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Question Number 173165 by mr W last updated on 07/Jul/22
what′s the largest number you can  form only using the digits 1,2,3,4?
$${what}'{s}\:{the}\:{largest}\:{number}\:{you}\:{can} \\ $$$${form}\:{only}\:{using}\:{the}\:{digits}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}? \\ $$
Answered by pnuvu last updated on 07/Jul/22
4^3^((2+1))
$$\mathrm{4}^{\mathrm{3}^{\left(\mathrm{2}+\mathrm{1}\right)} } \\ $$
Answered by mr W last updated on 08/Jul/22
2^3^(41)   or 3^2^(41)   or 4^2^(31)   ?  2^3^(41)  =4^(3^(41) /2) >3^(3^(41) /2) >3^(((1.5^(41) )/2)×2^(41) ) >3^2^(41)    2^3^(41)  =4^(3^(41) /2) >4^2^(31)    so the largest number is 2^3^(41)  .
$$\mathrm{2}^{\mathrm{3}^{\mathrm{41}} } \:{or}\:\mathrm{3}^{\mathrm{2}^{\mathrm{41}} } \:{or}\:\mathrm{4}^{\mathrm{2}^{\mathrm{31}} } \:? \\ $$$$\mathrm{2}^{\mathrm{3}^{\mathrm{41}} } =\mathrm{4}^{\frac{\mathrm{3}^{\mathrm{41}} }{\mathrm{2}}} >\mathrm{3}^{\frac{\mathrm{3}^{\mathrm{41}} }{\mathrm{2}}} >\mathrm{3}^{\frac{\mathrm{1}.\mathrm{5}^{\mathrm{41}} }{\mathrm{2}}×\mathrm{2}^{\mathrm{41}} } >\mathrm{3}^{\mathrm{2}^{\mathrm{41}} } \\ $$$$\mathrm{2}^{\mathrm{3}^{\mathrm{41}} } =\mathrm{4}^{\frac{\mathrm{3}^{\mathrm{41}} }{\mathrm{2}}} >\mathrm{4}^{\mathrm{2}^{\mathrm{31}} } \\ $$$${so}\:{the}\:{largest}\:{number}\:{is}\:\mathrm{2}^{\mathrm{3}^{\mathrm{41}} } . \\ $$

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