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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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26 Jul 2019, 11:02
1
Resolve by elimination, starting with the largest number in the answer options:
47: (9*3) + (20*1) -> Eliminate
46: (20*2) + (6*1) -> Eliminate
45: (9*5) ->Eliminate
44: (6*3)+(20*1) -> Eliminate
43: No combination

We can also solve by factorizing each of the answer options.

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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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26 Jul 2019, 11:39
1
For the second option we've got
44= 6×4 + 1×20
For the third option we've got
45= 9×5
For the fourth option we've got
46= 2×20 + 1×6
For the final optio we've got
47= 1×20 + 3×9

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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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26 Jul 2019, 11:40
1
Best way is to check with option

B. 44 (( 20 + 6*4 ))
C. 45 (( 5*9 ))
D. 46 (( 20 + 20 + 6))
E. 47 (( 20 + 6*3 + 9))
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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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26 Jul 2019, 12:07
1
IMO correct answer is A - Explanation is provided as attachment -
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IMG_20190727_003342.JPG [ 832.66 KiB | Viewed 309 times ]

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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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26 Jul 2019, 12:29
2
Here store can sell apples in a pack of (6k,6k+3,6k+2)i.e (6,9,20)
Now with these three nos. we can form any no.of the form

46(6m+4) : so any no of the form 6m+4 can be formed by taking two(6k+2) and rest 6k.
here 20+20+6

47(6m+5) : one 6k+3 and one 6k+2 rest 6k
here 9+20+6+6+6

45(6m+3) : one 6k+3 and rest 6k
here 9+6+6+6+6+6+6

44(6m+2) : one 6k+2 and rest 6k
here 20+ 6+6+6+6

43(6m+1) : it can be form if we take 2 (6k+2) and one (6k+3) i.e 20+20+9 which is equal to 49.
so the least no. of 6m+1 form that can be formed by 6,9,20 is 49 .
so 43 apples is not possible with these bags of 6,9,20 apples

hence A : 43

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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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26 Jul 2019, 12:40
1
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?

Let's see the numbers:

A. 43
Any combination of 6,9, and 20 won't give us 43

B. 44
4 of 6 pack = 24 + 1 pack of 20 = 44

C. 45
5 of 9 pack = 45

D. 46
2 of 20 pack = 40 + 1 pack of 6 pack = 46

E. 47
3 of 9 pack = 27 + 1 pack of 20 = 47

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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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26 Jul 2019, 13:20
1
To rephrase, one can only sell set of 6 apples or 9 apples or 20 apples and arrange that X number of set of 6 apples, Y number of set of 9 apples, and Z number of set of 20 apples

A. 43 =
B. 44 = 6*4 + 20*1
C. 45 = 6*6 + 9*1
D. 46 = 6*1 + 20*2
E. 47 = 9*3 + 20*1

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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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26 Jul 2019, 13:52
1
Quote:
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

As we are asked the greatest number of apples,
lets start from the options in the descending way.
47 = 20 + 27 = 20 + 9*3
hence E is out

46 = 20 +20 + 6
hence D out

45 = 9*5
hence C out.

44= 20 +24 = 20 + 6* 4
Hence B out

43 = 20 +23 or 9 + 32 or 6 +37
Hence combination of 6,9 and 20 is not possible.

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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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26 Jul 2019, 14:25
1
the 6-pack can produce packages of: 0,6,12,18,24,30,36,42,48
the 9-pack can produce packages of: 0,9,18,27,36,45
the 20-pack can produce packages of: 0,10,20,30,40

47 = 20 + 27 = $$(20*1)+(9*3)$$
46 = 40 + 6 = $$(20*2)+(6*1)$$
45 = 45 = $$(9*5)$$
44 = 20 + 24 = $$(20*1)+(6*4)$$

so 43 is left,
by trying combinations of the possible numbers, it is not possible

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

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Updated on: 27 Jul 2019, 00:27
1
The question basically asks which of the following CANNOT be the results of $$6x+9y+20z$$. $$x$$, $$y$$, and $$z$$ are non-negative integers. Let’s begin analyze one by one:

A. $$43$$. Let’s first begin from using 20 because it is the easiest to use.

1. $$43-2*20=3$$ out. $$43-20=23$$ can $$23$$ be the result of $$6x+9y$$? Let’s begin from the multiples of $$9$$ because we have less calculation with it. $$23-18=5$$ out. $$23-9=14$$ not divisible by $$6$$, so out. Now let’s use multiples of $$6$$. $$23-18=5$$ out. $$23-12=11$$ out. $$23-6=17$$ not divisible by $$9$$ out. So $$20$$ apple pack is not participating in $$43$$.

2. Let’s start from the multiples of $$9$$. $$43-36=7$$ out. $$43-27=16$$ out. $$43-18=25$$ out. $$43-9=34$$ out because all the results are not divisible by $$6$$.

