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Intern  Joined: 04 Mar 2012
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A person inherited few gold coins from his father. If he  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 42% (02:58) correct 58% (03:08) wrong based on 745 sessions

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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.
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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.

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). For more about this concept see: manhattan-remainder-problem-93752.html#p721341, when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552, when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654

$$c=63k+52$$ 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.

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GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: A person inherited a few gold coins from his father  [#permalink]

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If C = number of coins then C = 7m+3 = 9n + 7, where m and n are the number of bags used in the 7 coins per bag and 9 coins per bag case respectively.

Using statement (1), C lies between 50 and 120. However, 52 and 115 both satisfy the condition. Insufficient.

Using statement (2), C is less than 200. Only 52 satisfies all three conditions. Sufficient.

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

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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

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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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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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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). For more about this concept see: http://gmatclub.com/forum/manhattan-rem ... ml#p721341, http://gmatclub.com/forum/when-positive ... ml#p722552, http://gmatclub.com/forum/when-the-posi ... l#p1028654

$$c=63k+52$$ 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.

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:
$$c=9q+7$$
$$c=7p+3$$
Intern  B
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Re: A person inherited few gold coins from his father. If he  [#permalink]

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This question doesn't make any sense at all.
The "each bag" symbolizes that there is a FIXED amount of bags and that ALL bags are being used.

So, the fact that we get a smaller remainder by putting less coins in each bag is wrong.
What is the source? I very much doubt that this is an official GMAC question
Intern  Joined: 21 Mar 2019
Posts: 1
Re: A person inherited few gold coins from his father. If he  [#permalink]

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If the number of coins c is (9 x an integer) + 7, then (c+2) is simply a multiple of 9.

Since c is (7 x an integer) + 3, we know (c+2) is (7 x an integer) + 5. In other words, (c+2) is two less than a multiple of 7.

So look at the multiples of 9 until you find one that is 2 less than a multiple of 7: 9, 18, 27, 36, 45, 54. The next one will be 54 + 9(7)=117, and the next one will be 117 + 9(7) = 180.

That's the story for (c+2). For c, the list is shifted down 2 units, becoming 52, 115, 178, ...

Statement (1) says c is between 50 and 120. We don't know whether c is 52 or 115 , so (1) is insufficient.

Statement (2) says c is a multiple of 13 and is less than 200. In our list only 52 is a multiple of 13 under 200, so c must be 52. (2) is sufficient. Re: A person inherited few gold coins from his father. If he   [#permalink] 04 Apr 2019, 17:33
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