1970 AHSME Problems/Problem 34

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Problem

The greatest integer that will divide $13511$, $13903$ and $14589$ and leave the same remainder is

$\text{(A) } 28\quad \text{(B) } 49\quad \text{(C) } 98\quad\\ \text{(D) an odd multiple of } 7 \text{ greater than } 49\quad\\ \text{(E) an even multiple of } 7 \text{ greater than } 98$

Solution

We know that 13903 minus 13511 is equivalent to 392. Additionally, 14589 minus 13903 is equivalent to 686. Since we are searching for the greatest integer that divides these three integers and leaves the same remainder, the answer resides in the greatest common factor of 686 and 392. Therefore, the answer is 98, or $\fbox{C}$

"Credit: Skupp3"

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
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