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

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, which gives x=5.

At least two values of n are possible, 2 and 6. 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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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.
Answer (E)

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).

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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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

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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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

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.

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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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.

Bunuel

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

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??

Official Solution

Credit: Veritas Prep


Solution: 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.

Not an official question, jabhatta2, but we can still learn something from it. We know that the sum of any set is equal the product of the set's average * the number of terms in the set. Here, that product is 45, and we have an integer constraint for the number of terms.
Regarding the set's average: since it's a set of consecutive integers, the average must be either an integer (if the number of terms is odd) or halfway in between to integers (if the number of terms is even). So we don't have an integer constraint for the average, but we do have a constraint on the decimal of that average (it's either something.0 or something.5)
If the average DID have an integer constraint, we could say that the number of terms and the average each has to be a factor of 45.
To accommodate the possibility of the average=something.5, let's give the prime box of 45 a factor of 2, for a product of 90.
Now we can say that the number of terms in the set must be a factor of 90, or [2^1 * 3^2 * 5^1].
We haven't yet used the fact that the consecutive terms in the set are all positive. That fact gives an upper limit for the number of terms. Looks like 10 terms (for an average of 4.5) would already force one of our terms to be zero. So, the upper limit for the number of terms is 9. What are the factors of 90 between 1 and 9? 1, 2, 3, 5, 6, 9.
To answer your first question: yes, the number of terms could be 1.
I used reasoning above to narrow down the field of possible answers to a list of six possibilities. Now I'm ready for the statements.
Would I do all that work on an actual GMAT problem? Of course not. But, this isn't an actual GMAT problem, so let's just use it to advance our reasoning skills!

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

Hi avigutman - two quick questions

Q1) From the question stem (no the statements) - would you think n = 1 is a possible value for n ? I thought n =1 was possible from the question stem

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Q2) was wondering, how would you tackle this one.

I could test cases (n=1,2,3,4,5,6,7,8) but that will take me > 3 mins in addition to mistakes

Seems very manual laborious, this question.

Thank you !!

S1) n is even
22+23 or
5+6+7+8+9+10
Not suff

S2) n <9
There are multiple cases . See S1 above

S1+S2) n could be 2 or 6
Not suff

Posted from my mobile device

Hi avigutman



I think the above in red is an interesting take-away.
If consecutive set (D =1) - average is integer (if n = odd) and average is (something.5) (if n = even)

I dont believe the yellow decimal constraint would be there is perhaps d = 2 or d = 3 or d =5 for example ...




Based on the red - Interesting you doubled to 90. That way, (n) and (mean) are both integers. Thus the possibilities become

Number of terms x average =
1 x 90
2 x 45
3 x 30
4 x 22.5 = out because 22.5 is NOT an integers
5 x 16
9 x 10
10 x 9
15 x 6
30 x 3.
45 x 2
90 x 1



This is where doubling to 90 is getting dangerous.

Is number of terms = 10 an upper limit if your total sum = 90? I don’t believe so

Did you draw back down to a sum of 45, when figuring out that the number of terms <= 9 ?

That is why – I am not a fan of doubling to 90 for the first exercise above and then drawing back down to 45 for finding n <= 9.

Why not? Try to prove or disprove it, jabhatta2. Yes, I wanted to manufacture an integer constraint.Agreed. In my initial thinking (before I typed it up here) I had already found the upper limit of 9 before manufacturing the integer constraint. Also, I never worked with the number 90 -- I had a prime box for 45 [3^2 * 5^1] and then brought in a factor of 2 to manufacture the integer constraint. So I did all my work at the level of the prime box.

In consecutive numbers - considering n to be the middle number -

Sum of two consec num
n + (n+1) = 2n+1

Sum of three consec. num
n-1 + n + n+1 = 3n

Sum of four consec. num=sum of three consec num + the next num
=3n+n+2 = 2(2n+1)

Sum of five consec num = sum of four consec. num + the next num = 5n

The pattern then becomes -
- If n is odd, sum = no. of terms*middle number
- If n is even, sum = (2*middle number +1)(no. of terms/2)

Reminding ourselves that n, the middle number, must be an integer

S(1) = n = 45 [situation where number of terms is just 1]

S(2) = 2n+1 = 45 [possible as an integer value of n can be obtained]

S(3) = 3n = 45 [possible as an integer value of n can be obtained]

S(4) = 2(2n+1) = 45 [NOT possible as n cannot be an integer in such case]

S(5) = 5n = 45 [possible as an integer value of n can be obtained]

S(6) = 3(2n+1) =45 [possible as an integer value of n can be obtained]

S(7) = 7n [NOT possible as n cannot be an integer in such case]

S(8) = 4(2n+1) [NOT possible as n cannot be an integer in such case]


Condition 1 --> n is even
Possible cases are S(2), S(6). No definite answer obtained

Condition 2 --> n < 9
Possible cases are S(1), S(2), S(3), S(5) and S(6). No definite answer obtained


Combining Condition 1 and 2
Possible cases are S(2) and S(6). No definite answer obtained

Therefore answer ---> E - Both statements are not sufficient