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Sum survey challenge (made up)

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Sum survey challenge (made up)

by Mathsbuddy » Tue Nov 19, 2013 11:44 pm
If S is the sum of c consecutive integers where S < 100,c>1 and s * c = n^2 (where n is also an integer), can the following statement be true?

s*c is not equal to any multiple of 3.

(I would be interested in seeing your reasoning/logic/proof for your answer too.)

Made up while driving home last night, so I can't guarantee any GMat authenticity.
I would also be interested to know how you would rate how easy/challenging you found this question compared to real GMat questions. (It is not research, so no data will be stored; just personal curiosity for pitching questions right in future).

Rating scale:


1 = too easy
2 = as easy has the easiest GMat questions
3 = easier than most GMat questions
4 = about as easy or hard as the average thGMat questions
5 = harder than most GMat questions
6 = as hard has the hardest GMat questions
7 = much harder than the toughest GMat questions

Many thanks :)
Last edited by Mathsbuddy on Wed Nov 20, 2013 5:24 am, edited 1 time in total.
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by ganeshrkamath » Wed Nov 20, 2013 12:23 am
Mathsbuddy wrote:If S is the sum of c consecutive integers where S < 100,c>1 and s * c = n^2 (where n is also an integer), can the following statement be true?

s*c is not equal to 81.
S = (average of c consecutive integers) * c
S = (average of c consecutive integers) * c
S*c = (average of c consecutive integers) * c^2 = n^2

So average of c consecutive integers should be a perfect square.
Clearly, c should be an odd number.

Let c = 3 (smallest possible positive odd number after 1)
and average = 4 (smallest possible perfect square after 1)
The numbers are (3,4,5)
S = 12
S*c = 12*3 = 36 = 6^2

So the statement can be true.

Mathsbuddy wrote:(I would be interested in seeing your reasoning/logic/proof for your answer too.)

Made up while driving home last night, so I can't guarantee any GMat authenticity.
I would also be interested to know how you would rate how easy/challenging you found this question compared to real GMat questions. (It is not research, so no data will be stored; just personal curiosity for pitching questions right in future).

Rating scale:


1 = too easy
2 = as easy has the easiest GMat questions
3 = easier than most GMat questions
4 = about as easy or hard as the average thGMat questions
5 = harder than most GMat questions
6 = as hard has the hardest GMat questions
7 = much harder than the toughest GMat questions

Many thanks :)
I've never encountered such GMAT questions in any of my mocks or the actual test.
It's definitely hard. I would give it a 6. :)

Cheers,
Ganesh
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GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
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by Mathsbuddy » Wed Nov 20, 2013 5:28 am
ganeshrkamath wrote:
Mathsbuddy wrote:If S is the sum of c consecutive integers where S < 100,c>1 and s * c = n^2 (where n is also an integer), can the following statement be true?

s*c is not equal to 81.
S = (average of c consecutive integers) * c
S = (average of c consecutive integers) * c
S*c = (average of c consecutive integers) * c^2 = n^2

So average of c consecutive integers should be a perfect square.
Clearly, c should be an odd number.

Let c = 3 (smallest possible positive odd number after 1)
and average = 4 (smallest possible perfect square after 1)
The numbers are (3,4,5)
S = 12
S*c = 12*3 = 36 = 6^2

So the statement can be true.

Mathsbuddy wrote:(I would be interested in seeing your reasoning/logic/proof for your answer too.)

Made up while driving home last night, so I can't guarantee any GMat authenticity.
I would also be interested to know how you would rate how easy/challenging you found this question compared to real GMat questions. (It is not research, so no data will be stored; just personal curiosity for pitching questions right in future).

Rating scale:


1 = too easy
2 = as easy has the easiest GMat questions
3 = easier than most GMat questions
4 = about as easy or hard as the average thGMat questions
5 = harder than most GMat questions
6 = as hard has the hardest GMat questions
7 = much harder than the toughest GMat questions

Many thanks :)
I've never encountered such GMAT questions in any of my mocks or the actual test.
It's definitely hard. I would give it a 6. :)

Cheers,
Ganesh
Well done! As you beat the system, I have now made the question more interesting:
Can it be that s*c is not equal to any multiple of 3?

