A trader wants to measure all weights (integral) from 1 to 220 kg. using a simple balance in which weights are placed in one of the pans. What is the minimum number of weights needed?
Explanation:
We know any number can be written as sum of powers of 2.
Let the minimum weights required by n.
∴ 20 + 21 + 22 + … + 2(n-1) ≥ 220.
⇒ 2n – 1 ≥ 220
⇒ 2n ≥ 221
∴ least value of n = 8.
Hence, 8.
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