A trader wants measure (all integral weights) from 1 to 121 kg using a common balance where weights can be kept in both the pans. What is the minimum number of weights needed?
Explanation:
We know any number can be written as sum or different of powers of 3.
Let the minimum weights required by n.
∴ 30 + 31 + 32 + … + 3(n-1) ≥ 121.
⇒ (3n – 1)/2 ≥ 121
⇒ (3n – 1) ≥ 242
⇒ 3n ≥ 243
∴ least value of n = 5.
Hence, 5.
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