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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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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, 07:00
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6
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Difficulty:

75% (hard)

Question Stats:

66% (01:51) correct 34% (02:05) wrong based on 332 sessions

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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

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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, 07:24
3
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?

Given: 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.

Asked: What is the greatest number of apples that you CANNOT buy in the store?

Let 6x + 9y+ 20z = k where k = number of apples to be bought form the store

A. 43
43 is not divisible by 6 or 9
43 = 20 + 23 but 23 is also not divisible by 6 or 9
43 = 20*2 + 3 but 3 is also not divisible by 6 or 9
We can not purchase 43 items from the store

B. 44
44 is not divisible by 6 or 9
44 = 20 + 24 but 24 is divisible by 6
44 = 20 + 6*4
We can purchase 44 items from the store

C. 45
45 = 9*5
We can purchase 45 items from the store

D. 46
46 is not divisible by 6 or 9
46 = 20 + 26 but 26 is divisible by 6
46 = 20*2 + 6
We can purchase 46 items from the store

E. 47
47 is not divisible by 6 or 9
47 = 20 + 27 but 27 is divisible by 9
47 = 20 + 9*3
We can purchase 47 items from the store

IMO 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: 26 Jul 2019, 21:23
2
Let us go option by option to check which combination cannot be formed-

E. 47 - 9 * 3 + 20 -> can be formed.
D. 4620 * 2 + 6 -> can be formed.
C. 459 * 5 -> can be formed.
B. 446 * 4 + 20 -> can be formed.
A. 43 -> cannot be formed.

Originally posted by Sayon on 26 Jul 2019, 07:20.
Last edited by Sayon on 26 Jul 2019, 21:23, 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, 11:29
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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27 Jul 2019, 02:11
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--> correct :can't be bought: 43 can be make w/ any combination of 6,9 & 20
B. 44--> can be bought: 44=20+4*6
C. 45--> can be bought: 45 = 5*9
D. 46--> can be bought: 46=2*20+6
E. 47--> can be bought: 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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27 Jul 2019, 06:50
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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A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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Updated on: 26 Jul 2019, 07:35
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

All are possible except 43
9*5=45
9*3+20=47
20+6*4=44
6*6+20=46

Hence A

Originally posted by kitipriyanka on 26 Jul 2019, 07:13.
Last edited by kitipriyanka on 26 Jul 2019, 07:35, edited 3 times 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, 07:14
1
I thought of the combinations that you can buy, so disprove of the ones you can buy.

B (44) = 4(6) + 1(20)
C (45) = 5(9)
D (46) = 2(20) + 1(5)
E (47) = 1(20) + 3(9)

I did not find anything to sum to 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, 07:19
1
44 is a possible combination 6*6+20
45 is 5 packs of (9)
47 is 20+3*(9)
46 is 2*(20)+6

leaving us with 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, 07:25
1
A

All other no of apples can be bought using the combination below:

A. 43
B. 44 : 4*6+1*20
C. 45: 5*9
D. 46: 1*6+2*20
E. 47 : 3*9+1*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, 07:26
1
Using PoE
Start from Option E, since largest is asked

E) 47
20 + 27 = 20 + 9*3
Eliminated
D) 46
20*2 + 6
Eliminated
C) 45
9*5
Eliminated
B)44
20 + 6*4
Eliminated

So, 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, 07:28
1
To solve this problem, we should just check the answers - this is the easiest way.

A. 43

B. 44
20+4*6=44

C. 45
9*5=45

D. 46
2*20+6=46

E. 47
20+9*3=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, 07:30
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

B. 44=20+6(4)
C. 45=9(5)
D. 46=20(2)+6
E. 47=20+9(3)

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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, 07:30
1
Here you must find a number that is not a multiple of the possible combinations of 6,20 and/or 9.

A. 43 , hold
B. 44 = 20 + 6*4 - So discard it
C. 45 = 9*5 - So discard it
D. 46 = 20*2 + 6, so discard it
E. 47 = 20 + 9*3, so dicard it

So by POE, (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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26 Jul 2019, 07:36
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?

This is an easy question. Straight away get to POE.

A. 43
This is correct it is not possible.

B. 44
This is possible 20 and 24/6.

C. 45
This is possible 45/9.

D. 46
This is possible 20*2 and 6*1

E. 47
This is possible 20*1 & 27/9

Hence the correct answer 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, 07:38
1
4x6 + 1x20 = 44
5x9 =45
1x6 + 2x20 = 46
3x9 + 1x20 = 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, 07:44
1

45 = 27+18 --> multiples of 9 and 6

46 = 40 + 6 --> multiples of 20 and 6

44 = 6+18+20 --> multiples of 6,9 and 20

47 = 27 + 20 --> multiples of 9 and 20

only remaining answer is 43, which is not possible, so 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, 07:50
1
A. 43
NOT POSSIBLE

B. 44
44 = 6*4 + 20*1 - POSSIBLE

C. 45
45 = 9*5 - POSSIBLE

D. 46
46 = 6*1 + 20*2 - POSSIBLE

E. 47
47 = 6*3 + 9*1 + 20*1 - 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, 07:51
1
We are given that,
there are pack of 6, 9 or 20
We also know that if someone buys 2 pack of 6 =2*6=12 (he gets 12 but he cant get 13)
This means that the pack number is an Integer.

So we are going to use hit and trail method in this question.

1) 6*4+20=44
2) 6*6+9=45
3) 20*2+6=46
4) 9*3+20=47

This leaves us wtih ans =43=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, 07:59
1
Method - Eliminate possible combinations from answer choices

A. 43 - No Combination exists
B. 44 - (1*20 Packs) + (4*6 Packs)
C. 45 - (5*9 Packs)
D. 46 - (1*6 Packs) + (2*20 Packs)
E. 47 - (3*9 Packs) + (1*20 Packs)

IMO A
Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6   [#permalink] 26 Jul 2019, 07:59

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