Apologies for not being able to send the answer last weekend.

I got many correct answers for this one, and people sent different correct answers. Ones who sent correct answers include Pratik Poddar, Akash Tayal, Sandeep Singhal, Parag Dhanuka, Prakhar Prakash, Atul Tewari, Divya Mittal, Suman Saraf and Rahul Rane.

I am reproducing the answer from Sandeep, which I thought was the simplest (also sent by many others):

Weigh 50 coins against 50
If equal left out coin is odd and can be weighed against any coin.
If unequal, take any heap and 25 against 25.
If equal the other heap has the odd coin and if the other heap was heavier in the fist weighing, odd one is heavier and vice versa.
If unequal, odd one is in this heap and the odd one is heavier if this heap was heavier on the first weighing and vice versa…

Many people sent variations of this method, e.g. here is the answer from Pratik Poddar:

Divide the set of 101 coins in 3 sets:
Set A of 33 coins, Set B of 33 coins and Set C of the remaining 35 coins

Weight Set A and Set B. Without loss of generality, two cases are possible:
a) Set A == Set B
b) Set A > Set B

In the first case, we know that the counterfeit coin is in Set C, and Set A and B are all genuine coins. Using the coins of A and B, we create a 35 coin set which we know is genuine. Just weighing this against C tells us whether the counterfeit coin is heavier or lighter.

In the second case, we know that Set C is all genuine coins. Remove 2 coins from Set C. And weigh Set C against Set A, if Set A == Set C, Set B has counterfeit coin and its lighter. If Set A > Set C, Set A has counterfeit coin and is heavier.

In general, as long as you weight 26 or more coins on one side in the first weighing or (ceiling of x/2 for x coins), one can get to the answer.

Hope you enjoyed the puzzle!

Wonderful puzzle and I got many correct answers. Ones to send the answers were Pratik Poddar, Prakhar Prakash, Suman Saraf, Karan Sharma and Girish Tutakne.

The answer is that it can’t be done for 6 but can be done for 7 trees. Generally, can’t be done for even number of trees but can be done for odd number of trees. I am copying Girish’s answer verbatim (thanks Girish).
Explanation: Number trees 1 to 6 (or 7 or n). Each sparrow takes the tree number of the tree it is on. At the start, the total number achieved by adding the “sparrow-numbers” is 21 (or 28 or n.(n+1)/2).

With every move, the total sparrow number stays the same because for every sparrow increasing its sparrow number by +1 by moving to the right, a sparrow reduces it’s sparrow number by -1 by moving to the left.

If all sparrows need to be on the same tree, they need to be on the tree with number equal to the average of the 6 (or 7 or n) sparrows. This is only possible for n being odd because the average will be (n+1)/2, which is a natural number only for n being odd. And as it the central tree for all odd n, every odd n has a solution.

Hope you all enjoyed the puzzle!

Apologies for a long absence, was traveling abroad on the last two weekends and hence could not post.

This was a difficult puzzle, and I received answers from only 3 people – Pratik Poddar, Rahul Rane and Suman Saraf. Well done all!

I am replicating the original solution in Martin Gardner’s book. The answer is two trips, i.e. the electrician will need to go down once and the come back up, and with these trips he can figure out all the wires. Let’s see how!

Puzzle 120 answer graphic

On the top floor, the electrician shorted 5 pairs of wires (as shown in the figure above), leaving one free wire. Then he walked to the basement and identified the lower ends of the shorted pairs by means of his “continuity tester”. He labeled the ends as shown in the figure above, then shorted them in the manner indicated.

Back on the top floor, he removed all the shorts but left the wires twisted at insulated portions so that the pairs were still identifiable. He then checked for continuity between the free wire (F) and some other wire. When he found the other wire, he was able to label it as E2 and identify its original pair as E1. He then used the continuity tester between E1 and another wire, which can then be marked as D2 and also mark D1. Continuing in this fashion all ends can be identified.

As one will notice, this procedure will work for only odd number of wires. There are other more generic solutions, also captured within Martin Garder’s book and also contributed by Pratik Poddar. Interested reader may read on this link http://www.cseblog.com/2009/10/this-puzzle-was-asked-to-me-last-year.html

Hope you all enjoyed the puzzle.