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Question Number 122954 by mr W last updated on 21/Nov/20

Commented by Dwaipayan Shikari last updated on 21/Nov/20

Infinitely  many answers sir!  If they are in a relation  C=1+(n−1)  then 12 will cost 12 $   (for the people who.....)  If they are in a relation  n+(n−1)(n−2)(n−3)...(n−11)  If n=12  then price will be 12+11! $    :)  ......

$${Infinitely}\:\:{many}\:{answers}\:{sir}! \\ $$$${If}\:{they}\:{are}\:{in}\:{a}\:{relation}\:\:{C}=\mathrm{1}+\left({n}−\mathrm{1}\right) \\ $$$${then}\:\mathrm{12}\:{will}\:{cost}\:\mathrm{12}\:\$\:\:\:\left({for}\:{the}\:{people}\:{who}.....\right) \\ $$$${If}\:{they}\:{are}\:{in}\:{a}\:{relation} \\ $$$${n}+\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)...\left({n}−\mathrm{11}\right) \\ $$$${If}\:{n}=\mathrm{12} \\ $$$$\left.{then}\:{price}\:{will}\:{be}\:\mathrm{12}+\mathrm{11}!\:\$\:\:\:\::\right) \\ $$$$...... \\ $$

Commented by prakash jain last updated on 22/Nov/20

Information about 2 and 11 is  redundant.  12= 12 times ONE  So 12 will cost $12.

$$\mathrm{Information}\:\mathrm{about}\:\mathrm{2}\:\mathrm{and}\:\mathrm{11}\:\mathrm{is} \\ $$$$\mathrm{redundant}. \\ $$$$\mathrm{12}=\:\mathrm{12}\:\mathrm{times}\:\mathrm{ONE} \\ $$$$\mathrm{So}\:\mathrm{12}\:\mathrm{will}\:\mathrm{cost}\:\$\mathrm{12}. \\ $$

Commented by mr W last updated on 22/Nov/20

an other possibility:  say a letter O costs o $, a letter N  costs n $, etc.  o+n+e=1   ...(i)  t+w+o=2   ...(ii)  3e+l+v+n=11   ...(iii)  (iii)+(ii)−(i):  2e+l+v+t+w=12   ...(iv)    TWELVE costs 2e+t+w+l+v, that  is 12 $ from (iv).

$${an}\:{other}\:{possibility}: \\ $$$${say}\:{a}\:{letter}\:\boldsymbol{{O}}\:{costs}\:{o}\:\$,\:{a}\:{letter}\:\boldsymbol{{N}} \\ $$$${costs}\:{n}\:\$,\:{etc}. \\ $$$${o}+{n}+{e}=\mathrm{1}\:\:\:...\left({i}\right) \\ $$$${t}+{w}+{o}=\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{3}{e}+{l}+{v}+{n}=\mathrm{11}\:\:\:...\left({iii}\right) \\ $$$$\left({iii}\right)+\left({ii}\right)−\left({i}\right): \\ $$$$\mathrm{2}{e}+{l}+{v}+{t}+{w}=\mathrm{12}\:\:\:...\left({iv}\right) \\ $$$$ \\ $$$$\boldsymbol{{TWELVE}}\:{costs}\:\mathrm{2}{e}+{t}+{w}+{l}+{v},\:{that} \\ $$$${is}\:\mathrm{12}\:\$\:{from}\:\left({iv}\right). \\ $$

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