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Suggest-minimum-number-of-weights-two-peices-of-each-to-weigh-upto-at-least-60-kg-in-whole-kg-s-in-a-common-balance-

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Question Number 3205 by Rasheed Soomro last updated on 07/Dec/15
Suggest minimum number of weights ,two peices of each, to weigh upto at least 60 kg(in whole kg′s) in a common balance.
$$\mathcal{S}{uggest}\:{minimum}\:{number}\:{of}\:\:{weights}\:,{two}\:{peices}\:{of}\:{each},\: \\ $$$${to}\:{weigh}\:{upto}\:{at}\:{least}\:\mathrm{60}\:{kg}\left({in}\:{whole}\:{kg}'{s}\right)\:{in}\:{a}\:{common} \\ $$$${balance}. \\ $$
Commented by prakash jain last updated on 07/Dec/15
You mean you have two weight of the same measure?
$$\mathrm{You}\:\mathrm{mean}\:\mathrm{you}\:\mathrm{have}\:\mathrm{two}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{measure}? \\ $$
Commented by Rasheed Soomro last updated on 07/Dec/15
Exactly Sir.
$$\mathcal{E}{xactly}\:\mathcal{S}{ir}. \\ $$
Commented by 123456 last updated on 07/Dec/15
1:1 2:2,3 4:4,5,6,7 8:8,9,10,11,12,13,14,15 16:16,...,31 32:32,...,63
$$\mathrm{1}:\mathrm{1} \\ $$$$\mathrm{2}:\mathrm{2},\mathrm{3} \\ $$$$\mathrm{4}:\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7} \\ $$$$\mathrm{8}:\mathrm{8},\mathrm{9},\mathrm{10},\mathrm{11},\mathrm{12},\mathrm{13},\mathrm{14},\mathrm{15} \\ $$$$\mathrm{16}:\mathrm{16},…,\mathrm{31} \\ $$$$\mathrm{32}:\mathrm{32},…,\mathrm{63} \\ $$
Commented by prakash jain last updated on 07/Dec/15
I think 1 2 4 8 16 32 is solution which means that we use only one side of the balance to put weight.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{1}\:\mathrm{2}\:\mathrm{4}\:\mathrm{8}\:\mathrm{16}\:\mathrm{32}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{which}\:\mathrm{means} \\ $$$$\mathrm{that}\:\mathrm{we}\:\mathrm{use}\:\mathrm{only}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{balance}\:\mathrm{to} \\ $$$$\mathrm{put}\:\mathrm{weight}. \\ $$
Commented by prakash jain last updated on 08/Dec/15
Let us say you want to find minimum number of weights for 2 each, then as given in answer you will have the following Π_(i=a,b,c) ((1/x^(2i) )+(1/x^i )+1+x^i +x^(2i) )=Σ_(j=−62) ^(62) x^j factoring RHS taking: (1/x^(60) )((1/x^2 )+(1/x)+1+x+x^2 )+(1/x^(55) )((1/x^2 )+(1/x)+1+x+x^2 )+ .. +x^(55) ((1/x^2 )+(1/x)+1+x+x^2 )+x^(60) ((1/x^2 )+(1/x)+1+x+x^2 ) =((1/x^2 )+(1/x)+1+x+x^2 )((1/x^(60) )+(1/x^(55) )+..+x^(55) +x^(60) ) =((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) )((1/x^2 )+(1/x)+1+x+x^2 )[(1/x^(50) )((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) )+ +(1/x^(25) )((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) )+1((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) ) +x^(25) ((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) )+x^(50) ((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) ) =((1/x^2 )+(1/x)+1+x+x^2 )((1/x^(10) )+(1/x^5 )+1+x^5 +x^(10) )((1/x^(50) )+(1/x^(25) )+1+x^(25) +x^(50) ) so a=1, b=5, c=25
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{find}\:\mathrm{minimum} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{weights}\:\mathrm{for}\:\mathrm{2}\:\mathrm{each},\:\mathrm{then}\:\mathrm{as}\:\mathrm{given} \\ $$$$\mathrm{in}\:\mathrm{answer}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{the}\:\mathrm{following} \\ $$$$\underset{{i}={a},{b},{c}} {\prod}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}{i}} }+\frac{\mathrm{1}}{{x}^{{i}} }+\mathrm{1}+{x}^{{i}} +{x}^{\mathrm{2}{i}} \right)=\underset{{j}=−\mathrm{62}} {\overset{\mathrm{62}} {\sum}}{x}^{{j}} \\ $$$$\mathrm{factoring}\:\mathrm{RHS}\:\mathrm{taking}: \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{60}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{{x}^{\mathrm{55}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)+ \\ $$$$\:\:\:\:..