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# Gold chests

## Question

Hello,

I have this logical question:

## We have 3 chests A, B, C. We know that chest A have more coins than chest B. Also we know that one the chest contains 50 gold coins and next one contains 40 silver coins and the last one contains 10 gold coins and 10 silver coins. What is lowest number of coins blindly taken from chest which guarantees content determination of each chest?

Sorry english is not my native language.

Good Luck

## Recommended Posts

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I understood the OP that you can only remove X coins from ONE chest and deduce the contents of all chests by observing removed coins. What's the lowest X?

Spoiler

You need to remove 11 coins from chest C.

If they are all Gold then A = 40S, B = 10G+10S, C = 50G
If they are all Silver then A = 50G, B = 10G+10S, C = 40S
If they are a mix then A = 50G, B = 40S, C = 10G+10S

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On 12/20/2021 at 12:03 AM, Wynn said:

Hello,

I have this logical question:

## We have 3 chests A, B, C. We know that chest A have more coins than chest B. Also we know that one the chest contains 50 gold coins and next one contains 40 silver coins and the last one contains 10 gold coins and 10 silver coins. What is lowest number of coins blindly taken from chest which guarantees content determination of each chest?

Sorry english is not my native language.

Good Luck

22 coins in the worst scenary , 12 in the best.

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You could always be certain of each chest's identity after pulling 12 coins.

If you're lucky, you might be able to know them after pulling a single coin.

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I get 13.

Take one coin of each chest.

a) You get two Gold, one Silver:
Silver is identified, remains to distinguish all_Gold and Gold_Silver.
Take 10 more coins from the first G:
- if they are all gold, you identified all_Gold, the remaining is Gold_Silver
- if you get a silver coin, you identified Gold_Silver, the remaining is all_Gold

b) You get one Gold, two Silver:
Same proceeding, permute Gold/Silver.

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The thing that makes this problem different is

Spoiler

If you know that chest A has more coins than chest B, then chest A can't have the fewest coins (10 gold and 10 silver). So you can pull one coin from chest A and be certain of whether it has 50 gold or 40 silver.

If chest A has 40 silver, and it has more coins than chest B, then chest B must have 10 gold and 10 silver so you're done.

If chest A has 50 gold, I think you're stuck having to pick either chest B or chest C and pull up to 11 coins from it. If all 11 are silver then that's the chest with 40 silver; if any of them are gold then you can stop at that point and know it's the 10 gold and 10 silver chest.

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are they simply "mystery  trunks" - not revealing count?   & not the Letter?!

only that there are 50 G and 40 S

Eleven would determine  10-10  ----> ran through all the S  or all  the G

if it is different - then it is the "10-10"

one would then need to move to a different box ------> & there you have it!!

where the next draw reveals itself as either all G or all S  [if G then it is A];  [if S, then it is B]

## We know that chest A have more coins than chest B.

one chest contains 50 gold coins and next one contains 40 silver coins and the last one contains 10 gold coins and 10 silver coins.

BUT WHAT IF the 11th draw was the same?? i

one would then need to move to a different box ------> & either the mix or the other "all"

what then. lift the box 50  or 20?  i know that's lame....  gonna go think some more...

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suppose instead, take 1 from each chest (COUNT is 3) you will have 2 coins the same & the odd box would either the gold 50 (now 49) or the silver 40 (now 39).

the matching, let's say "silver" coins one is the 39 the other 19 - sadly you may not be lucky enough to draw gold from the 20 er 19 chest until the 10th coin  from that chest is drawn- when a gold coin is found (or not) we know each of those 2.  So without a merciful reveal the total will have been 3 + 10.

But maybe , if y're fortunate... the 4th draw will be the reveal.

My previous consideration would have come to 11  plus 1 = 12

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