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A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6

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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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New post 27 Jul 2019, 06:21
1
6+9*2+20*1=44
6*3+9*3=45
6+20*2=46
6*3+9+20=47

43 is not possible .Hence A
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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New post 27 Jul 2019, 07:44
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?

A. 43
B. 44
C. 45
D. 46
E. 47

Lets start with highest number in the option 47. It can be bought. 1 pack of 20 and 3 packs of 9.
46 cannot be bought - hence answer is D
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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New post 27 Jul 2019, 07:50
2
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?

A. 43
B. 44
C. 45
D. 46
E. 47

hence bananas can be of the form 6x or 9x or 20 x or 6 x+9y+20z
here x y and z are >=0
Try out various combinations
6x+9y+20z
47 = 6*0 +9*3+20*1
46 = 6*1 +20*2
45 = 6*0+9*5 + 20*0
44 = 6*4 +20*1+ 9*0
43 .. no combination can not form it
Hence 43 A is the answer
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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New post 27 Jul 2019, 09:15
1
backsolve and get 43
option a
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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New post 04 Dec 2019, 05:06
Prasannathawait wrote:
we cannot have 43 and 47. 47 is the largest.
E is the answer.


E option is not correct as:

47=20*1+9*3
Among the given numbers, only 43 is not possible to be obtained from any combination
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6   [#permalink] 04 Dec 2019, 05:06

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