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

N! is completely divisible by 13^(52) . What is the sum of the digits of the smallest such number N?

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

Commented by MJS_new last updated on 11/Sep/20

N!=13^(52) ×k 13! has 1 factor 13 (2×13)! has 2 (13k)! has k for k<13 (13^2 )! has 14 (13k)! has k+1 for k<26 (13×26)! has 28 (13k)! has k+2 for k<39 (13×39)! has 42 (13k)! has k+3 for k<52 ⇒ (13×49)! id divisible by 13^(52) but I can′t calculate the sum of the digits of 637!

$${N}!=\mathrm{13}^{\mathrm{52}} ×{k} \\ $$$$\mathrm{13}!\:\mathrm{has}\:\mathrm{1}\:\mathrm{factor}\:\mathrm{13} \\ $$$$\left(\mathrm{2}×\mathrm{13}\right)!\:\mathrm{has}\:\mathrm{2} \\ $$$$\left(\mathrm{13}{k}\right)!\:\mathrm{has}\:{k}\:\mathrm{for}\:{k}<\mathrm{13} \\ $$$$\left(\mathrm{13}^{\mathrm{2}} \right)!\:\mathrm{has}\:\mathrm{14} \\ $$$$\left(\mathrm{13}{k}\right)!\:\mathrm{has}\:{k}+\mathrm{1}\:\mathrm{for}\:{k}<\mathrm{26} \\ $$$$\left(\mathrm{13}×\mathrm{26}\right)!\:\mathrm{has}\:\mathrm{28} \\ $$$$\left(\mathrm{13}{k}\right)!\:\mathrm{has}\:{k}+\mathrm{2}\:\mathrm{for}\:{k}<\mathrm{39} \\ $$$$\left(\mathrm{13}×\mathrm{39}\right)!\:\mathrm{has}\:\mathrm{42} \\ $$$$\left(\mathrm{13}{k}\right)!\:\mathrm{has}\:{k}+\mathrm{3}\:\mathrm{for}\:{k}<\mathrm{52} \\ $$$$\Rightarrow\:\left(\mathrm{13}×\mathrm{49}\right)!\:\mathrm{id}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{13}^{\mathrm{52}} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{637}!\: \\ $$

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