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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] New post 04 Apr 2008, 19:08
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E

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-sum-of-n-consecutive-positive-integers-is-45-what-is-126861.html
[Reveal] Spoiler: OA

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Re: DS: Sum of consecutive numbers [#permalink] New post 04 Apr 2008, 19:48
Question Tells us sum is of consecutive integers.

Statement 1:
Tells us number of element is even. So it can 22+23 (2 even) = 45 or 1+ 2 + 3+ ...+ 9 (9 odd)= 45 or 5+6+7+...+10 (6 even) = 45,
So insufficient.

Statement 2:
Tells us total number of terms is less than 9. In all the cases considered above n is less than 9. So this statement alone is also insufficient.

Combining them together is not going to give any results as we are using same set of example for both the statements.

Answer E.
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Re: DS: Sum of consecutive numbers [#permalink] New post 04 Apr 2008, 21:12
My approach is:

taking them together:

let's say n =2, sum = 2m+1 = 45, so numbers are 22 and 23
n=6, numbers are 5,6,7,8,9,10

still insuff

Last edited by vshaunak@gmail.com on 05 Apr 2008, 16:24, edited 1 time in total.
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Re: DS: Sum of consecutive numbers [#permalink] New post 05 Apr 2008, 05:37
wait I am a abit confused..

we know N is positive..

so the sum of the n consecutive integers is n(n+1)/2=sum

if we plug that in we get n=9

but then I see your point as well..

so how do we know when to use this formula?
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Re: DS: Sum of consecutive numbers [#permalink] New post 05 Apr 2008, 05:43
The formula n(n+1)/2 is only valid if you have series starting from 1.
So it will be valid for 1+2+3+...+9 But it will not be valid for 4+5+6+7 or for that matter any series that does not start with 1.
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Re: DS: Sum of consecutive numbers [#permalink] New post 05 Apr 2008, 06:36
Excellent point abhijit_sen...key word is "FIRST N integers"...when looking to see which formula to use..
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DS: sum of n consecutive ints [#permalink] New post 25 May 2008, 19:00
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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
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Re: DS: sum of n consecutive ints [#permalink] New post 25 May 2008, 21:07
Expert's post
E

Thanks for a good problem. +1

sum=45.
So, construct all sequences:

n is odd.

For odd n we can say that ((n+1)/2)th number is median and equal sum/n.
In our case, 45 is divisible by odd n only if n is {3,5,9,15,45} (odd factors of 45)
n=3: 45/3=15. 14,15,16
n=5: 45/5=9. 7,8,9,10,11
n=9: 45/5=5. 1,2,3,4,5,6,7,8,9
n=15,45: sequences will be include negative numbers that contradicts the problem.


n is even.
For even n we can say that sum/n is median and has to be equidistant between (n/2)th and (n/2+1)th numbers. In other words, sum has to be decimal number like xxx,5

In our case, 45 is divisible by even n and result xxx.5 only if n is {2,6,10,18,30,90}
n=2: 45/2=22.5. 22,(22.5),23
n=6: 45/6=7.5. 5,6,7,(7.5),8,9,10
n=10,18,30,90: sequences will be include negative numbers that contradicts the problem.

1st condition: n=2,6 - insufficient
2st condition: n=2,3,5,6 - insufficient
1&2st conditions: n=2,6 - insufficient

E
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Re: DS: sum of n consecutive ints [#permalink] New post 25 May 2008, 23:14
Let the series be 1,2,3,....,m,.......m+n

Now sum of all positive integers from m+1 to m+n = (m+n)(m+n+1)/2 - m(m+1)/2
= 45

After simplification Implies: n(2m+n+1)= 90

90 = 9X10 = 6X15 = 5X18 = 3X30 = 2X 45

Now accordingly we can find n & m by substituting values above with n= 9, 6,5,3,2 and then we get m= 0, 4, 6, 13, 21

Statement1: n is even : n can take values of 6 and 2 so insufficient

Statement 2: N<9 : n can take 9, 6, 5, 3, 2 hence insufficient

Bothe together: N can take 6 and 2 thus Ans is E
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 00:21
chineseburned wrote:
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9



The sum of arthimetic progression (here with diff = 1) needs is

S = n(a1 + aN)/2 OR S = N [2*a1 + (N-1)d] /2

Thus if you know sum, you still have 2 unknowns,

(1) & (2) individually or together do not give any definite value of N,

So E.
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 00:31
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alpha_plus_gamma wrote:

The sum of arthimetic progression (here with diff = 1) needs is

S = n(a1 + aN)/2 OR S = N [2*a1 + (N-1)d] /2

Thus if you know sum, you still have 2 unknowns,

(1) & (2) individually or together do not give any definite value of N,

So E.


But how can you use your approach for slightly modified problem: "The sum of n consecutive positive integers is 47. What is the value of n?"
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 01:04
I agree with alpha_plus_gamma.

na+ n(n-1)/2 = 45

1) n=2,a=22 ; n=6,a=5 satisfies
2) above two values still hold
combine - above two values still hold

so E.

same can be worked out with 47 as well.
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 01:07
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farend wrote:
same can be worked out with 47 as well.