\:+{x}^{\mathrm{55}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)+{x}^{\mathrm{60}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right) \\ $$$$=\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{60}} }+\frac{\mathrm{1}}{{x}^{\mathrm{55}} }+..+{x}^{\mathrm{55}} +{x}^{\mathrm{60}} \right) \\ $$$$=\left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)\left[\frac{\mathrm{1}}{{x}^{\mathrm{50}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right)+\right. \\ $$$$\:\:\:\:\:\:\:+\frac{\mathrm{1}}{{x}^{\mathrm{25}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right)+\mathrm{1}\left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right) \\ $$$$\:\:\:\:\:\:\:+{x}^{\mathrm{25}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right)+{x}^{\mathrm{50}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right) \\ $$$$=\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{50}} }+\frac{\mathrm{1}}{{x}^{\mathrm{25}} }+\mathrm{1}+{x}^{\mathrm{25}} +{x}^{\mathrm{50}} \right) \\ $$$$\mathrm{so}\:{a}=\mathrm{1},\:{b}=\mathrm{5},\:{c}=\mathrm{25} \\ $$
Commented by Rasheed Soomro last updated on 08/Dec/15
Q3143 and this question may seem to be closely related with number base system.Former is related base3 system and the latter is related base5 system. The difference between number base system and these Questions is as under. Base3 system digits are 0,1,2 Wheras in Q3143 −1,0,1 are used instead. −1 is because of subtraction possibility. Similarly base5 system digits are 0,1,2,3,4 but this question uses −2,−1,0,1,2 instead We can see that if there be three peices of every weight the ′ digits ′ will be −3,−2,−1,0,1,2,3 and the related number system will be base7 system. Different weights will be 7^0 ,7^1 ,7^2 ,... If we use only one side of balance to place weights these type of questions will clmpletely be examples of base system numbers.
$$\mathcal{Q}\mathrm{3143}\:{and}\:\:{this}\:\:{question}\:{may}\:{seem}\:{to}\:{be}\:{closely} \\ $$$${related}\:{with}\:{number}\:{base}\:{system}.{Former}\:{is}\:{related} \\ $$$${base}\mathrm{3}\:\:{system}\:{and}\:\:{the}\:{latter}\:{is}\:{related}\:{base}\mathrm{5} \\ $$$${system}. \\ $$$$\:\:\:\:\:\mathcal{T}{he}\:{difference}\:\:{between}\:{number}\:{base}\:{system} \\ $$$${and}\:\:{these}\:\:{Questions}\:{is}\:{as}\:{under}. \\ $$$$\:\:\:\:\:\mathcal{B}{ase}\mathrm{3}\:\:\:{system}\:\:\:{digits}\:{are}\:\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\mathcal{W}{heras}\:{in}\:{Q}\mathrm{3143}\:\:−\mathrm{1},\mathrm{0},\mathrm{1}\:{are}\:{used}\:{instead}. \\ $$$$−\mathrm{1}\:\:{is}\:{because}\:{of}\:{subtraction}\:{possibility}. \\ $$$$\mathcal{S}{imilarly}\:{base}\mathrm{5}\:\:{system}\:{digits}\:{are}\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$$\boldsymbol{{but}}\:{this}\:{question}\:{uses}\:−\mathrm{2},−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2}\:{instead} \\ $$$$\mathcal{W}{e}\:{can}\:{see}\:{that}\:{if}\:{there}\:{be}\:{three}\:{peices}\:{of}\:{every}\:{weight} \\ $$$${the}\:'\:\boldsymbol{{digits}}\:'\:\:{will}\:{be}\:−\mathrm{3},−\mathrm{2},−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\:\:{and}\:\:{the}\:{related} \\ $$$${number}\:{system}\:{will}\:{be}\:{base}\mathrm{7}\:\:{system}.\:\mathcal{D}{ifferent} \\ $$$${weights}\:{will}\:{be}\:\:\mathrm{7}^{\mathrm{0}} ,\mathrm{7}^{\mathrm{1}} ,\mathrm{7}^{\mathrm{2}} ,… \\ $$$$\mathcal{I}{f}\:\:{we}\:{use}\:{only}\:{one}\:{side}\:{of}\:{balance}\:{to}\:{place}\:{weights}\:{these} \\ $$$${type}\:{of}\:{questions}\:{will}\:{clmpletely}\:{be}\:\:\:{examples}\:{of}\: \\ $$$${base}\:{system}\:{numbers}. \\ $$
