Different weights but least weighing!

We have 32 balls that look identical but each of them has a different weight. We have a balance in which we can weigh one ball against another ball. You have to find out the minimum number of weighing required to find the second heaviest ball in the lot.

Can you guys please help?

#Logic

Easy Math Editor

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Comments

You will need 31 weighing to determine the heaviest ball:

First step you pair every two balls and weighing them to each other, they take 16 weighing,

Second step you pair every two balls that heavier in first step, they take 8 weighing

Until the fifth step, you weight the two remain ball to determine the heaviest. Let's call this ball A.

Now to find the second heaviest: Let's call the ball you weight with ball A in each step A1,A2,A3,A4,A5. Easy to figure out that the second heaviest ball is in those 5 balls. You will need 4 weighing more to see which one (in the five) is the heaviest.

Total it will take 35 weighing.

Tran Hieu - 5 years ago

CouʟԀ somє oňє posţ ţһє soʟuţıoň

Rushabh Zambad - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$