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Number system

by harsh.champ » Sun Feb 07, 2010 7:39 am
A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase using some combination of these boxes?

(A)43
(B)199
(C)99
(D)86
(E)None of the above.
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by ajith » Sun Feb 07, 2010 7:52 am
harsh.champ wrote:A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase using some combination of these boxes?

(A)43
(B)199
(C)99
(D)86
(E)None of the above.
199 = 2*6+3*9+ 8*20
99 =2*6+3*9+ 3*20
86 = 20*4+6
43 ??( to have a last digit 3,the options are ... 9*7, 3(9+2*6), 6*4+9, .. but I could not make 43)
So I am gonna stick my head out and say 43 is not possible
Last edited by ajith on Sun Feb 07, 2010 7:52 am, edited 1 time in total.
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by shashank.ism » Sun Feb 07, 2010 7:52 am
harsh.champ wrote:A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase using some combination of these boxes?

(A)43
(B)199
(C)99
(D)86
(E)None of the above.
The ans is [spoiler](B)[/spoiler]
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by ajith » Sun Feb 07, 2010 8:22 am
shashank.ism wrote:
harsh.champ wrote:A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase using some combination of these boxes?

(A)43
(B)199
(C)99
(D)86
(E)None of the above.
The ans is [spoiler](B)[/spoiler]
See if you have put some efforts to solve the problem why not post it?
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by onedayi'll » Sun Feb 07, 2010 1:33 pm
agreed with 43,
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by shashank.ism » Sun Feb 07, 2010 10:53 pm
ajith wrote:
harsh.champ wrote:A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase using some combination of these boxes?

(A)43
(B)199
(C)99
(D)86
(E)None of the above.
199 = 2*6+3*9+ 8*20
99 =2*6+3*9+ 3*20
86 = 20*4+6
43 ??( to have a last digit 3,the options are ... 9*7, 3(9+2*6), 6*4+9, .. but I could not make 43)
So I am gonna stick my head out and say 43 is not possible
Ajith actually in haste I tried to solve the problem and thought it could be B. But in between you posted your solution which let me think in ur direction. You have posted a good solution and surely its a nice approach.. kudos.

Then also I am not exactly getting this problem after giving a day thought. If we go through this question again, We find that it is asking for largest number. so whether we have to check for the largest number among the given option or any arbitrary number. If we go for any arbitrary number then there can be other numbers possible(It is just my view as I m in lots of confusion regarding this problem.)

Also there is an option "None of These" which makes it more confusing.
Ajith I would like to have further suggestion on this post. It would be great if some instructor help us out with detailed solution. Harsh do you have the exact solution of the problem, if any, please post it.
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by kvamsy » Sun Feb 07, 2010 11:25 pm
harsh.champ wrote:A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase using some combination of these boxes?

(A)43
(B)199
(C)99
(D)86
(E)None of the above.

I agree with Harsh, Because If you could see the option B in which, it is possible to split in to above combinations as below

i:e 199 = First 100 can be split in to 5 cases consists of 20 = 100
Second 99 can be split into 11 cases consists of 9 = 99
Total = 199 Hence the option B is not not valid.

43 in which it is not possible to split the cases in to any one of the combination....
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