Question Number 202648 by liuxinnan last updated on 31/Dec/23
$${how}\:{to}\:{use}\:{the}\:{fewest}\:“\mathrm{2}''\:{to}\:{make}\:\mathrm{2024} \\ $$$${you}\:{only}\:{can}\:{use}\:“\mathrm{2}''\:{and}\:−\:+\:×\:/\:\left(\:\right)\:!\:\:\hat {\:}\:… \\ $$$${as}\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\mathrm{2}×\mathrm{2}×\mathrm{2}\left(\left(\mathrm{2}×\mathrm{2}\right)!−\mathrm{2}/\mathrm{2}\right)\left(\mathrm{2}×\left(\mathrm{2}+\mathrm{2}/\mathrm{2}\right)!−\mathrm{2}/\mathrm{2}\right)=\mathrm{2024} \\ $$$${totally}\:{use}\:{thirteen}\:“\:\mathrm{2}'' \\ $$
Commented by Frix last updated on 31/Dec/23
![You can use the subfactorial !2=1 2024=2^(2+!2) (2^2^(2+!2) −2−!2) [9] btw 2023=(2^(2+!2) −!2)(2^2^2 +!2)^2 [9] 2025=(2+!2)^2^2 (2^2 +!2)^2 [8]](https://www.tinkutara.com/question/Q202669.png)
$$\mathrm{You}\:\mathrm{can}\:\mathrm{use}\:\mathrm{the}\:\mathrm{subfactorial}\:!\mathrm{2}=\mathrm{1} \\ $$$$\mathrm{2024}=\mathrm{2}^{\mathrm{2}+!\mathrm{2}} \left(\mathrm{2}^{\mathrm{2}^{\mathrm{2}+!\mathrm{2}} } −\mathrm{2}−!\mathrm{2}\right)\:\:\:\:\:\left[\mathrm{9}\right] \\ $$$$\mathrm{btw} \\ $$$$\mathrm{2023}=\left(\mathrm{2}^{\mathrm{2}+!\mathrm{2}} −!\mathrm{2}\right)\left(\mathrm{2}^{\mathrm{2}^{\mathrm{2}} } +!\mathrm{2}\right)^{\mathrm{2}} \:\:\:\:\:\left[\mathrm{9}\right] \\ $$$$\mathrm{2025}=\left(\mathrm{2}+!\mathrm{2}\right)^{\mathrm{2}^{\mathrm{2}} } \left(\mathrm{2}^{\mathrm{2}} +!\mathrm{2}\right)^{\mathrm{2}} \:\:\:\:\:\left[\mathrm{8}\right] \\ $$
Commented by liuxinnan last updated on 31/Dec/23
$${real}? \\ $$
Commented by MathematicalUser2357 last updated on 06/Jan/24
$$\mathrm{Yes}!\:\mathrm{The}\:\mathrm{excalmation}\:\mathrm{mark}\:\left(!\right)\:\mathrm{is}\:\mathrm{subfactorial}\:\mathrm{or}\:\mathrm{factorial}! \\ $$
Commented by liuxinnan last updated on 06/Jan/24
Commented by liuxinnan last updated on 06/Jan/24
$${I}\:{get}\:{it} \\ $$