The sum of n consecutive positive integers is 45. What is : GMAT Data Sufficiency (DS) - Page 2
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# The sum of n consecutive positive integers is 45. What is

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09 Aug 2009, 09:19
22 and 23 = 45
n=2
5 6 7 8 9 10 =45
n=6
insufficient

They both work so insufficient

14,15,16 =45
+ already we have n=2 and n=6
So insufficient

Together

we have atleast 2 option where n is even and less than 9

so E
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09 Aug 2009, 21:32
sher676 wrote:
22 and 23 = 45
n=2
5 6 7 8 9 10 =45
n=6
insufficient

They both work so insufficient

14,15,16 =45
+ already we have n=2 and n=6
So insufficient

Together

we have atleast 2 option where n is even and less than 9

so E

thank you for the expanation. Do you know any shortcut for picking numbers? it took me some time to check on different combinations and we cannot afford it during the test. thanks!
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09 Aug 2009, 21:54
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We can pick numbers as follows

Case 1) divide the number by 2 and play around with the numbers around it to get the sum
Case 2) divide the number by 3 and play around with the numbers around it to get the sum
Case 3) divide the number by 4 and play around with the numbers around it to get the sum
Case 4) divide the number by 5 and play around with the numbers around it to get the sum
and go on ... till you realize that sum of the n numbers exceeds 45

In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46
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09 Aug 2009, 22:29
gmanjesh wrote:
We can pick numbers as follows

Case 1) divide the number by 2 and play around with the numbers around it to get the sum
Case 2) divide the number by 3 and play around with the numbers around it to get the sum
Case 3) divide the number by 4 and play around with the numbers around it to get the sum
Case 4) divide the number by 5 and play around with the numbers around it to get the sum
and go on ... till you realize that sum of the n numbers exceeds 45

In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46

thank you gmanjesh. it seems like a quicker approach. +1 for you
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14 Aug 2009, 17:05
gmanjesh wrote:
We can pick numbers as follows

Case 1) divide the number by 2 and play around with the numbers around it to get the sum
Case 2) divide the number by 3 and play around with the numbers around it to get the sum
Case 3) divide the number by 4 and play around with the numbers around it to get the sum
Case 4) divide the number by 5 and play around with the numbers around it to get the sum
and go on ... till you realize that sum of the n numbers exceeds 45

In the above case we see that 45 will arise when you play around with numbers in case 1,2,4 and not in 3 as it will arise in either 9+10+11+12 = 42 or 10+11+12+13 = 46

well said.. thanks..
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05 Apr 2011, 18:16
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
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Re: Consecutive Integers: 45 Caliber [#permalink]

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05 Apr 2011, 19:36
1) + 2)

n = -10. But it not legal. So is the question alluding that since n does not exist, it is indeterminate hence E. Is this correct?
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Re: Consecutive Integers: 45 Caliber [#permalink]

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05 Apr 2011, 21:21
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45 = 22 + 23

or 45 = 5 + 6 + 7 + 8 + 9 + 10

So n = 2 or n = 6 and n < 9

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Re: Consecutive Integers: 45 Caliber [#permalink]

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05 Apr 2011, 22:47
sum of consecutive integers=n/2(2a+n-1)
where a is first term of series
in the above equation we know all the terms except a and we need to find value of n
S1 and S2 do not give value of "a" directly or indirectly.
even after combining them we cant find value of "a".
hence both statements are insufficient.
Ans=E
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Re: Consecutive Integers: 45 Caliber [#permalink]

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05 Apr 2011, 22:50
gmat1220 wrote:
1) + 2)

n = -10. But it not legal. So is the question alluding that since n does not exist, it is indeterminate hence E. Is this correct?

think u misread the statement 2
n<9 (not < -9)
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Re: Consecutive Integers: 45 Caliber [#permalink]

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06 Apr 2011, 10:07
since we dont know whether the first term of the sequence is 1 , we cant use the formula n(n+1)/2 formula bcoz that is sum of n consecutive natural numbers starting from 1

so using the formula n/2 (a+l), where l is the last term and a is the first term
we get n/2(a+l) = 45
n(a+l)=90.....A1.
10*9
18*5
6*15
2*45
30*3
now statement 1 says n is even so all the above cases are possible but only 2*45 actually satisfies the condition.
then A should be good enough.

