Question Number 92188 by ~blr237~ last updated on 05/May/20 | ||
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$${let}\:{a},{b},{c}\:{be}\:{three}\:{digits}\:{all}\:{different}\:{of}\:{zero} \\ $$$${Prove}\:{that}\:\frac{{ac}}{{cb}}=\frac{{a}}{{b}}\:\Leftrightarrow\:\forall\:{n}\geqslant\mathrm{1}\:\:\:\:\frac{{accc}...{cc}}{{ccc}...{ccb}}\:=\frac{{a}}{{b}}\:\:\: \\ $$$${the}\:{number}\:{accc}...{cc}\:\:\:{has}\:{the}\:{digit}\:{c}\:\:{n}\:{times} \\ $$ | ||