
One of nine identical-looking gold coins in a safe was swapped for a fake. The fake is slightly lighter than a real coin. There is no other difference in weight.
All you have is a balance scale that compares two pans. You cannot read any weight; you only learn one of three results: "left is heavier, right is heavier, or equal."
What is the fewest weighings that is always enough to identify the fake for certain?
If you get stuck, open the hints one at a time, in order.
Try splitting the nine coins into three groups of three.
A weighing has three outcomes. How far can the first weighing reduce the candidates?
How many more weighings does it take to find one coin among three?
What is the fewest weighings that is always enough to identify the fake for certain?
Choose the answer closest to the truth.
2 times
The answer is 2 times.
Weighing 1: split the nine coins into three groups of three (A, B, C) and weigh A against B. If one side is lighter, the fake is in it; if they balance, the fake is in C. Either way, the candidates are narrowed to three coins.
Weighing 2: put one of the three candidates on each pan. The lighter pan holds the fake. If they balance, the coin left off is the fake.
Since there are three possible results, one weighing cuts the candidates to a third. Doing that twice always narrows 9 to 3 to 1.
You tend to compare two coins at a time or test them one by one. Unless you think of using all three outcomes, you overestimate the number of weighings.
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