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How-many-artificially-made-weights-are-required-minimally-to-weigh-2-4-6-80-kg-and-what-should-be-they-




Question Number 183617 by Rasheed.Sindhi last updated on 27/Dec/22
How many artificially made weights are   required (minimally) to weigh   2,4,6,...,80 kg   and what should be they?
$$\mathcal{H}{ow}\:{many}\:{artificially}\:{made}\:{weights}\:{are}\: \\ $$$$\boldsymbol{{required}}\:\left({minimally}\right)\:{to}\:{weigh} \\ $$$$\:\mathrm{2},\mathrm{4},\mathrm{6},…,\mathrm{80}\:{kg}\: \\ $$$${and}\:{what}\:{should}\:{be}\:{they}? \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/22
Commented by mr W last updated on 27/Dec/22
let′s say 1 Tom=2 kg, then you want  to weigh 1, 2, 3, ..., 27 Toms.  As you can see from Q183583,  with 4 weights of 1, 3, 9, 27 Toms, you  can weigh a weight upto 40 Toms.  that means you need at least 4 weights   (2, 6, 18, 54 kg) to weigh 2, 4, 6, ..., 54 kg.
$${let}'{s}\:{say}\:\mathrm{1}\:{Tom}=\mathrm{2}\:{kg},\:{then}\:{you}\:{want} \\ $$$${to}\:{weigh}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…,\:\mathrm{27}\:{Toms}. \\ $$$${As}\:{you}\:{can}\:{see}\:{from}\:{Q}\mathrm{183583}, \\ $$$${with}\:\mathrm{4}\:{weights}\:{of}\:\mathrm{1},\:\mathrm{3},\:\mathrm{9},\:\mathrm{27}\:{Toms},\:{you} \\ $$$${can}\:{weigh}\:{a}\:{weight}\:{upto}\:\mathrm{40}\:{Toms}. \\ $$$${that}\:{means}\:{you}\:{need}\:{at}\:{least}\:\mathrm{4}\:{weights}\: \\ $$$$\left(\mathrm{2},\:\mathrm{6},\:\mathrm{18},\:\mathrm{54}\:{kg}\right)\:{to}\:{weigh}\:\mathrm{2},\:\mathrm{4},\:\mathrm{6},\:…,\:\mathrm{54}\:{kg}. \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/22
ThanX sir! your answer is very right  but my question is not very right. I  am going to edit it.
$$\mathbb{T}\boldsymbol{\mathrm{han}}\mathbb{X}\:\boldsymbol{\mathrm{sir}}!\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{very}\:\mathrm{right} \\ $$$$\mathrm{but}\:\mathrm{my}\:\mathrm{question}\:\mathrm{is}\:\mathrm{not}\:\mathrm{very}\:\mathrm{right}.\:\mathrm{I} \\ $$$$\mathrm{am}\:\mathrm{going}\:\mathrm{to}\:\mathrm{edit}\:\mathrm{it}. \\ $$

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