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Yeah. The article says this.

Solution: We have 100 = 2² × 5²

Now to find the odd number of factors, just ignore 2 and the powers of 2 as multiplying with 2 will make the factor even. So we are left with 5² only. Hence the number of odd factors are 2 + 1 = 3 (the odd factors of 100 are 1, 5 and 25). It means that the total number of ways to write 100 as the sum of two or more consecutive natural numbers is 3 – 1 = 2.

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Not sure why it cannot be universally applied
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hD13
How many ways can 720 be expressed as a sum of n Consecutive positive Integers ? ( n > 1 )

A. 3
B. 2
C. 6
D. 4
E. None of the above

We can use the property of sum of n consecutive integers=n*average of the numbers.
Also, average of consecutive integers is the middle or median of the numbers.


Average when n is even
The average of the integers must be a decimal of the form xy.5, as it is middle of two consecutive numbers.
And when do we have that?: When we have just one 2 extra in the divisor
\(720=2^4*3^2*5\)
1) We have \(2^4\) in 720, so \(2^5\) will get a decimal of the form xy.5
\(\frac{720}{2^5}=22.5\)
32*22.5….n=32
7,8,9,….,21,22,23,24,…..36,37,38

2) We have \(2^4*3^2\) in 720, so \(2^5*3^2\) will get a decimal of the form xy.5
\(\frac{720}{2^5*3^2}=2.5\)
But this will get negative integers too.

3) We have \(5*2^4\) in 720, so \(5*2^5\) will get a decimal of the form xy.5
\(\frac{720}{5*2^5}=4.5\)
But this too will get negative integers.

Average when n is odd
The average of the integers will be an integer.
And when do we have that?: When we divide by an odd factor.
\(720=2^4*3^2*5\)
1) We have \(3\) in 720, so
\(\frac{720}{3}=240\)
3*240….n=3
239,240,241

2) We have \(3^2\) in 720, so
\(\frac{720}{3^2}=80\)
9*80….n=9
76,77,78,79,80,81,82,83,84

3) We have \(5\) in 720, so
\(\frac{720}{5}=144\)
5*144….n=5
142,143,144,145,146

4) We have \(3^2*5\) in 720, so
\(\frac{720}{3^2*5}=45\)
16*45….n=45
If average is 16, we will require 22 integers lesser than 16, and it will result in negative integers too.

Thus, total 4


D

Isn't there another way?
720/15 = 48
41,42,43,44,45,46,47,48,49,50,51,52,53,54,55
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chetan2u
hD13
How many ways can 720 be expressed as a sum of n Consecutive positive Integers ? ( n > 1 )

A. 3
B. 2
C. 6
D. 4
E. None of the above

We can use the property of sum of n consecutive integers=n*average of the numbers.
Also, average of consecutive integers is the middle or median of the numbers.


Average when n is even
The average of the integers must be a decimal of the form xy.5, as it is middle of two consecutive numbers.
And when do we have that?: When we have just one 2 extra in the divisor
\(720=2^4*3^2*5\)
1) We have \(2^4\) in 720, so \(2^5\) will get a decimal of the form xy.5
\(\frac{720}{2^5}=22.5\)
32*22.5….n=32
7,8,9,….,21,22,23,24,…..36,37,38

2) We have \(2^4*3^2\) in 720, so \(2^5*3^2\) will get a decimal of the form xy.5
\(\frac{720}{2^5*3^2}=2.5\)
But this will get negative integers too.

3) We have \(5*2^4\) in 720, so \(5*2^5\) will get a decimal of the form xy.5
\(\frac{720}{5*2^5}=4.5\)
But this too will get negative integers.

