Fake Coin

There are 10 identical coins: 9 are genuine and 1 is fake. What is the minimum number of weighing to find the fake coin? Is it possible to find fake coin in just 1 attempt (i.e. weighing only once) ?

#Logic

Easy Math Editor

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Comments

The minimum chances needed are 9.There is only a 10% chance,that is,1 in a 10 chance of finding the fake one in one try.

Kuldeep Sachan - 5 years, 6 months ago

The minimum no of ways required are 2 First distribute all 10 coins in equal groups of 5 and weigh them obviously 1 side is going to be heavier So remove the top coins presnt at the top grom both the pairs,if the weihhts now become equal implies that one of removed pair consist of fake coin If the weight doesnot become equal implies the top coins removed are real nd the fake coin is still in the lot Continue the process till u get the required pair after that when u get d required pair remove all the coins from weighing slot nd put those two coins along ith a pair of genuine coin One genuine coin nd one coin of required pai nd d other pair with d same Weigh both Obviously dey will not be equalnow remove d genuine coin and one coin from different weighing bare if weight comes equal implies d removes coin which is not genuine is fake Or if not comes equal implies d coin present in bar is d fake coin D weighing machine i have used is d one which was used in barter system not d presnt electronic or other weighing machine Sorry for gramatical errors Feel free to give ur response Its my frst tym in briiliant

Aanchal Shahi - 5 years, 5 months ago

What is your maximum number of weighing ?

Jaimin Pandya - 5 years, 5 months ago

Aanchal Shahi - 5 years, 5 months ago

@Aanchal Shahi – in every cases you will get fake coin by weighing just 2 times ? or this is for the best case only !?

Jaimin Pandya - 5 years, 5 months ago

@Jaimin Pandya – In my opinion this is the best way As to get the required solution

Aanchal Shahi - 5 years, 5 months ago

@Aanchal Shahi – if you take any coin out of pile it simply makes another weighing as per your solution if fake coin in at the bottom it will need to weigh 6 times I think this is not the best solution what I found is 1st -->Divide into 2 groups each having 5 coins and weigh them , discard the higher weighted pile 2nd -->Remaining 5 , divide in 2 , 2 and 1 and compare 2 with 2 , if they are same weighted then remaining one is the fake coin if they are not same weighted then 3rd --> again discard higher weighted and weigh the lower one by doing this at the end you can find fake one

Jaimin Pandya - 5 years, 5 months ago

@Jaimin Pandya – I am using it only twice I am using the weight bar one system In which i put the two piles on two parts of the bar I am not changing the wieghts i am simply removing the top ones therefore in this process i have used it for once only Hope u may have understood

Aanchal Shahi - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$