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Number theory Difficulty 6.3 National olympiad Prove it

[12.2] 设 a,b,na, b, n 是给定的自然数,且都大于 1 . 再设 An−1A_{n-1} 和 AnA_{n} 在 aa 进位制中可表示为
An−1=xn−1xn−2⋯x0,An=xnxn−1⋯x0A_{n-1}=x_{n-1} x_{n-2} \cdots x_{0}, \quad A_{n}=x_{n} x_{n-1} \cdots x_{0}
Bn−1B_{n-1} 和 BnB_{n} 在 bb 进位制中可表示为
Bn−1=xn−1xn−2⋯x0,Bn=xnxn−1⋯x0B_{n-1}=x_{n-1} x_{n-2} \cdots x_{0}, \quad B_{n}=x_{n} x_{n-1} \cdots x_{0}

这里 xn−1≠0,xn≠0x_{n-1} \neq 0, x_{n} \neq 0. 证明: 当 a>ba>b 时,有 An−1/An<Bn−1/BnA_{n-1} / A_{n}<B_{n-1} / B_{n}.

Solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.