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Sum of numbers

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by harsh.champ » Tue Feb 09, 2010 5:03 am
How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers"Œ?

(A)11
(B)9
(C)10
(D)5
(E)7

The OA is E.
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by shashank.ism » Tue Feb 09, 2010 5:10 am
harsh.champ wrote:How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers"Œ?

(A)11
(B)9
(C)10
(D)5
(E)7

The OA is E.
The nine consecutive integers can be from (1 to 9) (2 to 10) (3 to 11) (4 to 12) (2 to 10) (5 to 13) (6 to 14) (7 to 15).

Here we haven't included (8 to 16) as they sum up to 108. Now the sums are 45, 54, 63, 72, 81, 90, 99. Thus there are 7 numbers less than 100 which can be expressed as a sum of 9 consecutive integers. The correct choice is [spoiler]option (E)[/spoiler]
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by ajith » Tue Feb 09, 2010 6:44 am
harsh.champ wrote:How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers"Œ?

(A)11
(B)9
(C)10
(D)5
(E)7

The OA is E.
1+...9 = 45
2+ ...10 = 55-1 =54
3....+11 = 66 -3 =63
4....+12 = 78 -6 =72
5......... 13 = 91 -10 = 81
6.........14 = 105 - 15 = 90
7...........15 = 120 - 21 = 99

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by ldoolitt » Tue Feb 09, 2010 7:03 am
...

Or to add another spin, I see this as a consecutive number problem plus an inequality. The sum of eight consecutive numbers can be represented as

n + (n+1) + (n+2) + ... + (n+8) = 9n + 35

We also know that this sum must be less than 100. Therefore we have

9n + 35 < 100
n < 65/9

what we really want is floor(n) since n must be an integer.

floor(n) = floor(65/9) = 7
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by harsh.champ » Fri Feb 19, 2010 11:27 am
ldoolitt wrote:...

Or to add another spin, I see this as a consecutive number problem plus an inequality. The sum of eight consecutive numbers can be represented as

n + (n+1) + (n+2) + ... + (n+8) = 9n + 35

We also know that this sum must be less than 100. Therefore we have

9n + 35 < 100
n < 65/9

what we really want is floor(n) since n must be an integer.

floor(n) = floor(65/9) = 7
I didn't understand the floor function.
Is this some mathematical function??If it is can you give me any related link to it ??
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by shashank.ism » Fri Feb 19, 2010 11:34 am
ldoolitt wrote:...

Or to add another spin, I see this as a consecutive number problem plus an inequality. The sum of eight consecutive numbers can be represented as

n + (n+1) + (n+2) + ... + (n+8) = 9n + 35

We also know that this sum must be less than 100. Therefore we have

9n + 35 < 100
n < 65/9

what we really want is floor(n) since n must be an integer.

floor(n) = floor(65/9) = 7
what is this floor (n) . I am not able to understand this concept. is it some new concept??????
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by ldoolitt » Sun Feb 21, 2010 3:05 pm
The floor function can be defined as follows:

Floor(n) is the greatest integer not larger than n.

For example

Floor(7) = 7
Floor(7.1) = 7
Floor(7.999) = 7

Help?
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by harsh.champ » Sun Feb 21, 2010 3:15 pm
ldoolitt wrote:The floor function can be defined as follows:

Floor(n) is the greatest integer not larger than n.

For example

Floor(7) = 7
Floor(7.1) = 7
Floor(7.999) = 7

Help?
Hey idolitt,
Thanks for explaining it.
Can you give me some link where I can find the concept questions??
Where it is used??
Also,does the knowledge of the function necessary for the GMAT /Is it some advanced maths??

Also can you give me the examples of the question where we can use floor functions and it helps to save time??

Also,is there any corollary function:- "greatest integer larger than n."


This was the 1st time I came across such a function.
Now,I reviewed the soln. poasted by you.
It does make the soln. very very simple.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by shashank.ism » Sun Feb 21, 2010 5:00 pm
ldoolitt wrote:The floor function can be defined as follows:

Floor(n) is the greatest integer not larger than n.

For example

Floor(7) = 7
Floor(7.1) = 7
Floor(7.999) = 7

Help?
Thanks Idoolitt , The floor function is really a very new concept for me .. I was not knowing about it,..
I would try to search some of the similar concept on net..
I f you have any document in details please provide as an attachment...
I think we call it "least inetger function"
[7.999] = 7
[7.2]= 7
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by ldoolitt » Tue Feb 23, 2010 6:50 pm
harsh.champ wrote:
Hey idolitt,
Thanks for explaining it.
Can you give me some link where I can find the concept questions??
Where it is used??
Also,does the knowledge of the function necessary for the GMAT /Is it some advanced maths??

Also can you give me the examples of the question where we can use floor functions and it helps to save time??

Also,is there any corollary function:- "greatest integer larger than n."


This was the 1st time I came across such a function.
Now,I reviewed the soln. poasted by you.
It does make the soln. very very simple.
Hey Champ,

I don't know if its a topic that is covered on the GMAT but it can come in handy (like bayes rule; not tested but super useful for those tricky probability problems) So I can't really guide you to example problems. I can say it can come in handy in "counting" problems where the endpoints don't line up, like the problem above. You know that n must be an integer so you can solve an equation and simply "round down", which is all the floor function does.

There is a corollary, the ceiling function, which is the smallest integer larger than n, not the largest integer larger than n (which would obviously be infinity all the time)
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by ldoolitt » Tue Feb 23, 2010 6:52 pm
shashank.ism wrote:
Thanks Idoolitt , The floor function is really a very new concept for me .. I was not knowing about it,..
I would try to search some of the similar concept on net..
I f you have any document in details please provide as an attachment...
I think we call it "least inetger function"
[7.999] = 7
[7.2]= 7
You have the concept correct.

A link to the wikipedia page might be helpful (although not very exhaustive mathematically)...

https://en.wikipedia.org/wiki/Floor_and_ ... _functions

Luke
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by Mom4MBA » Wed Feb 24, 2010 2:26 pm
nine consecutive numbers will be n, n+1, ...... (n+8)
sum of 9 consecutive positive numbers will be [n+(n+8)]9/2 = 9(n+4)

this term should be a multiple of 9
9(n+4) = 99
:
:
9(n+4) = 45
9(n+4) = 36 this will make n=0, which is not a positive integer. the terms below this will make n negative

so 9(n+4) goes from 9x5 to 9x11, giving 7 values for 9(n+4)
or
so on solving n goes from 1 to 7, that makes 7 values of n
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by Fiver » Fri Feb 26, 2010 12:34 am
harsh.champ wrote:How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers"Œ?

(A)11
(B)9
(C)10
(D)5
(E)7

The OA is E.
Let 'a' be the first term
Avg. = (a+4)
Sum = 9(a+4)
9(a+4) < 100
Max a+4 = 11 (because 9 * 11 = 99)
Therefore Max 'a' = 7
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