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Question Number 202648 by liuxinnan last updated on 31/Dec/23

$$howtousethefewest‘‘2^{″}tomake2024$$$$youonlycanuse‘‘2^{″}and-+\times /\left(\right)!\hat{}...$$$$asthefollowing$$$$2\times 2\times 2(\left(2\times 2\right)!-2/2)(2\times (2+2/2)!-2/2)=2024$$$$totallyusethirteen‘‘2^{″}$$

Commented by Frix last updated on 31/Dec/23

$$\mathrm{You}\mathrm{can}\mathrm{use}\mathrm{the}\mathrm{subfactorial}!2=1$$$$2024=2^{2+!2}(2^{2^{2+!2}}-2-!2)\left[9\right]$$$$\mathrm{btw}$$$$2023=(2^{2+!2}-!2){(2^{2^2}+!2)}^2\left[9\right]$$$$2025={(2+!2)}^{2^2}{(2^2+!2)}^2\left[8\right]$$

Commented by liuxinnan last updated on 31/Dec/23

$$real?$$

Commented by MathematicalUser2357 last updated on 06/Jan/24

$$\mathrm{Yes}!\mathrm{The}\mathrm{excalmation}\mathrm{mark}(!)\mathrm{is}\mathrm{subfactorial}\mathrm{or}\mathrm{factorial}!$$

Commented by liuxinnan last updated on 06/Jan/24

Commented by liuxinnan last updated on 06/Jan/24

$$Igetit$$

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