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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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