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An eccentric casino owner decides that his casino should

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TGC
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An eccentric casino owner decides that his casino should [#permalink] New post   Updated on: 19 Mar 2013, 00:26
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An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21
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D

Originally posted by TGC on 19 Mar 2013, 00:12.
Last edited by Bunuel on 19 Mar 2013, 00:26, edited 1 time in total.
Edited the question.
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Re: An eccentric casino owner decides [#permalink] New post   19 Mar 2013, 00:13
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I have a query in this question?

As there it is not mentioned that there can be no chips for particular denomination.

Eqn: 5x + 7y

My query is can we have either x = 0 or y=0 or both?
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Bunuel
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Re: An eccentric casino owner decides [#permalink] New post   19 Mar 2013, 00:33
Expert Reply
targetgmatchotu wrote:
An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21

I have a query in this question?

As there it is not mentioned that there can be no chips for particular denomination.

Eqn: 5x + 7y

My query is can we have either x = 0 or y=0 or both?


Yes, the number of 5 or/and 7 dollar chips can be zero, however both being zero mean that the casino is paying out $0, which is not realistic.

Each option but D can be represented as the sum of a multiple of 5 and a multiple of 7:

A. $31 = 7*3 + 5*2
B. $29 = 7*2 + 5*3
C. $26 = 7*3 + 5*1
E. $21 = 7*3

Answer: D.

Hope it's clear.
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ArunpriyanJ
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Re: An eccentric casino owner decides that his casino should [#permalink] New post   09 Jul 2016, 13:20
Bunuel wrote:
targetgmatchotu wrote:
An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21

I have a query in this question?

As there it is not mentioned that there can be no chips for particular denomination.

Eqn: 5x + 7y

My query is can we have either x = 0 or y=0 or both?


Yes, the number of 5 or/and 7 dollar chips can be zero, however both being zero mean that the casino is paying out $0, which is not realistic.

Each option but D can be represented as the sum of a multiple of 5 and a multiple of 7:

A. $31 = 7*3 + 5*2
B. $29 = 7*2 + 5*3
C. $26 = 7*3 + 5*1
E. $21 = 7*3

Answer: D.

Hope it's clear.


Bunel,

I understood the above explanation. But is there is any other logic we can use to get the answer quick?

Thanks,
Arun
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Bunuel
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Re: An eccentric casino owner decides that his casino should [#permalink] New post   09 Jul 2016, 23:30
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ArunpriyanJ wrote:
Bunuel wrote:
targetgmatchotu wrote:
An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21

I have a query in this question?

As there it is not mentioned that there can be no chips for particular denomination.

Eqn: 5x + 7y

My query is can we have either x = 0 or y=0 or both?


Yes, the number of 5 or/and 7 dollar chips can be zero, however both being zero mean that the casino is paying out $0, which is not realistic.

Each option but D can be represented as the sum of a multiple of 5 and a multiple of 7:

A. $31 = 7*3 + 5*2
B. $29 = 7*2 + 5*3
C. $26 = 7*3 + 5*1
E. $21 = 7*3

Answer: D.

Hope it's clear.


Bunel,

I understood the above explanation. But is there is any other logic we can use to get the answer quick?

Thanks,
Arun


Trial and error is pretty much the only way to solve this question.
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Re: An eccentric casino owner decides that his casino should [#permalink]
09 Jul 2016, 23:30
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