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The sum of n consecutive positive integers is 45. What is
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Updated on: 12 Jun 2013, 02:33
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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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Originally posted by enigma123 on 31 Jan 2012, 16:42.
Last edited by Bunuel on 12 Jun 2013, 02:33, edited 2 times in total.
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Re: Value of n
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31 Jan 2012, 16:55
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
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23 Sep 2012, 21:51
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
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24 Sep 2012, 03:12
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
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24 Sep 2012, 07:30
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
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06 Oct 2012, 04:52
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+n90=0 ....> n9=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
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06 Oct 2012, 07:09
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=45Therefore n(n+1)=90 ...> n^2+n90=0 ....> n9=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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The sum of n consecutive positive integers is 45
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18 May 2013, 22:50
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
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18 May 2013, 23:06
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
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18 May 2013, 23:34
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\(\frac{n}{2}[2a+(n1)]\) = 45 From F.S 1, we know that n=even, thus 2a+(n1) = 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. What is
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17 Oct 2016, 20:17
Hello Friends, Bunuel, @VertiasPrepKarishma, Vyshak, Skywalker18I 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



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Re: The sum of n consecutive positive integers is 45. What is
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17 Oct 2016, 20:34
manishtank1988 wrote: Hello Friends, Bunuel, @VertiasPrepKarishma, Vyshak, Skywalker18I 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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The sum of n consecutive positive integers is 45. What is
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10 Dec 2016, 02:02
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
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11 Dec 2016, 10:11
Found a Great Solution to this problem.
Follow the link > statisticsmadeeasyallinonetopic203966.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
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12 Dec 2016, 02:32
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+(n1)*1) is applicable
If the number is 45 then it follows that 90=n(2a+ (n1))
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.



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Re: The sum of n consecutive positive integers is 45. What is
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12 Dec 2016, 02:32
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+(n1)*1) is applicable
If the number is 45 then it follows that 90=n(2a+ (n1))
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.



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Re: The sum of n consecutive positive integers is 45. What is
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12 May 2019, 10:41
enigma123 wrote: The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is even (2) n < 9 Official Solution Credit: Veritas PrepSolution: First I will give the solution of this question and then discuss the logic used to solve it. In how many ways can you write n consecutive integers such that their sum is 45? Let’s see whether we can get such numbers for some values of n. n = 1 > Numbers: 45 n = 2 > Numbers: 22 + 23 = 45 n = 3 > Numbers: 14 + 15 + 16 = 45 n = 4 > No such numbers n = 5 > Numbers: 7 + 8 + 9 + 10 + 11 = 45 n = 6 > Numbers: 5 + 6 + 7 + 8 + 9 + 10 = 45 Let’s stop right here. Statement I: n must be even. n could be 2 or 6. Statement I alone is not sufficient. Statement II: n < 9 n can take many values less than 9 hence statement 2 alone is not sufficient. Both statements together: Since n can take values 2 or 6 which are even and less than 9, both statements together are not sufficient. Logic Now, the interesting thing is how do we get these numbers for different values of n. How do we know the values that n can take? It’s pretty easy really. Follow my thought here. Of course, n can be 1. In that case we have only one number i.e. 45. n can be 2. Why? When we divide 45 by 2, we get 22.5. Since 2*22.5 is 45, we have to find 2 consecutive integers such that their arithmetic mean is 22.5. The integers are obviously 22 and 23. n can be 3. When we divide 45 by 3, we get 15. So we need 3 consecutive integers such that their mean is 15. They are 14, 15, 16. When we divide 45 by 4, we get 11.25. Do we have 4 consecutive integers such that their mean is 11.25? No, because mean of even number of consecutive integers is always of the form x.5. n can be 5. When we divide 45 by 5, we get 9 so we need 5 consecutive integers such that their mean is 9. They must be 7, 8, 9, 10, 11. n can be 6. When we divide 45 by 6, we get 7.5. We need 6 consecutive integers such that their mean is 7.5. The integers are 5, 6, 7, 8, 9, 10 Obviously, we just need to focus on getting 2 even values of n which are less than 9. So we check for 2, 4 and 6 and we immediately know that the answer is (E). We don’t have to do this process for all numbers less than 9 and we don’t have to do it for odd values of n.



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Re: The sum of n consecutive positive integers is 45. What is
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02 Jun 2020, 10:20
Bunuel wrote: 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. How do you know to test N as 6 and that it will work? For example, if I wanted to test 4, (x)+(x+1)+(x+2)+(x+3) wouldn't result in an even X but I would waste my time testing this number, 4. How would I know to test 6 first to save time? Thank you!




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