Question

find the greatest number which is divisible 62 132 237 to leave the same remainder in each case​

Answer 1

Answer:

This concludes that (62-r), (132-r) & (237-r) are exactly divisible by d as remainder now = 0. Or we can say that d is gcd of all these three numbers . The largest number which divides 62, 132 and 237 is 35 leaving remainder 0.