Average when n is odd
The average of the integers will be an integer.
And when do we have that?: When we divide by an odd factor.
\(720=2^4*3^2*5\)
1) We have \(3\) in 720, so
\(\frac{720}{3}=240\)
3*240….n=3
239,240,241

2) We have \(3^2\) in 720, so
\(\frac{720}{3^2}=80\)
9*80….n=9
76,77,78,79,80,81,82,83,84

3) We have \(5\) in 720, so
\(\frac{720}{5}=144\)
5*144….n=5
142,143,144,145,146

4) We have \(3^2*5\) in 720, so
\(\frac{720}{3^2*5}=45\)
16*45….n=45
If average is 16, we will require 22 integers lesser than 16, and it will result in negative integers too.

Thus, total 4


D

Isn't there another way?
720/15 = 48
41,42,43,44,45,46,47,48,49,50,51,52,53,54,55

Yes, you are correct
I checked for 3, 3^2, 5 and 3*5^2, but missed out on checking for 3*5.
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I had read,

No. of ways to write a number into sum of 2 or more consecutive positive int = no. of odd factors -1.

(Just found a link for this : https://cat.wordpandit.com/cat-number-system-trick/ )

720 = (3^2) * (2^4) * (5^1)

So, no. of odd factors = (2+1) * (1+1) = 6
So, for this question, required answer = 6-1 =5

Am I missing something ? Is there any caveat to the formula, which I am not aware of ?

Bunuel TestPrepUnlimited avigutman GMATinsight EgmatQuantExpert chetan2u

Hi

That seems to be a very valid formula. The reasoning is the same - that is n*average.
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hD13
How many ways can 720 be expressed as a sum of n Consecutive positive Integers ? ( n > 1 )

A. 3
B. 2
C. 6
D. 4
E. None of the above

We can use the property of sum of n consecutive integers=n*average of the numbers.
Also, average of consecutive integers is the middle or median of the numbers.


Average when n is even
The average of the integers must be a decimal of the form xy.5, as it is middle of two consecutive numbers.
And when do we have that?: When we have just one 2 extra in the divisor
\(720=2^4*3^2*5\)
1) We have \(2^4\) in 720, so \(2^5\) will get a decimal of the form xy.5
\(\frac{720}{2^5}=22.5\)
32*22.5….n=32
7,8,9,….,21,22,23,24,…..36,37,38

2) We have \(2^4*3^2\) in 720, so \(2^5*3^2\) will get a decimal of the form xy.5
\(\frac{720}{2^5*3^2}=2.5\)
But this will get negative integers too.

3) We have \(5*2^4\) in 720, so \(5*2^5\) will get a decimal of the form xy.5
\(\frac{720}{5*2^5}=4.5\)
But this too will get negative integers.

Average when n is odd
The average of the integers will be an integer.
And when do we have that?: When we divide by an odd factor.
\(720=2^4*3^2*5\)
1) We have \(3\) in 720, so
\(\frac{720}{3}=240\)
3*240….n=3
239,240,241

2) We have \(3^2\) in 720, so
\(\frac{720}{3^2}=80\)
9*80….n=9
76,77,78,79,80,81,82,83,84

3) We have \(5\) in 720, so
\(\frac{720}{5}=144\)
5*144….n=5
142,143,144,145,146

4) We have \(3*5\) in 720, so
\(\frac{720}{3*5}=48\)
15*48….n=15
41, 42,….48,…..54, 55

5) We have \(3^2*5\) in 720, so
\(\frac{720}{3^2*5}=16\)
16*45….n=45
If average is 16, we will require 22 integers lesser than 16, and it will result in negative integers too.

Thus, total 5




When n is 45. We get negative integers. QUESTION STEM asks for positive Integers only.

4 solutions.
n= 3,5,9,15
Which is the 5th solution you're referring to?

Posted from my mobile device
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chetan2u
hD13
How many ways can 720 be expressed as a sum of n Consecutive positive Integers ? ( n > 1 )

A. 3
B. 2
C. 6
D. 4
E. None of the above

We can use the property of sum of n consecutive integers=n*average of the numbers.
Also, average of consecutive integers is the middle or median of the numbers.