Please, give me two examples with different n and that satisfy both conditions in the case of sum=47
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 01:31
walker wrote:
farend wrote:
same can be worked out with 47 as well.


Please, give me two examples with different n and that satisfy both conditions in the case of sum=47


I didn't mean to say that the answer will be same in case of sum 47.
Rather the approach holds good for 47 as well.

I see only n=2,a=23 satisfying the case of 47.
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 01:50
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Maybe I've understood incorrectly, Alpha_plus_gamma's approach in based on S = N [2*a1 + (N-1)d] /2 formula and he chose E because too many unknowns. So, if we substitute 45 for 47 his approach is insensitive to this substitution, isn't it?
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 01:59
walker wrote:
Maybe I've understood incorrectly, Alpha_plus_gamma's approach in based on S = N [2*a1 + (N-1)d] /2 formula and he chose E because too many unknowns. So, if we substitute 45 for 47 his approach is insensitive to this substitution, isn't it?


- I wouldn't say E is chosen bcoz of too many unknowns; rather given the two infos in original questions i.e 1) n is even and 2) n<9, the formula nicely gives E.
- You are right; substitute 45 for 47 is insensitive to the substitution.
- On a side note even the method you gave for counting 5s in 4500! question is an offshoot of this method.
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 02:02
When dealing with the sum of evenly spaced integers (also known as arithmetic progression or sequence), just remember the
simple, universal formula: Sum = Average* (# of values). [* represents the multiplication sign.] The space is 1 here.
Applied to this problem, Sum = n*(a1+aN)/2. a1 and aN are the first term and the last term of the sequence respectively, (a1+ aN)/2 being the average, but also the median or the midpoint. They all coincide when dealing with an arithmetic progression. n is of course the number of terms.
So, it follows that n(a1 + aN)/2 = 45 and (a1 + aN) = 90/n. Since a1 + a2 is an integer, so must be 90/n. It then boils down to finding the positive factors of 90. Let's find the prime factorization of 90. 90 = 9*10 = 3*3*2*5 = 3^2 * 2 * 5.
We can easily list all the positive factors of 90; there are 12 of them. I actually have a formula to find them fast but that's another topic.
1/Since n is even, let's focus on the even values of n; they are 2, 6, 10, 18, 30, and 90.
But if n = 10 then the median of the sequence will be (a1 + a2)/2 = 45/10 = 4.5 and some of the terms will be negative (NOT Allowed).
Any n > 10 will have to be rejected for the same reason. Therefore n must be 2 or 6: UNSUFF.
2/ n < 9 includes more than one n (2, 3, 5, 6): UNSUFF.
Together, we are still stuck with 2 and 6.
Note that when n = 2, the average is 45/2 = 22.5; then a1 = 22 and aN = a2 = 23. 22+23 = 45
When n = 6, then a1 = 5; a2 = 6; a3 =7; a4 = 8, a5 = 9, a6 = 10. 5+6+7 +8+9+10 = 45.
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 02:03
farend wrote:
walker wrote:
Maybe I've understood incorrectly, Alpha_plus_gamma's approach in based on S = N [2*a1 + (N-1)d] /2 formula and he chose E because too many unknowns. So, if we substitute 45 for 47 his approach is insensitive to this substitution, isn't it?


- I wouldn't say E is chosen bcoz of too many unknowns; rather given the two infos in original questions i.e 1) n is even and 2) n<9, the formula nicely gives E.
- You are right; substitute 45 for 47 is insensitive to the substitution.
- On a side note even the method you gave for counting 5s in 4500! question is an offshoot of this method.


I too meant
>> given the two infos in original questions i.e 1) n is even and 2) n<9, the formula nicely gives E.
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Re: DS: sum of n consecutive ints [#permalink] New post 26 May 2008, 08:57
chineseburned wrote:
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9


to formulas are handy here..

sum=N/2 * {2a+(n-1)d} where d is the difference, in our case= 1, a=the first number in the series..we dont what that is..)

second formula..
n*(na+nd)/2=sum where na is the first number in the series and nd=last number in the series..i.e avg*number of terms=sum..

{na+nd}n=90..

1)n is even..

na+nd=90/n now..lets see what n can divide 90 evenly..n=2, 6..lets stick with that

if n=2..na+nd=45.. lets see what those numbers..23+22=45.ok great!
if n=6 then lets (2*a+n-1}=90
suppose a=5.. n{10+n-1}=n{n+9}=90..n=6..works!

you can also check via.. {na+nd}=15...if na=5 and Nd=10...works..

Insuff

2) N<9

well i just showed 2 examples of n=2 and 6...Insuff
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CONSECUTIVE INTEGERS.. [#permalink] New post 09 Aug 2009, 08:57
The sum of n consecutive positive integers is 45. What is the value of n?

(1) n is even

(2) n < 9

OA:
[Reveal] Spoiler:
E


Explain please.. looks harder for me..
CONSECUTIVE INTEGERS..   [#permalink] 09 Aug 2009, 08:57
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