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# Numbers 9

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Manager
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Numbers 9 [#permalink]  06 Nov 2009, 06:00
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Q. If you hav a subset of integers to chosen bet. 1 to 3000, such that no integers add up to multiple of nine, what can be the max number of elements in the subset?

1. 1668
2. 1332
3. 1333
4. 1334
Manager
Joined: 04 Nov 2009
Posts: 66
Schools: London Business School (int)
WE 1: Research
WE 2: Corporate Strat
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Kudos [?]: 12 [1] , given: 1

Re: Numbers 9 [#permalink]  06 Nov 2009, 06:50
1
KUDOS
What is the OA? I get 1336 as the answer which isn't in the options.

The set is {1,2,3,4,9,10,11,12,13,19,20,21,22,28,...2998,2999,3000)

Working:
So all numbers between 1-3000 are of the form 9m, 9m+1, 9m+2, 9m+3, 9m+4, 9m+5, 9m+6, 9m+7, 9m+8

To avoid having any pair of integers at all that add up to 9 or a multiple of 9, we can't have numbers where the remainders add up to 9.
e.g. if you had a number that was 9m+2, then you can't have a number that is of the form 9m+7 as they will add up to be 18m + 9 i.e. a multiple of 9.

To have the maximum number of elements, we take 9m+1, 9m+2, 9m+3, 9m+4 and avoid 9m+5, 9m+6, 9m+7, 9m+8.
Also, we can take one and only one number of the form 9m into the set.

So the total number is no. of integers of the form
9m - we can take 1 such number
9m+1 - there are 334 such numbers below 3000
9m+2 - also 334 such numbers
9m+3 - also 334 such numbers
9m+4 - 333 such numbers (as the last one is >3000)

So total = 1+ 334 + 334 + 334 + 333 = 1336.

Is that the OA ?
Manager
Joined: 19 Aug 2009
Posts: 84
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Kudos [?]: 19 [0], given: 46

Re: Numbers 9 [#permalink]  06 Nov 2009, 12:21
4test1 wrote:
What is the OA? I get 1336 as the answer which isn't in the options.

The set is {1,2,3,4,9,10,11,12,13,19,20,21,22,28,...2998,2999,3000)

Working:
So all numbers between 1-3000 are of the form 9m, 9m+1, 9m+2, 9m+3, 9m+4, 9m+5, 9m+6, 9m+7, 9m+8

To avoid having any pair of integers at all that add up to 9 or a multiple of 9, we can't have numbers where the remainders add up to 9.
e.g. if you had a number that was 9m+2, then you can't have a number that is of the form 9m+7 as they will add up to be 18m + 9 i.e. a multiple of 9.

To have the maximum number of elements, we take 9m+1, 9m+2, 9m+3, 9m+4 and avoid 9m+5, 9m+6, 9m+7, 9m+8.
Also, we can take one and only one number of the form 9m into the set.

So the total number is no. of integers of the form
9m - we can take 1 such number
9m+1 - there are 334 such numbers below 3000
9m+2 - also 334 such numbers
9m+3 - also 334 such numbers
9m+4 - 333 such numbers (as the last one is >3000)

So total = 1+ 334 + 334 + 334 + 333 = 1336.

Is that the OA ?

Again your approach is perfect, logic is also perfect...though answer in the book is 1333, i cant see a flaw in your solving!! Awesome logic!
CEO
Joined: 29 Aug 2007
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Kudos [?]: 422 [0], given: 19

Re: Numbers 9 [#permalink]  06 Nov 2009, 19:41
virtualanimosity wrote:
Q. If you hav a subset of integers to chosen bet. 1 to 3000, such that no integers add up to multiple of nine, what can be the max number of elements in the subset?

1. 1668
2. 1332
3. 1333
4. 1334

What if you add up 9 (9m+1), which is divisible by 9. 3(9m+1) + 2(9m+3) = 9(5m+1) is also divisible by 9.

The list goes long and the OA should be so small that only few such numbers can exist. Answer choices for possible OA look incorrect....

"4test1" did a nice effort.
"virtualanimosity" has a question of 800+ level. What is the source?

virtualanimosity wrote:
4test1 wrote:
What is the OA? I get 1336 as the answer which isn't in the options.

The set is {1,2,3,4,9,10,11,12,13,19,20,21,22,28,...2998,2999,3000)

Working:
So all numbers between 1-3000 are of the form 9m, 9m+1, 9m+2, 9m+3, 9m+4, 9m+5, 9m+6, 9m+7, 9m+8

To avoid having any pair of integers at all that add up to 9 or a multiple of 9, we can't have numbers where the remainders add up to 9.
e.g. if you had a number that was 9m+2, then you can't have a number that is of the form 9m+7 as they will add up to be 18m + 9 i.e. a multiple of 9.

To have the maximum number of elements, we take 9m+1, 9m+2, 9m+3, 9m+4 and avoid 9m+5, 9m+6, 9m+7, 9m+8.
Also, we can take one and only one number of the form 9m into the set.

So the total number is no. of integers of the form
9m - we can take 1 such number
9m+1 - there are 334 such numbers below 3000
9m+2 - also 334 such numbers
9m+3 - also 334 such numbers
9m+4 - 333 such numbers (as the last one is >3000)

So total = 1+ 334 + 334 + 334 + 333 = 1336.

Is that the OA ?

Again your approach is perfect, logic is also perfect...though answer in the book is 1333, i cant see a flaw in your solving!! Awesome logic!

_________________
Manager
Joined: 04 Nov 2009
Posts: 66
Schools: London Business School (int)
WE 1: Research
WE 2: Corporate Strat
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Kudos [?]: 12 [0], given: 1

Re: Numbers 9 [#permalink]  06 Nov 2009, 19:52
I've assumed the question means that no two integers in the set add up to a multiple of 9.

If you instead take that no subset of integers should add up to 9 then the answer will be much smaller.
CEO
Joined: 29 Aug 2007
Posts: 2510
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Kudos [?]: 422 [0], given: 19

Re: Numbers 9 [#permalink]  06 Nov 2009, 20:04
I know however the question did not say anything about how to add up. The question is so vague......

You are right on the possible asnswer - it should be somewhere arround, I guess, 100.

4test1 wrote:
I've assumed the question means that no two integers in the set add up to a multiple of 9.

If you instead take that no subset of integers should add up to 9 then the answer will be much smaller.

_________________
Re: Numbers 9   [#permalink] 06 Nov 2009, 20:04
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# Numbers 9

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