Commented by prakash jain last updated on 08/Dec/15
Suppose you are allowed to put weight only on one side of balance then the number of combinations that you can generate with n weights are 2^n (choose or not choose) suppose you want 63 weights than you need 6 weight such that Π_(i=a,b,c,d,e,f) (1+x^i )=Σ_(j=0) ^(63) x^j = 1+x+x^2 +x^4 +...+x^(63) =(1+x)+x^2 (1+x)+...+x^(60) (1+x)+x^(62) (1+x) ... =(1+x)(1+x^2 )(1+x^4 )(1+x^8 )(1+x^(16) )(1+x^(32) ) weights you will need are 1 2 4 8 16 32. I think you should treat these problems as combination problems and use those techniques to solve.
$$\mathrm{Suppose}\:\mathrm{you}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{put}\:\mathrm{weight} \\ $$$$\mathrm{only}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{balance}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{combinations}\:\mathrm{that}\:\mathrm{you}\:\mathrm{can} \\ $$$$\mathrm{generate}\:\mathrm{with}\:{n}\:\mathrm{weights}\:\mathrm{are}\:\mathrm{2}^{{n}} \:\left(\mathrm{choose}\right. \\ $$$$\left.\mathrm{or}\:\mathrm{not}\:\mathrm{choose}\right)\:\mathrm{suppose}\:\mathrm{you}\:\mathrm{want}\:\mathrm{63}\:\mathrm{weights} \\ $$$$\mathrm{than}\:\mathrm{you}\:\mathrm{need}\:\mathrm{6}\:\mathrm{weight}\:\mathrm{such}\:\mathrm{that} \\ $$$$\underset{{i}={a},{b},{c},{d},{e},{f}} {\prod}\left(\mathrm{1}+{x}^{{i}} \right)=\underset{{j}=\mathrm{0}} {\overset{\mathrm{63}} {\sum}}{x}^{{j}} = \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…+{x}^{\mathrm{63}} \\ $$$$=\left(\mathrm{1}+{x}\right)+{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+…+{x}^{\mathrm{60}} \left(\mathrm{1}+{x}\right)+{x}^{\mathrm{62}} \left(\mathrm{1}+{x}\right) \\ $$$$… \\ $$$$=\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{8}} \right)\left(\mathrm{1}+{x}^{\mathrm{16}} \right)\left(\mathrm{1}+{x}^{\mathrm{32}} \right) \\ $$$${weights}\:{you}\:{will}\:{need}\:{are}\:\mathrm{1}\:\mathrm{2}\:\mathrm{4}\:\mathrm{8}\:\mathrm{16}\:\mathrm{32}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{should}\:\mathrm{treat}\:\mathrm{these}\:\mathrm{problems}\:\mathrm{as} \\ $$$$\mathrm{combination}\:\mathrm{problems}\:\mathrm{and}\:\mathrm{use}\:\mathrm{those}\:\mathrm{techniques} \\ $$$$\mathrm{to}\:\mathrm{solve}. \\ $$
Commented by Rasheed Soomro last updated on 08/Dec/15
The problem is ALSO related to base system. Suppose we have four weights(one peice of each) 1kg,2 kg,4 kg and 8 kg and we place these on one side of balance. Suppose we have placed 1kg,2kg and 8kg.Total weight is 11 kg.Couldn′t we write 1011_2 kg? Couldn′t we write every combination weights in binary system. We have one peice of a weight when we place it its coefficient is 1 and when we don′t its coefficient is 0.If these coefficients are c_0 ,c_1 ,c_2 ,c_3 of 1,2,2^2 ,2^3 respectively. then the combination of weights can be written in the form c_0 (1)+c_1 (2)+c_2 (4)+c_3 (8) c_0 (2^0 )+c_1 (2^1 )+c_2 (2^2 )+c_3 (2^3 ) ( c_0 c_1 c_2 c_3 )_2 where c_i =0 or 1 Now we can say combination of 1,2,4,8 kg can weigh upto 1111_2 kg Similarly If we take 1,3,9 kg, two peices of each, If we place 2 peices of 1 kg 0 peice of 3 kg 1 peice of 9 kg T he combination can be represented by base3 system 102_3 So I think the problem is also directly related to base system. For weights on both sides I ′ll write in next comment.
