Henry Ernest Dudeney/Modern Puzzles/97 - A Common Divisor/Solution

by : $97$

 * A Common Divisor

Solution

 * $79$.

The remainder in each case is $51$.

Proof
Let $n$ be the divisor sought.

Let $m$ be the remainder.

We have that:

Hence we see that $n$ is a common divisor for each of $28 \, 203$, $242 \, 609$ and $214 \, 406$.

So we establish the prime factors of these numbers:

and it is seen by inspection that $n = 79$.

Thus we have: