You
have
8
Sholay coins
which
are
all
the
same
weight,
except
for
one
which
is
slightly
heavier
than
the
others
(you
don’t
know
which
coin
is
heavier).
You
also
have
an
old‐style
balance,
which
allows
you
to
weigh
two
piles
of
coins
to
see
which
one
is
heavier
(or
if
they
are
of
equal
weight).
What
is
the
fewest
number
of
weighings
that
you
can
make
which
will
tell
you
which
coin
is
the
heavier
one?
3rd time , we’ll get the heavier coin . is it rgt ???
Good attempt Navaneetha! Please optimize the solution.
7
One possible solution. Please optimize the solution.
Twice. Put 3/3 each coin on each balance.
Condition-1: if the coins placed on the weighing pan are equal, then the heavier coin lies in the remaining two, weight the two, you’ll get the answer.
Condition-2: If one side of the pan containing 3 coins is slightly heavier, then remove the 3 coins and put 2 on the pans. If both show equal balance, the remaining coin is the heavier one, or the one heavier will show deflection during weighing.
You hit the bull eye! A neat and clean solution!
thanks a lot.!! I am looking forward, for the upcoming questions!
f the coins placed on the weighing pan are equal, then the heavier coin lies in the remaining two, weight the two, you’ll get the answer.
”it is similar 2 the 8 marble puzzle”
Good One!
in 2 steps
as explained in the 2nd comment