1970 AMC 12 第 34 题

Question

除

13{,}511

、

13{,}903

和

14{,}589

后所得余数相同的最大整数是：

The greatest integer that will divide

13{,}511,

13{,}903,

and

14{,}589

and leave the same remainder is:

28

49

98

大于

49

的

7

的奇数倍

an odd multiple of

7

greater than

49

大于

98

的

7

的偶数倍

an even multiple of

7

greater than

98

余数相同意味着除数能整除每两个数之差

A common remainder means the divisor divides every pairwise difference

求

13903-13511

与

14589-13903

的最大公因数

Find the GCD of

13903-13511

and

14589-13903

一个除数除这三个数所得余数相同，当且仅当它能整除这些数之间的差。相邻两个差为

\begin{gathered} 13903-13511=392,\\ 14589-13903=686 \end{gathered}\text{。}

因此最大的可能除数为

\begin{gathered} \gcd(392,686)\\ =\gcd(392,294)\\ =\gcd(294,98)=98 \end{gathered}\text{。}

因此，正确答案是 C。

A divisor leaves the same remainder on all three numbers exactly when it divides their differences. The two successive differences are

\begin{gathered} 13903-13511=392,\\ 14589-13903=686. \end{gathered}

Therefore the greatest possible divisor is

\begin{gathered} \gcd(392,686)\\ =\gcd(392,294)\\ =\gcd(294,98)=98. \end{gathered}

Therefore, the correct answer is C.

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