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Question Number 202203 by SEKRET last updated on 22/Dec/23

$$$$$$$$$$$$$$2^{2023}=\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{\mathbf{abc}.............}$$$$\mathbf{a}+\mathbf{b}+\mathbf{c}=?$$$$$$$$$$$$$$$$$$

Answered by Frix last updated on 22/Dec/23

$$\mathrm{With}\mathrm{a}\mathrm{calculator}\mathrm{not}\mathrm{capable}\mathrm{of}\mathrm{exact}$$$$\mathrm{numbers}\mathrm{as}\mathrm{great}\mathrm{as}\approx {10}^{609}:$$$$\log 2^{2023}=2023\log 2$$$$\log 2\approx .301029995664$$$$2023\log 2\approx 608.983681228$$$${10}^{.983681228}\approx 9.63121833$$$$2^{2023}=963...$$$$9+6+3=18$$

Commented by SEKRET last updated on 23/Dec/23

$$$$Hello, thank you sir for your reply. Can it be done without a calculator? Is there an easier way?

Commented by Frix last updated on 23/Dec/23

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{any}\mathrm{other}\mathrm{way}.$$

Commented by SEKRET last updated on 23/Dec/23

$$\mathbf{thank}\mathbf{you}\mathbf{sir}\mathbf{Frix}$$

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