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Out of 9 equal weights, one is faulty (lighter). Find the least number of weighing required to identify a faulty weight using a two pan balance.

I came across this similar type of question posted here long back and I used the relation it mentioned :-

"In my module they have given some sort shortcut to evaluate the result which is for $n$ bags: In first case if $3^a \lt n \lt 3^{a+1}$, then the minimum number of weighing required is $a+1$. In the second case, we need $2^a \lt n \lt 2^{a+1}$ which gives the minimum of $a+1$ weighing."

when I put 9 in the equation $3^a \lt 9 \lt 3^{a+1}$ and solve it, I get a<2 and a>1 which does not give a reasonable answer because it should be an integer answer right? Where I am going wrong?

Ganit
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1 Answers1

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The answer is obviously two weighings for nine weights:

  • weigh three against three, revealing which batch of three has the light weight
  • from that batch weigh one against one, revealing which one is the light weight

The expression should have been that you need at least $a$ weighings when $3^{a-1} \lt n \le 3^a$, i.e. $a=\lceil \log_3(n) \rceil$

It gets a little more complicated when you do not know whether the faulty weight is lighter or heavier than the others

Henry
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