Amazing GRE math problem:101 gold coins puzzle - Videos

An evil warden holds you prisoner, but offers you a chance to earn your freedom.
You are given 101 coins, of which 51 are impure and 50 are pure. Each impure coin is identical. And each pure coin is identical to a impure coin, except that it differs in weight by exactly 1 gram (all are lighter or all are heavier, only the warden knows).
The warden gives you a randomly selected coin from the 101 coins. You have to guess whether it is impure or pure. If you are incorrect, you are imprisoned forever. If you are correct, you are set free.
You are given a weighing balance that displays the difference in weight between its left and right pans (for example, if the left pan has 8.3 grams and right pan has 10.3 grams, the display would show -2 grams). You can weigh any of the 101 coins, but you only get to use the weighing balance once before you have to guess.
What is your best strategy to identify your coin and escape the prison?
Watch the video for a solution.
An evil warden holds you prisoner, but offers you a chance to earn your freedom.
You are given 101 coins, of which 51 are impure and 50 are pure. Each impure coin is identical. And each pure coin is identical to a impure coin, except that it differs in weight by exactly 1 gram (all are lighter or all are heavier, only the warden knows).
The warden gives you a randomly selected coin from the 101 coins. You have to guess whether it is impure or pure. If you are incorrect, you are imprisoned forever. If you are correct, you are set free.
You are given a weighing balance that displays the difference in weight between its left and right pans (for example, if the left pan has 8.3 grams and right pan has 10.3 grams, the display would show -2 grams). You can weigh any of the 101 coins, but you only get to use the weighing balance once before you have to guess.
What is your best strategy to identify your coin and escape the prison?
Watch the video for a solution.
Answer To Identify If The Coin Is Pure In One Weighing
The solution is remarkable for a number of reasons. First, there is a strategy that is guaranteed to work. Second, the strategy works without knowing the weight of the impure coin or whether the pure coin is lighter or heavier. And third, you never actually weigh the coin you need to identify–you figure it out by weighing all of the other coins!
Here is the procedure. Place your coin aside. Split the remaining 100 coins into left and right pans with 50 coins in each. Now the balance will always display an integer (explained below). If the integer is even, then you initially received a impure coin. If the integer is odd, then you initially received a pure coin.
By a clever mathematical scheme, you have outsmarted the evil warden and guaranteed your freedom!
But why does this work? Let’s investigate the reason.
Proof
Number the coins in the left pan from 1 to 50, and number the coins in the right pan from 51 to 100. Write ci to denote the weight of coin i. The weighing balance displays the difference in the weight between the coins on the left pan and the right pan, which is the following expression:
(c1 + c2 + … c50) – (c51 + c52 + … c100)
Let w be the weight of a impure coin. Since w – w = 0, we can add and subtract w in pairs without changing the result of the above expression. We will subtract w to each term on the left pan, and we will add w to each term on the right pan. Notice that adding w is the same thing as subtracting –w. So the weighing scale display is equivalent to the following expression:
(c1 – w) + (c2 – w) + … (c50 – w) – (c51 – w) – (c52 – w) – … (c100 – w)
The above expression sums 50 terms of (cj – w) and then subtracts 50 terms of (ck – w).
If a coin is impure, its weight is exactly w, so (cimpure – w) = 0 and those terms vanish.
If a coin is pure, its weight differs by 1 gram from w, so (cpure – w) = 1 if the coin is heavier and -1 if the coin is lighter.
The weighing scale display is equal to an expression with 100 terms, each of which is 0, or +1 (or -1). As each term is an integer, the ultimate result will always be an integer.
Furthermore, the display’s parity (oddness or evenness) depends on the number of pure coins in those 100 coins.
Recall the expression for the value on the weighing scale balance:
(c1 – w) + (c2 – w) + … (c50 – w) – (c51 – w) – (c52 – w) – … (c100 – w)
The oddness or evenness of the result depends only on the number of pure coins on the two pans, which can either be 50 or 49.
As proven below, the result is even for 50 and odd for 49. Thus, if the result is even, you must have weighed all 50 pure coins, so you initially received a impure coin. And if the result is odd, then you must have weighed 49 of the pure coins, meaning you received the last pure coin.

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