Question

Prove that there is no natural number for which 9n ends with digit 0.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Any}\mathrm{positive}\mathrm{integer}\mathrm{ending}\mathrm{with}\mathrm{the}\mathrm{digit}0\mathrm{is}\mathrm{divisible}\mathrm{by}5\mathrm{and}2,\mathrm{so}\mathrm{its}\mathrm{prime}\mathrm{factorization}\mathrm{mus} \\ \mathrm{contain}2\mathrm{and}5. \\ {9}^{\mathrm{n}}={\left(3^2\right)}^{\mathrm{n}}={\left(3\times 3\right)}^{\mathrm{n}} \\ \mathrm{The}\mathrm{prime}\mathrm{numbers}\mathrm{in}\mathrm{factorization}\mathrm{of}{9}^{\mathrm{n}}\mathrm{does}\mathrm{not}\mathrm{contain}2\mathrm{and}5. \\ \mathrm{So},9^{\mathrm{n}}\mathrm{does}\mathrm{not}\mathrm{ends}\mathrm{with}\mathrm{the}\mathrm{digit}0. \end{gathered}$$

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