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Question Number 113001 by Aina Samuel Temidayo last updated on 10/Sep/20

$$\mathrm{N}!\mathrm{is}\mathrm{completely}\mathrm{divisible}\mathrm{by}{13}^{52}.$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{digits}\mathrm{of}\mathrm{the}$$$$\mathrm{smallest}\mathrm{such}\mathrm{number}\mathrm{N}?$$

Commented by MJS_new last updated on 11/Sep/20

$$N!={13}^{52}\times k$$$$13!\mathrm{has}1\mathrm{factor}13$$$$\left(2\times 13\right)!\mathrm{has}2$$$$\left(13k\right)!\mathrm{has}k\mathrm{for}k<13$$$$\left({13}^2\right)!\mathrm{has}14$$$$\left(13k\right)!\mathrm{has}k+1\mathrm{for}k<26$$$$\left(13\times 26\right)!\mathrm{has}28$$$$\left(13k\right)!\mathrm{has}k+2\mathrm{for}k<39$$$$\left(13\times 39\right)!\mathrm{has}42$$$$\left(13k\right)!\mathrm{has}k+3\mathrm{for}k<52$$$$\Rightarrow \left(13\times 49\right)!\mathrm{id}\mathrm{divisible}\mathrm{by}{13}^{52}$$$$\mathrm{but}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{calculate}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{digits}\mathrm{of}637!$$

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