Average when n is even
The average of the integers must be a decimal of the form xy.5, as it is middle of two consecutive numbers.
And when do we have that?: When we have just one 2 extra in the divisor
\(720=2^4*3^2*5\)
1) We have \(2^4\) in 720, so \(2^5\) will get a decimal of the form xy.5
\(\frac{720}{2^5}=22.5\)
32*22.5….n=32
7,8,9,….,21,22,23,24,…..36,37,38

2) We have \(2^4*3^2\) in 720, so \(2^5*3^2\) will get a decimal of the form xy.5
\(\frac{720}{2^5*3^2}=2.5\)
But this will get negative integers too.

3) We have \(5*2^4\) in 720, so \(5*2^5\) will get a decimal of the form xy.5
\(\frac{720}{5*2^5}=4.5\)
But this too will get negative integers.

Average when n is odd
The average of the integers will be an integer.
And when do we have that?: When we divide by an odd factor.
\(720=2^4*3^2*5\)
1) We have \(3\) in 720, so
\(\frac{720}{3}=240\)
3*240….n=3
239,240,241

2) We have \(3^2\) in 720, so
\(\frac{720}{3^2}=80\)
9*80….n=9
76,77,78,79,80,81,82,83,84

3) We have \(5\) in 720, so
\(\frac{720}{5}=144\)
5*144….n=5
142,143,144,145,146

4) We have \(3*5\) in 720, so
\(\frac{720}{3*5}=48\)
15*48….n=15
41, 42,….48,…..54, 55

5) We have \(3^2*5\) in 720, so
\(\frac{720}{3^2*5}=16\)
16*45….n=45
If average is 16, we will require 22 integers lesser than 16, and it will result in negative integers too.

Thus, total 5




When n is 45. We get negative integers. QUESTION STEM asks for positive Integers only.

4 solutions.
n= 3,5,9,15
Which is the 5th solution you're referring to?

Posted from my mobile device

The 4 cases are when n is odd, but 5th is the case when n is even.
The average is 22.5 and the numbers are 7 to 38
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sting8
I had read,

No. of ways to write a number into sum of 2 or more consecutive positive int = no. of odd factors -1.

(Just found a link for this : https://cat.wordpandit.com/cat-number-system-trick/ )

720 = (3^2) * (2^4) * (5^1)

So, no. of odd factors = (2+1) * (1+1) = 6
So, for this question, required answer = 6-1 =5

Am I missing something ? Is there any caveat to the formula, which I am not aware of ?

Bunuel TestPrepUnlimited avigutman GMATinsight EgmatQuantExpert chetan2u

Thanks for your insights chetan2u .

Putting this out for everybody's reference,
There's one more corollary to the formula.

No. Of ways to express a number as sum of 2 or more consecutive integers (includes positive, negative, 0) = 2*(no. Of odd factors) - 1.

I don't have proof for this, I just read it somewhere so sharing it. :)

Posted from my mobile device
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sting8
I had read,

No. of ways to write a number into sum of 2 or more consecutive positive int = no. of odd factors -1.

(Just found a link for this : https://cat.wordpandit.com/cat-number-system-trick/ )

720 = (3^2) * (2^4) * (5^1)

So, no. of odd factors = (2+1) * (1+1) = 6
So, for this question, required answer = 6-1 =5

Am I missing something ? Is there any caveat to the formula, which I am not aware of ?

Bunuel TestPrepUnlimited avigutman GMATinsight EgmatQuantExpert chetan2u

Thanks for your insights chetan2u .

Putting this out for everybody's reference,
There's one more corollary to the formula.

No. Of ways to express a number as sum of 2 or more consecutive integers (includes positive, negative, 0) = 2*(no. Of odd factors) - 1.

I don't have proof for this, I just read it somewhere so sharing it. :)

Posted from my mobile device
Much appreciated. If you were to get this question in exam, you know your score
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