$$\mathcal{T}{he}\:{problem}\:{is}\:\mathcal{ALSO}\:\:{related}\:{to}\:{base}\:{system}. \\ $$$$\mathcal{S}{uppose}\:{we}\:{have}\:{four}\:{weights}\left({one}\:{peice}\:{of}\:{each}\right)\:\mathrm{1}{kg},\mathrm{2}\:{kg},\mathrm{4}\:{kg}\:{and} \\ $$$$\mathrm{8}\:{kg}\:\:{and}\:{we}\:{place}\:{these}\:{on}\:{one}\:{side}\:{of}\:{balance}. \\ $$$$\mathcal{S}{uppose}\:{we}\:{have}\:{placed}\:\mathrm{1}{kg},\mathrm{2}{kg}\:{and}\:\mathrm{8}{kg}.\mathcal{T}{otal}\:{weight} \\ $$$${is}\:\mathrm{11}\:{kg}.{Couldn}'{t}\:{we}\:{write}\:\mathrm{1011}_{\mathrm{2}} {kg}?\:{Couldn}'{t}\:{we}\:{write} \\ $$$${every}\:{combination}\:{weights}\:{in}\:{binary}\:{system}. \\ $$$$\mathcal{W}{e}\:{have}\:{one}\:{peice}\:{of}\:{a}\:{weight}\:{when}\:{we}\:{place}\:{it} \\ $$$${its}\:{coefficient}\:{is}\:\mathrm{1}\:{and}\:{when}\:{we}\:{don}'{t}\:{its}\:{coefficient} \\ $$$${is}\:\mathrm{0}.\mathcal{I}{f}\:{these}\:{coefficients}\:{are}\:{c}_{\mathrm{0}} ,{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} ,{c}_{\mathrm{3}} \:\:{of}\:\mathrm{1},\mathrm{2},\mathrm{2}^{\mathrm{2}} ,\mathrm{2}^{\mathrm{3}} \: \\ $$$${respectively}.\:{then}\:{the}\:{combination}\:{of}\:{weights} \\ $$$${can}\:{be}\:{written}\:{in}\:{the}\:{form}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}_{\mathrm{0}} \left(\mathrm{1}\right)+{c}_{\mathrm{1}} \left(\mathrm{2}\right)+{c}_{\mathrm{2}} \left(\mathrm{4}\right)+{c}_{\mathrm{3}} \left(\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}_{\mathrm{0}} \left(\mathrm{2}^{\mathrm{0}} \right)+{c}_{\mathrm{1}} \left(\mathrm{2}^{\mathrm{1}} \right)+{c}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}} \right)+{c}_{\mathrm{3}} \left(\mathrm{2}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:{c}_{\mathrm{0}} {c}_{\mathrm{1}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} \right)_{\mathrm{2}} \\ $$$$\:\:{where}\:{c}_{{i}} =\mathrm{0}\:{or}\:\mathrm{1} \\ $$$$\:\:\mathcal{N}{ow}\:{we}\:{can}\:{say}\:{combination}\:{of}\:\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{8}\:{kg}\:{can}\:{weigh} \\ $$$${upto}\:\mathrm{1111}_{\mathrm{2}} \:{kg} \\ $$$$\mathcal{S}{imilarly} \\ $$$$\:\:\:\:\mathcal{I}{f}\:\:{we}\:{take}\:\mathrm{1},\mathrm{3},\mathrm{9}\:{kg},\:{two}\:{peices}\:{of}\:{each}, \\ $$$$\:\:\:\:\:\mathcal{I}{f}\:{we}\:{place}\:\mathrm{2}\:{peices}\:{of}\:\mathrm{1}\:{kg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:{peice}\:{of}\:\mathrm{3}\:{kg} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:{peice}\:{of}\:\mathrm{9}\:{kg} \\ $$$$\mathcal{T}\:{he}\:{combination}\:{can}\:{be}\:{represented}\:{by}\:{base}\mathrm{3}\:{system} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{102}_{\mathrm{3}} \\ $$$$\mathcal{S}{o}\:\mathcal{I}\:{think}\:{the}\:{problem}\:{is}\:{also}\:{directly}\:{related}\:{to} \\ $$$${base}\:{system}. \\ $$$$\mathcal{F}{or}\:{weights}\:{on}\:{both}\:{sides}\:\mathcal{I}\:'{ll}\:{write}\:{in}\:{next}\:{comment}. \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 08/Dec/15
