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A person inherited few gold coins from his father. If he [#permalink]

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13 May 2012, 08:13

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A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father.

(1) The number of coins lies between 50 to 120. (2) If he put 13 coins in each bag then no coin is left over and number of coins being lesser than 200.

A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father.

If he puts 9 coins in each bag then 7 coins are left over --> \(c=9q+7\), so # of coins can be: 7, 16, 25, 34, 43, 52, 61, ... If he puts 7 coins in each bag then 3 coins are left over --> \(c=7p+3\), so # of coins can be: 3, 10, 17, 24, 31, 38, 45, 52, 59, ...

\(c=63k+52\) means that # of coins can be: 52, 115, 178, 241, ...

(1) The number of coins lies between 50 to 120 --> # of coins can be 52 or 115. Not sufficient.

(2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200 --> # of coins is a multiple of 13 and less than 200: only 52 satisfies this condition. Sufficient.

Re: A person inherited few gold coins from his father. If he [#permalink]

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06 Jun 2013, 17:25

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gmihir wrote:

A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father.

(1) The number of coins lies between 50 to 120. (2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200.

Please correct the question the 2nd statement should mention each bag. ((2) If he put 13 coins in each bag then no coin is left over and number of coins being lesser than 200.[/quote])

Re: A person inherited few gold coins from his father. If he [#permalink]

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05 Aug 2014, 21:01

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Re: A person inherited few gold coins from his father. If he [#permalink]

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11 Aug 2014, 05:59

HI Bunuel:

I understand that this theorem is " Chinese Theorem". C=63K + 52 is obtained by C = (9*7) K + 52. 52 is found by looking at both the lists. However, I remember reading a shortcut to arrive at 52 as well. But not able to recall now.

Can u please describe the shortcut here. (This will help saving extra seconds in gmat).

Thanks
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Re: A person inherited few gold coins from his father. If he [#permalink]

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11 Aug 2014, 22:03

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Bunuel wrote:

A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father.

If he puts 9 coins in each bag then 7 coins are left over --> \(c=9q+7\), so # of coins can be: 7, 16, 25, 34, 43, 52, 61, ... If he puts 7 coins in each bag then 3 coins are left over --> \(c=7p+3\), so # of coins can be: 3, 10, 17, 24, 31, 38, 45, 52, 59, ...

General formula for \(c\) based on above two statements will be: \(c=63k+52\) (the divisor should be the least common multiple of above two divisors 9 and 7, so 63 and the remainder should be the first common integer in above two patterns, hence 52). .

Awesome!!! I have never known about the formula before. Thanks a ton, Bunuel!!!!!!
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Re: A person inherited few gold coins from his father. If he [#permalink]

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03 Oct 2015, 16:53

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Re: A person inherited few gold coins from his father. If he [#permalink]

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13 Nov 2016, 16:38

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Re: A person inherited few gold coins from his father. If he [#permalink]

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02 Apr 2017, 10:08

Bunuel wrote:

A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father.

If he puts 9 coins in each bag then 7 coins are left over --> \(c=9q+7\), so # of coins can be: 7, 16, 25, 34, 43, 52, 61, ... If he puts 7 coins in each bag then 3 coins are left over --> \(c=7p+3\), so # of coins can be: 3, 10, 17, 24, 31, 38, 45, 52, 59, ...

\(c=63k+52\) means that # of coins can be: 52, 115, 178, 241, ...

(1) The number of coins lies between 50 to 120 --> # of coins can be 52 or 115. Not sufficient.

(2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200 --> # of coins is a multiple of 13 and less than 200: only 52 satisfies this condition. Sufficient.

Answer: B.

Bunuel Would not the no of bag be same for both the scenarios "If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over" . If that be the case: \(c=9q+7\) \(c=7p+3\) p and q above would be same constant. Please help in understanding this

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