Third time successful

Logic Level 2

I have $n$ coins one of which is lighter than all the other $(n-1)$ coins. The other $(n-1)$ coins are all of exactly the same mass.

I have to use a weighing balance to find out the light coin.

I am sure that I will only need at most 3 weighings to determine the lighter coin. What is the maximum possible value of $n$ ?

The answer is 27.

3 solutions

Abhay Tiwari
Jul 20, 2015

In this question the n is to be divided into groups of three so that two of the groups can be balanced, after which two cases arrive.

1) The two groups are equal, which means the coin is in the third group.

2) One of them is lighter. That means the coin is in the lighter group.

Now let's start from the last. At last I will have three coins remaining at last then two of them can be balanced, and from there we could identify which one is lighter.

Now, since at last weighing had three coins, then second last weighing would have 3×3=9 coins.

And the third weighing will have 9×3=27 coins, which is the answer.

The question stumped me because I thought three weighings meant that the scale could only be used three times!

Tan Ho - 5 years, 10 months ago

Thats what I was thinking too.

Ryan Berg - 5 years, 6 months ago

That's exactly what it means .. if you have 27 coins you can find the lighter coin by using the scale 3 times only

1-Group them in 9's .. weigh two gruops together if they balance take the third if not choose the lighter

2-Then divide the 9 into 3 groups of 3 .. repeat as 1

3-Now weigh 2 of the lighter group against each other .. if balanced the third is your coin if not the lighter is the one

1.. 2 .. 3 .. weighings only

Ahmed Obaiedallah - 5 years, 8 months ago

Well then isn't 8 an acceptable answer aswell?

Split 8 into two 4s, weigh both,choose the lighter pile.

Split the 4 into two 2s, weigh, choose lighter pile again.

Weigh both, lighter coin is lighter.

Ian Ashcroft - 5 years, 5 months ago

@Ian Ashcroft – Well in fact 8 coins can be solved with using scale 2 times only as minimum number, by making them into 2 piles of 3 coins and a pile of 2 coins

And to answer your question .. Actually no it's not an acceptable answer as the problem needs "The maximum possible value for $\space\textbf{n}\space$ when you are limited only by 3 weighing" which is 27

To solve this kind of problem you need to know the maximum number of coins that you can differentiate between them by using scale "1" time only and that number is 3, as you always can -(with a pile of 3)- weigh 2 together and face 3 possibilities

$\textbf{1-The right hand of the scale}\space\space\space\space$ $\textbf{2-The left hand of the scale}$ $\textbf{3-They balance .. which will be the 3rd coin}$

Meaning those -3 coins- are the "cell/1st brick/base" -of such problems- that you'll use to build with and upon

It's a base 3 problem and mathematically represented as $\Large n=3^x$ where

x=1,2,3,4 is the number of weighing

n: is the maximum number of coins at that "x"

creating an intervals of $\space\LARGE[\large(n_{._{(x-1)}}+1),n_x\LARGE]$

Ahmed Obaiedallah - 5 years, 5 months ago

The semantics of this question is wrong. I also understood that the scale had to be used three times at most.

Bernardo Tavora - 5 years, 2 months ago

Yeah ,the semantics are wrong

Deep Vora - 1 year, 8 months ago

The question does mean you can only use the scale 3 times.

Gabriel Kropf - 4 years, 11 months ago

So simply put it, if you need 4 weighing, its 3×3×3×3?

Davking Davking - 4 years, 9 months ago

Noel Lo
Jul 19, 2015

When we have $n$ coins, we can split them into 3 equal groups of $\frac{n}{3}$ coins each. Let's call the groups A, B and C. We place group A on one side of the balance, group B on the other side and group C off the balance.

There are 3 possible outcomes:

the pan carrying group A goes up - which means the lighter coin is in group A.
the pan carrying group B goes up - the lighter coin is in group B.
the two pans are level - the lighter coin is neither in group A nor B otherwise there will be a tilt, so the lighter coin must be in group C.

In each outcome, we can pinpoint which group of $\frac{n}{3}$ coins contains the lighter coin and proceed to divide that group into 3 subgroups and place these 3 subgroups as we have done for the first weighing. This will be for the second weighing.

Repeat this process until we have 3 coins for the last weighing. Then we place one coin on each side of the balance and the third coin off the balance. Whatever the outcome, we can pinpoint the lighter coin.

If this entire process takes us 3 weighings, it means that we have to divide $n$ coins into 3 equal groups 3 times - that is dividing $n$ coins into $3^3=27$ subgroups. Since each subgroup ultimately has one coin, the largest possible value of $n$ is $\boxed{27}$ .

Niranjan Khanderia
Jul 21, 2015

With n\3 out of balance and other two in balance, we can identify the group n\3 that has the light coin. Similarly in second weighting we can identify (n/3)/3=n/9 group that has light coin. on THIRD weighing we can identify (n/9)/3=n/27 group that has the light coin. But this must be the light coin itself to be 100% sure. So n/27=1 and n=27.

Third time successful

The answer is 27.

3 solutions

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