the minimum number of wieghtings needed is just 5.as shown below

(1) 80->30-30

(2) 15-15

(3) 7-7

(4) 3-3

(5) 1-1

SAMEERA

Jun 13th, 2006

4 WEIGHS

Nitin

Jun 29th, 2006

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

rajitha

Aug 22nd, 2006

hi,

can anyone explain in detail.

thanks,

Rajitha.

gauravkanwar

Sep 12th, 2006

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

chaitanya bijur

Jun 23rd, 2007

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

caa0601

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

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

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

Ik

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 :)

Tu Nguyen

May 28th, 2017

8

Anjali

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

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

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