find the least number which when divided by 37 and 7 leaves a remainder of 2 in each case .
the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 37 , 7 ) + 2 = 261 . answer : b
add(2, lcm(37, 7))