1970 AMC 12 Problem 34

Question

The greatest integer that will divide

13{,}511,

13{,}903,

and

14{,}589

and leave the same remainder is:

28

49

98

an odd multiple of

7

greater than

49

an even multiple of

7

greater than

98

A common remainder means the divisor divides every pairwise difference

Find the GCD of

13903-13511

and

14589-13903

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.

Problem 34 in other years