Thank you.
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by ganeshrkamath » Wed Nov 20, 2013 6:04 am
Mathsbuddy wrote:Well done! As you beat the system, I have now made the question more interesting:
Can it be that s*c is not equal to any multiple of 3?

Thank you.
For S*c to be not equal to a multiple of 3:
c shouldn't be a multiple of 3
Let c = 5 (the next smallest odd number)
Again, average of the c consecutive numbers should be a perfect square:
Let average = 1 (the smallest possible perfect square)
The numbers are (-1,0,1,2,3)
S = 5
c = 5
S*c = 25 = 5^2

So, S*c need not be a multiple of 3.

Cheers,
Ganesh
Every job is a self-portrait of the person who did it. Autograph your work with excellence.

Kelley School of Business (Class of 2016)
GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494
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by Mathsbuddy » Wed Nov 20, 2013 9:28 am
ganeshrkamath wrote:
Mathsbuddy wrote:Well done! As you beat the system, I have now made the question more interesting:
Can it be that s*c is not equal to any multiple of 3?

Thank you.
For S*c to be not equal to a multiple of 3:
c shouldn't be a multiple of 3
Let c = 5 (the next smallest odd number)
Again, average of the c consecutive numbers should be a perfect square:
Let average = 1 (the smallest possible perfect square)
The numbers are (-1,0,1,2,3)
S = 5
c = 5
S*c = 25 = 5^2

So, S*c need not be a multiple of 3.

Cheers,
Ganesh
Well done, you found the solution with ease! I was trying to "hide" the fact that there was a negative value in the list, but you got it. I think its the only solution, or at least that was the intention. Thank you for trying it out!
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by rahul.sehgal@btgchampion » Sun Dec 01, 2013 4:21 am
ganeshrkamath wrote:
Mathsbuddy wrote:If S is the sum of c consecutive integers where S < 100,c>1 and s * c = n^2 (where n is also an integer), can the following statement be true?

s*c is not equal to 81.
S = (average of c consecutive integers) * c
S = (average of c consecutive integers) * c
S*c = (average of c consecutive integers) * c^2 = n^2

So average of c consecutive integers should be a perfect square.
Clearly, c should be an odd number.

Let c = 3 (smallest possible positive odd number after 1)
and average = 4 (smallest possible perfect square after 1)
The numbers are (3,4,5)
S = 12
S*c = 12*3 = 36 = 6^2

So the statement can be true.

Mathsbuddy wrote:(I would be interested in seeing your reasoning/logic/proof for your answer too.)

Made up while driving home last night, so I can't guarantee any GMat authenticity.
I would also be interested to know how you would rate how easy/challenging you found this question compared to real GMat questions. (It is not research, so no data will be stored; just personal curiosity for pitching questions right in future).

Rating scale:


1 = too easy
2 = as easy has the easiest GMat questions
3 = easier than most GMat questions
4 = about as easy or hard as the average thGMat questions
5 = harder than most GMat questions
6 = as hard has the hardest GMat questions
7 = much harder than the toughest GMat questions

Many thanks :)
I've never encountered such GMAT questions in any of my mocks or the actual test.
It's definitely hard. I would give it a 6. :)

Cheers,
Ganesh

Ganesh - I have been following your solutions and I must say you are very helpful. Could you help me understand when you say - 'Clearly, c should be an odd number' in your solution above..

I might be missing a trick here.. Thanks in advance and I wish you best for your ISB application :)
Best Regards,
Rahul Sehgal
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by naseem_iqbal » Sun Dec 01, 2013 11:20 am
If c was an even number, their average would not be an integer. To avoid confusion, imagine all numbers on a number line. Since they are all consecutive, the average should be exactly on the center. If there were an even number of numbers, avg would be something point 5.
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