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The largest number, which divides 25, 73 and 97 to leave the same remainder In each case, is
- A24
- B23
- C21
- D6
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0
CCommunity
option 1 is the answer
0
CCommunity
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 $72 = 24$
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 $72 = 24$