202. Invariant remainder. Find a number that, when dividing the numbers $1108, 1453, 1844$, and 2281, gives the same remainder.
202. Since the remainder does not change, the divisor must be odd. Further, $1453-1108=345,1844-1453=391,2281-1844=$ 437. Now notice that $437-391=391-345=46=2 \cdot(23)$. Since it is always true that

$$
\left(N_{1} d+r\right)-\left(N_{2} d+r\right)=d\left(N_{1}-N_{2}\right)
$$

the desired number is 23, and the corresponding remainder is 4.

$[$ [ $2 .$, 3. 35, 62 (January 1962).]