Let weights are 1kg,3kg,9kg (one peice of each) and let these can be placed on both sides of balance. (If we have two peices of a weight and we can place this only on one side then it may have coefficients 0,1,2 But if we can place on both sides then it may have coefficients −2,−1,0,1,2 ) So each weight may have coefficients −1,0,1. Let coefficients of 1,3,9 kg are respectively c_0 , c_1 ,c_2 . Any combination of these weights can take the form c_0 (1)+c_1 (3)+c_2 (9) Or c_0 (3^0 )+c_1 (3^1 )+c_2 (3^2 ) Or (c_0 c_1 c_2 )_3 where c_i =−1,0,1 Hence I think that these problems are ALSO directly related to base systems. If we think these problems relating to base system numbers they seem simpler!
$$\mathcal{L}{et}\:{weights}\:{are}\:\mathrm{1}{kg},\mathrm{3}{kg},\mathrm{9}{kg}\:\left({one}\:{peice}\:{of}\:{each}\right) \\ $$$${and}\:\:{let}\:{these}\:{can}\:{be}\:{placed}\:{on}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{sides}}\:{of}\:{balance}. \\ $$$$\left(\mathcal{I}{f}\:{we}\:{have}\:{two}\:{peices}\:{of}\:{a}\:{weight}\:{and}\:{we}\:{can}\:{place}\right. \\ $$$${this}\:{only}\:{on}\:{one}\:{side}\:{then}\:{it}\:{may}\:{have}\:\:{coefficients}\:\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\mathcal{B}{ut}\:{if}\:{we}\:{can}\:{place}\:{on}\:{both}\:{sides}\:{then}\:{it}\:{may}\:{have} \\ $$$$\left.{coefficients}\:−\mathrm{2},−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2}\:\right) \\ $$$$\mathcal{S}{o}\:\:{each}\:{weight}\:{may}\:{have}\:{coefficients}\:−\mathrm{1},\mathrm{0},\mathrm{1}. \\ $$$$\mathcal{L}{et}\:{coefficients}\:{of}\:\mathrm{1},\mathrm{3},\mathrm{9}\:{kg}\:{are}\:{respectively}\:{c}_{\mathrm{0}} , \\ $$$${c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \:. \\ $$$$\:\:\:\:\mathcal{A}{ny}\:{combination}\:{of}\:{these}\:{weights}\:{can}\:{take}\:{the} \\ $$$${form}\:{c}_{\mathrm{0}} \left(\mathrm{1}\right)+{c}_{\mathrm{1}} \left(\mathrm{3}\right)+{c}_{\mathrm{2}} \left(\mathrm{9}\right) \\ $$$${Or}\:\:\:\:\:\:{c}_{\mathrm{0}} \left(\mathrm{3}^{\mathrm{0}} \right)+{c}_{\mathrm{1}} \left(\mathrm{3}^{\mathrm{1}} \right)+{c}_{\mathrm{2}} \left(\mathrm{3}^{\mathrm{2}} \right) \\ $$$${Or}\:\:\:\:\:\left({c}_{\mathrm{0}} {c}_{\mathrm{1}} {c}_{\mathrm{2}} \right)_{\mathrm{3}} \:\:\:\:\:{where}\:{c}_{{i}} =−\mathrm{1},\mathrm{0},\mathrm{1} \\ $$$$\mathcal{H}{ence}\:{I}\:{think}\:{that}\:{these}\:{problems}\:{are}\:\mathcal{ALSO}\:{directly} \\ $$$${related}\:{to}\:{base}\:{systems}. \\ $$$$\mathcal{I}{f}\:{we}\:{think}\:{these}\:{problems}\:{relating}\:{to}\:{base}\:{system} \\ $$$${numbers}\:{they}\:{seem}\:{simpler}! \\ $$