3. Let’s use the multiples of the $$6$$ now. $$43-42=1$$ out. $$43-36=7$$ out. $$43-30=13$$ out. $$43-24=19$$ out. $$43-18=25$$ out. $$43-12=31$$ out. $$43-6=37$$ out because all the results are not divisible by $$9$$. Hence $$6x+9y+20z\neq{43}$$.

B. $$44 = 20 + 4*6$$

C. $$45 = 5*9$$

D. $$46 = 2*20 + 6$$

E. $$47 = 20 + 3*9$$

Hence A
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Originally posted by JonShukhrat on 26 Jul 2019, 15:52.
Last edited by JonShukhrat on 27 Jul 2019, 00:27, edited 1 time in total.
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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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26 Jul 2019, 17:17
1
We need to find out the answer choices which are sum of the multiples of 6,9, and 20 and eliminate them. The highest of the remaining answer choices is our Answer.

47=20(1)+9(3)=20+27. Hence E is out.
46=20(2)+6=40+6, hence D is out as well.
45 =9(5), hence C is out.
44=20+6(4)=20+24, hence B is out.

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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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26 Jul 2019, 17:52
1
A. 43 —> Not Possible
B. 44 = 20 + 24 = 20x1 + 6x4 —> Possible
C. 45 = 36 + 9 = 6x6 + 9x1 —> Possible
D. 46 = 40 + 6 = 20x2 + 6x1 —> Possible
E. 47 = 20 + 27 = 20x1 + 9x3 —> Possible

IMO Option A

Pls Hit kudos if you like the solution

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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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26 Jul 2019, 17:52
1
We can buy 5 packs of 9 apples = 45.........Eliminate C

We can buy 3 packs of 9 apples and 1 pack of 20 apples= 27 + 20=47.............Eliminate E

We can buy 2 packs of 20 apples and 1 pack of 6 apples = 40 + 6 =46.............Eliminate D

We can buy 1 pack of 20 apples and 4 packs of 6 apples = 20 +24 =44.............Eliminate B

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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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26 Jul 2019, 18:04
1
Since answer choices are small, it's easy to check some numbers

A) 43 not possible
B) 44 = 20(1) + 6(4)
C) 45 = 9(5)
D) 46 = 20(2) + 6(1)
E) 47 = 20(1) + 9(3)

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

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26 Jul 2019, 19:54
1
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?

lets solve one by one using combination

B. 44 - 4 packs of 6 apples and 1 pack of 20 apples = 44
C. 45 - 6 packs of 6 apples and 1 pack of 9 apples = 45
D. 46 1 pack of 6 apples and 2 packs of 20 apples = 46
E. 47 3 packs of 9 apples and 1 pack of 20 apples = 47

so i can buy above 44,45,46,and 47 no of apples as per the above combinatons explanined

but 43 apples cannot be bought by any combination, Hence that is the greatest number not possible out of the choices given

so out of the choices
A. 43 cannot be bought
B. 44 - can be bought
C. 45 can be bought
D. 46 can be bought
E. 47 can be bought

Hence ans is 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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26 Jul 2019, 20:34
1
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

20 = 2*2*5
6 = 2*3
9 = 3*3

So any multiple of the above combinations will be possible to buy

I tried to backsolve using the answer choices

Taking 43, 43 can't be factorized into multiples of 20, 6 or 9. So, this should be the correct answer.
Other choices can be verified for confirmation.
44 = 20 + 6*4
45 = 9*5
46 = 20*2 + 6
47 = 20 + 3*9

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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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26 Jul 2019, 20:51
1
POE
B- 44: 20+ (6x4)= 44 Eliminate
C- 45: (9x5)= 45 Eliminate
D- 46: (20x2)+ +6= 46 Eliminate
E- 47: 20+ (9x3)= 47 Eliminate

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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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26 Jul 2019, 20:54
1
A cant be bought.
B 6 + 2×9 + 20
C 9x5
D 20x2 + 6
E 9x3 + 20

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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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26 Jul 2019, 21:20
1
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?

Eliminating options,
6*4 + 20 = 44, eliminate option B
9*5 = 45, eliminate option C
6 + 20*2 = 46, eliminate option D
9*3 + 20 = 47, eliminate option E

So, 43 apples cannot be bought.

Hence 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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26 Jul 2019, 21:22
1
IMO A.

Given: store sells apples in pack of 6, 9, or 20 and can be sold in whole packs only.

To find: Greatest number of apples that cannot be bought from the store.

A. 43--> Cannot be summed. Correct answer.
B. 44--> Can be summed as = $$6 X 4+ 20 X 1.$$
C. 45--> Can be summed as =$$6 X 6+ 9 X 1.$$
D. 46 --> Can be summed as =$$20 X 2+ 6 X 1.$$
E. 47 --> Can be summed as = $$20 X 2+ 9 X 3.$$
Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6   [#permalink] 26 Jul 2019, 21:22

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