**Question:** You are given 8 identical looking balls. One of them is heavier than the rest of the 7 (all the others weigh exactly the same). You a provided with a simple mechanical balance and you are restricted to only 2 uses. Find the heavier ball.

**Answer:** For convenience sake, let’s name the balls 1-8. First we weigh {1,2,3} on the left and {4,5,6} on the right. There are three scenarios which can arise from this.

If the left side is heavier, then we know that one of 1, 2 or 3 is the heavier ball. Weigh {1} on the left and {2} on the right. By doing this, we will know if 1 or 2 is heavier. If they balance out, then 3 is the heavier one.

If the right side is heavier, then we know that either 4, 5 or 6 is the heavier ball. Weigh {4} on the left and {5} on the right. By doing this we will know if 4 or 5 is heavier. If they balance out, then 6 is the heavier one.

If {1,2,3} and {4,5,6} balance out, then we know either 7 or 8 is the heavier one. Weigh both of them to find out which one is heavier.

Confused yet? or was it too easy? This is one of the basic identical ball problems. Here is a more complex problem involving 12 balls with one fake.

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Please help me. I have a cue epic and line is very small. Palade help me

Thanks for the good work…It helps.

This solutions works for 9 balls as well…

Good question for efficiency measurement

The first weighing is 3 vs 3 but then we go on to use the other 2 balls in the second weighing. That makes it 8 in total.

The question should be reframed to indicate 6 balls instead of 8 as you are only weighing 3 and 3….