In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin

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prateek nayak

  • Feb 26th, 2006

minimum- 5 weighingmaximum-6

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u.kalyan chakravarthy

  • May 16th, 2006

the minimum number of wieghtings needed is just shown below

            (1)    80->30-30

            (2)      15-15

             (3)       7-7  

             (4)      3-3               

              (5)      1-1

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  • Jun 13th, 2006


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  • Jun 29th, 2006

can u b a little more specific how u got the answer..cause its hard to understand in such less words....


  • Aug 22nd, 2006


   can anyone explain in detail.



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  • Sep 12th, 2006

yes it is 5 (min) and max (6)40 -4020-2010 105 -52 -2 1

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chaitanya bijur

  • Jun 23rd, 2007

Since there are two possibilities 1. counterfeited and 2. uncounterfeited and 2^6<80<2^7

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  • Jul 30th, 2007

first we divide the 80 coins into two halves -----> 40+40  (step 1)

                  so we will be weighing its weight so one set will have less weight then
 again same ----------------> 20+20   again         (step 2)
                    ----------------> 10+10 again           (step 3)
                    ----------------> 5+5 again                (step 4)
                      ----------------> 2+2+1                   (step 5) 

weighing the two sets of two coins if they found equal  the remained one is the required. else again------------------------->1+1      (step 6)
then we will be getting.       so minimum 5, max 6


  • Sep 4th, 2007

How can you find the one which is counterfiet., coz, they didnt tell that is lighter than original or heavier., then how can you find the difference,

ex suppose you take lighter one., but your CF coin is heavier than original coin? means what will you do?
Reply me

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  • May 27th, 2008

make group of 30 30 and 20 coins
in first weight 30 30
if they are same then coin is in group of 20 if not then it is in lighter one
after one weight u got 30 coins in which lighter is present

in second make three group of 10 10 10 and weight any two group
if both are same then coin is in third one else in a lighter one

after two weight u find 10 coins in which lighter is present

in third
make group of 3 3 4
and weight  group of 3 3
if they are differ then u get the three else u get the group of 4

in fourth
if u get three weight any two and find the lighter one
but if u get four there is a need for fifth weight

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  • Jul 20th, 2015

Why not 3?
80---35-35 10
10---3-3 4
3---1 1 1
Hence 3 steps will do.

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  • Nov 14th, 2016

We can say the minimum number is 2.
80 >> 39:39:2
2 >> 1:1 since it is the minimum number of weightings :)

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Tu Nguyen

  • May 28th, 2017


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  • Aug 18th, 2017

1. Assume each coin is with 1 gram weight except that counterfeit coin
step-1: 80 in first step divide into 2 halves i.e 40+40
step-2: take the 40 which is not getting exact 40 weight and then divide into 2 half 40 --> 20+20
step-3: 20 = 10+10
step-4: 10=5+5
step-5: 2+2+1 , in this step we got 2 sets are with same weight, coin which is left is counterfeit
we got 2 different weights in the above step go to divide the set as
2=1+1 ==>
so, max steps = 6
min steps = 5

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  • Jan 16th, 2021

What if the coin is in among 35 coins?

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