is a sum of consecutive terms we just need to rearrange the possibilities :

1. 90/2=45=[(3^2*5)/2]*2=22.5*2 where 22.5 is the avg and 2 the number of terms . The series has 2 integers with an average of 22.5 and this is possible only when the series is 22+23.

2. 45=(15/2)*6 =7.5*6 i.e 6 terms with an average of 7.5. write 7,8 first which give an average of 7.5 and then add 2 terms on each side to generate the series 5,6,7,8,9,10

3. 90/2=45=[5/2]*18 =2.5*18 where 2.5 is the average and 18 are the number of terms. Thus the series possible is -6 ,-5,-4,-3…., 10, 11. (a simple tip here is to write the 2 consecutive integers that give an average of 2.5 i.e 2,3 and then write the 16 numbers symmetrically on either sides of 2 and 3)


4. 90/2=45=[3^2/2]*10=4.5*10 ie 10 terms with average of terms as 4.5. Hence the series 0,1,2,3,4,5,6,7,8,9

5. 5. 45=[9*2/2]*5=9*5 i.e average 5 with 9 terms . Hence the series is 1,2,3,4,5,6,7,8,9. (write the center term 5 which is the average and then add four terms to the left and right side to get the balance 8 terms)


6. 45=[30/2]*3 =15*3 average 15 with 3 terms . I.e 14,15,16

By checking the combination of its prime factors we can thus generate the series that can represent a number as a sum of its consecutive terms.
Now considering each of the above conditions or combining both the conditions ( n is even and n<9) we cannot get a unique solution to the problem. The correct answer has to be E.
Re: The sum of n consecutive positive integers is 45. What is   [#permalink] 12 Dec 2016, 03:32

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