## Problem based on coins

Hi all, Did anyone has a solution for this question? The maximum number of identical coins from which a faulty one (weighing less than all others) can be found in three weighings on a simple balance is: Thanks in advance. Reference: It is given that the answer is 27.

## The simple trick behind all

The simple trick behind all such coin problem is the concept of 3 0 , 3 1, 3 2. 3 3 & etc.

If you consider any coin problem carefully with above funda then it will not take much time to solve such problems.Using weights 1, 3, 9, 27, 81, etc. you can add and subtract them to make any whole number, for example 21 is 27+3-9. So you could weigh a 21-ounce object by placing it in one pan along with the 9-ounce weight, and putting 3 and 27 in the other pan.

The way to approach this type of problem is to start with just a single coin, and work up towards the maximum. With just one coin, there is nothing you can do with just a pan balance. However, if you have two coins then you can find out which one is lighter and if you have three coins then also you can find out which one is lighter on one check. Put one coin on left pan and one one right, the pan which keeps low contains the lighter weight. If the pan remains on level then the third coin is lighter

Similarly if you have 3 2 coins you can fig out which one is of lesser weight in 2 steps Step 1: Divide into 3 groups of coins each containing 3 i.e G1- 3 , G2- 3 , G3-3 . Put G1 and G2 on pans and with the argument described earlier you can find the lighter group of coins After first step you will know which group of 3 has the lighter coin. Step 2: Do as stated earlier to find out which coin is lighter.

So with 3 steps you can find out among 3 3 coins â€“ which is the faulty one.

## Coins Problem

Thanks for the concept as well as your time. Simple and good method to solve the problem.

## No problem, my

No problem, my pleasure. Please come with more such problems we will love to solve.