Answered by prakash jain last updated on 07/Dec/15
1 1 weighs up to 2 kg 5 5 weighs upto 12 kg 25 25 weighs upto 62 kg 1k=1 2k=1+1 3k=5−1−1 4k=5−1 5k=5 6k=5+1 7k=5+1+1 8k=5+5−1−1 9k=5+5−1−1 10k=5+5 11k=5+5+1 12k=5+5+1+1 similarly you can weigh upto 37 kg with one 25kg weight and upto 62 kg with extra 25 kg weight. so you need 6 weights: 1 1 5 5 25 25
$$\mathrm{1}\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{weighs}\:\mathrm{up}\:\mathrm{to}\:\mathrm{2}\:\mathrm{kg} \\ $$$$\mathrm{5}\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\mathrm{weighs}\:\mathrm{upto}\:\mathrm{12}\:\mathrm{kg} \\ $$$$\mathrm{25}\:\mathrm{25}\:\:\:\:\:\:\mathrm{weighs}\:\mathrm{upto}\:\mathrm{62}\:\mathrm{kg} \\ $$$$\mathrm{1k}=\mathrm{1}\:\:\mathrm{2k}=\mathrm{1}+\mathrm{1}\:\mathrm{3k}=\mathrm{5}−\mathrm{1}−\mathrm{1}\:\:\mathrm{4k}=\mathrm{5}−\mathrm{1}\:\mathrm{5k}=\mathrm{5} \\ $$$$\mathrm{6k}=\mathrm{5}+\mathrm{1}\:\mathrm{7k}=\mathrm{5}+\mathrm{1}+\mathrm{1}\:\mathrm{8k}=\mathrm{5}+\mathrm{5}−\mathrm{1}−\mathrm{1}\:\mathrm{9k}=\mathrm{5}+\mathrm{5}−\mathrm{1}−\mathrm{1} \\ $$$$\mathrm{10k}=\mathrm{5}+\mathrm{5}\:\mathrm{11k}=\mathrm{5}+\mathrm{5}+\mathrm{1}\:\mathrm{12k}=\mathrm{5}+\mathrm{5}+\mathrm{1}+\mathrm{1} \\ $$$$\mathrm{similarly}\:\mathrm{you}\:\mathrm{can}\:\mathrm{weigh}\:\mathrm{upto}\:\mathrm{37}\:\mathrm{kg}\:\mathrm{with} \\ $$$$\mathrm{one}\:\mathrm{25kg}\:\mathrm{weight}\:\mathrm{and}\:\mathrm{upto}\:\mathrm{62}\:\mathrm{kg}\:\mathrm{with}\:\mathrm{extra}\:\mathrm{25}\:\mathrm{kg} \\ $$$$\mathrm{weight}. \\ $$$$\mathrm{so}\:\mathrm{you}\:\mathrm{need}\:\mathrm{6}\:\mathrm{weights}:\:\mathrm{1}\:\mathrm{1}\:\mathrm{5}\:\mathrm{5}\:\mathrm{25}\:\mathrm{25} \\ $$
Commented by prakash jain last updated on 07/Dec/15
Suppose you have following weights a a b b generating function for weights is the following ((1/x^(2a) )+(1/x^a )+1+x^a +x^(2a) )((1/x^(2b) )+(1/x^b )+1+x^b +x^(2b) ) You can get maximum 25 results (incl 0). The exponents of x gives the weight generated. With 3 weights (each 2) you can 125 results and 62 +ve results. So you will need a minimum of 6 weights. Let us say weights are a a b b c c The generating function gives the following: ((1/x^(2a) )+(1/x^a )+1+x^a +x^(2a) )((1/x^(2b) )+(1/x^b )+1+x^b +x^(2b) )((1/x^(2c) )+(1/x^c )+1+x^c +x^(2c) ) There a total of 125 terms. Since we want result upto 60 we try for −62 to +62. a=1, b=5, c=25 satisfies.
