. A person has nine coins, all looking alike but one of them slightly lighter than the others. Using a balance without weights, can he discover the lighter coin in just two weighings? If the number of coins is 12, how many minimum weighings will be required to find the lighter coin?
Answers
Given : A person has nine coins, all looking alike but one of them slightly lighter than the others. Using a balance without weights,
To find : can he discover the lighter coin in just two weighings
Solution:
YES , he can
Makes 3 heap of 3 coins each
A , B & C
Weigh
A & B with each other
if A is lighter then lighter coin is in A
if B is lighter then lighter coin is in B
if A = B then lighter coin is in C
After on weigh we know lighter coin is in which heap
After that
Take 3 coins of that lighter heap
X , Y & Z
Weigh
X & Y with each other
if X is lighter then lighter coin is X
if Y is lighter then lighter coin is Y
if X = Y then lighter coin is Z
Hence we got lighter coin in 2 Weight
if there are tweleve coins then
We can make 4 heaps of 3 coins
and after two weigh we will get surely lighter heap
and total 3 weigh for lighter coin
or
3 heap of 4 coins each
and one weight for lighter heap
& maximum two weigh for lighter coin
Hence we need 3 Weighings for 12 coins
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