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

This question is related to TCS Interview

Showing Answers 1 - 17 of 17 Answers

prateek nayak

  • Feb 26th, 2006

minimum- 5 weighingmaximum-6

  Was this answer useful?  Yes

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

  Was this answer useful?  Yes


  • Jun 13th, 2006


  Was this answer useful?  Yes


  • 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.



  Was this answer useful?  Yes


  • Sep 12th, 2006

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

  Was this answer useful?  Yes

chaitanya bijur

  • Jun 23rd, 2007

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

  Was this answer useful?  Yes


  • 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

  Was this answer useful?  Yes


  • 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

  Was this answer useful?  Yes


  • Jul 20th, 2015

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

  Was this answer useful?  Yes


  • 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 :)

  Was this answer useful?  Yes

Tu Nguyen

  • May 28th, 2017


  Was this answer useful?  Yes


  • 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

  Was this answer useful?  Yes

Give your answer:

If you think the above answer is not correct, Please select a reason and add your answer below.


Related Answered Questions


Related Open Questions