$$\mathrm{Suppose}\:\mathrm{you}\:\mathrm{have}\:\mathrm{following}\:\mathrm{weights}\: \\ $$$${a}\:{a}\:{b}\:{b} \\ $$$$\mathrm{generating}\:\mathrm{function}\:\mathrm{for}\:\mathrm{weights}\:\mathrm{is}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}{a}} }+\frac{\mathrm{1}}{{x}^{{a}} }+\mathrm{1}+{x}^{{a}} +{x}^{\mathrm{2}{a}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}{b}} }+\frac{\mathrm{1}}{{x}^{{b}} }+\mathrm{1}+{x}^{{b}} +{x}^{\mathrm{2}{b}} \right) \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{get}\:\mathrm{maximum}\:\mathrm{25}\:\mathrm{results}\:\left(\mathrm{incl}\:\mathrm{0}\right).\:\mathrm{The} \\ $$$$\mathrm{exponents}\:\mathrm{of}\:{x}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{weight}\:\mathrm{generated}. \\ $$$$\mathrm{With}\:\mathrm{3}\:\mathrm{weights}\:\left(\mathrm{each}\:\mathrm{2}\right)\:\mathrm{you}\:\mathrm{can}\:\mathrm{125}\:\mathrm{results} \\ $$$$\mathrm{and}\:\mathrm{62}\:+\mathrm{ve}\:\mathrm{results}.\:\mathrm{So}\:\mathrm{you}\:\mathrm{will}\:\mathrm{need}\:\mathrm{a}\:\mathrm{minimum} \\ $$$$\mathrm{of}\:\mathrm{6}\:\mathrm{weights}. \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{weights}\:\mathrm{are}\:{a}\:\:{a}\:\:{b}\:\:{b}\:{c}\:{c} \\ $$$$\mathrm{The}\:\mathrm{generating}\:\mathrm{function}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{following}: \\ $$$$\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}{a}} }+\frac{\mathrm{1}}{{x}^{{a}} }+\mathrm{1}+{x}^{{a}} +{x}^{\mathrm{2}{a}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}{b}} }+\frac{\mathrm{1}}{{x}^{{b}} }+\mathrm{1}+{x}^{{b}} +{x}^{\mathrm{2}{b}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}{c}} }+\frac{\mathrm{1}}{{x}^{{c}} }+\mathrm{1}+{x}^{{c}} +{x}^{\mathrm{2}{c}} \right) \\ $$$$\mathrm{There}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:\mathrm{125}\:\mathrm{terms}.\:\mathrm{Since}\:\mathrm{we}\: \\ $$$$\mathrm{want}\:\mathrm{result}\:\mathrm{upto}\:\mathrm{60}\:\mathrm{we}\:\mathrm{try}\:\mathrm{for}\:−\mathrm{62}\:\mathrm{to}\:+\mathrm{62}. \\ $$$${a}=\mathrm{1},\:{b}=\mathrm{5},\:{c}=\mathrm{25} \\ $$$$\mathrm{satisfies}. \\ $$
Commented by Rasheed Soomro last updated on 07/Dec/15
eXCELlent!
$${e}\mathcal{XCEL}{lent}! \\ $$

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