maths

We have 12 Coins.

One of them (we dont know which one?) has different weight from the others.

With three time scaling (the scale has two pans) find the differrent coin.

source:mathlinks

16 Answers

11
virang1 Jhaveri ·

IS the odd coin more in weight?

1
only_chem ·

Does 3 time scaling mean that we can use the scale only 3 times???

1
Arshad ~Died~ ·

man this question is for the lucky person

supposing i am lucky

just divide the coins into 3 groups of 4 each
weigh the first two groups.....and u find them equal

my luck comes into play here

then u know that the faulty coin is in the third group
now weigh two-two coins from that group together and from that we can find out the faulty coin.........
but its not zaruri that i get the wrong coin in three steps.......

11
Gone.. ·

is this question from the Russian book they give on qualifying in RMO ??

correct solution is i think what arshad has said ..

1
Arshad ~Died~ ·

but my solution depends on luck and since when has mathematics started depending on luck???

23
qwerty ·

@ arshad elaborate the last part ........
if we weigh 2 -2 coins and suppose LHS wala 2 coins are heavier .......so wat would u say wich set has the abnormal coin??here it isnt given that the abnormal coin is heavy or light

23
qwerty ·

@ injun i think u didnt get wat i m trying to say ....
see suppose the 2 grps u take first are unequal in weight ...so wat will u conclude .... faulty coin is in the heavier group or lighter group??? faulty coin is heavy or light ye diya nahi hai ...

btw i m getting it in 4 steps [102]

1357
Manish Shankar ·

Actually here it is not mentioned that the faulty coin is heavier or lighter.

So its difficult in 3 scalings

Ok let's find out how many minimum scaling required for finding the faulty one

1
injun joe ·

Err..I took it to be heavier. My mistake.

1357
Manish Shankar ·

I m getting minimum 4 scalings to find the faulty

39
Dr.House ·

no manish, 3 are enough..

this was one from a old forum which i particularly found interesting while i was going through the old posts,, so gave it here.

here`s the link:

http://www.mathlinks.ro/viewtopic.php?search_id=717924636&t=49152

1
bakshisubhomoy ·

supposing the odd one to be heavier

frst divide into two groups.......6 each..... nw take the heavier grp and divide into 2 grp of 3 and weigh.....so we hav already used the scale twice..... nw ttake the heavier grp and divide it into 3 grps with one ball each.....take any two balls and weigh them thus using the scale for the 3rd time...... if either of the ball is heavy then, then we have found out the faulty ball......if both are of same wt then the 3rd ball is the faulty one......

so where m i wrong?????????????

1
Arshad ~Died~ ·

u r wrong in assuming any coin to be heavier

23
qwerty ·

i m getting in 4 steps

23
qwerty ·

1
Arshad ~Died~ ·

nice work dude.....quite systematical

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