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