2 CASE ) n < 9 only bottom three cases are possible
But again here 2*45 is the only option which gives n=2 and terms will be 22 +23.
no other soluton satisfies condition A1 ...hence B is also sufficient

I KNOW THAT I AM MAKING SOME MISTAKE BUT WHERE....PLS HELP
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Re: Consecutive Integers: 45 Caliber [#permalink]

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06 Apr 2011, 17:51
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Expert's post
HelloKitty wrote:
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

In how many ways can you write n consecutive integers such that their sum is 45?
n = 1 -> 45
n = 2 -> 22, 23
n = 3 -> 14, 15, 16
n = 4 X
n = 5 -> 7, 8, 9, 10, 11
n = 6 -> 5, 6, 7, 8, 9, 10
n = 7 X
n = 8 X
n = 9 -> 1, 2, 3, 4, 5, 6, 7, 8, 9

n can be 2 or 6 hence statement 1 is not sufficient.
n can take many values less than 9 hence statement 2 is not sufficient.
Since n can take values 2 or 6 which are even and less than 9, both statements together are not sufficient.

Now, the interesting thing is how to get these values of n. How do I know which values can n take and which can it not?
Its pretty easy really. Follow my thought here.

"Of course n can be 1.

n can be 2 since when I divide 45 by 2, I get 22.5. So 2*22.5 is 45 so I have to find 2 consecutive integers whose mean is 22.5. The integers are obviously 22 and 23.

When I divide 45 by 3, I get 15. So I need 3 consecutive integers whose mean is 15. They are 14, 15, 16

When I divide 45 by 4, I get 11.25. Do I have 4 consecutive integers such that their mean is 11.25? No because mean of consecutive integers is always an integer or of the form n.5

When I divide 45 by 5, I get 9 so I need 5 consecutive integers whose mean is 9. They must be 7, 8, 9, 10, 11

and so on....."

Obviously, I just need to focus on getting 2 even values of n. So I check for 2, 4, 6 and I would know that answer is (E).
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Re: Consecutive Integers: 45 Caliber [#permalink]

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08 Apr 2011, 08:56
Wow Karishma, i love u! great explanation. +1
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18 Sep 2011, 09:06
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9
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18 Sep 2011, 13:36

n(n+1)
--------- = 45 => n = 9.

Hence, I and II individually are insufficient and together also they are not sufficient.
2
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18 Sep 2011, 14:12
dsmishra1981 wrote:

n(n+1)
--------- = 45 => n = 9.

Hence, I and II individually are insufficient and together also they are not sufficient.
2

Could you pls explain a little bit more?
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18 Sep 2011, 18:30
aeros232 wrote:
dsmishra1981 wrote:

n(n+1)
--------- = 45 => n = 9.

Hence, I and II individually are insufficient and together also they are not sufficient.
2

Could you pls explain a little bit more?

Approach:

If we can find two different even nos satisfying I & II individually we are done.

45= x + (x+1) ; x=22. The two nos are 22 & 23

Checking 4 . Failed .

Check 6. Ok. Nos are 5,6,7,8,9,10.

A, D,B,C are out. Only E is left.

Hope it helps.
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18 Sep 2011, 22:43
aeros232 wrote:
dsmishra1981 wrote:

n(n+1)
--------- = 45 => n = 9.

Hence, I and II individually are insufficient and together also they are not sufficient.
2

Could you pls explain a little bit more?

Sorry mate, I got the question wrong and took the integers from 1 to n and applied the formula n(n+1)/2 = 45, which is wrong.

The explanation above is the correct one. For at least 2 even numbers (viz n = 2 and n = 6), we can get the sum of 45. Hence E.
Re: Consecutive Numbers   [#permalink] 18 Sep 2011, 22:43

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