The largest number, which divides 25, 73 and 97 to leave the same remainder In each case, is
1). 24
2). 23
3). 21
4). 6
Let x be the remainder. Then, $(25 – x ), (73 – x ),$ and $(97 – x )$ Will be exactly divisible by the required number.
Required number = HCF of $(73 –x ) – (25 –x ), (97 –x ) – (73 –x )$ and $(97 –x ) – (25 –x ) =$ HCF of $(73 –25), (97 –73),$ and $(97 –25)$
$= $HCF of $48, 24$ and $72 = 24$