the least number which should be added to 2400 so that the sum is exactly divisible by 5 , 6,4 and 3 is :
solution l . c . m . of 5 , 6,4 and 3 = 60 . on dividing 2400 by 60 , the remainder is 40 . ∴ number to be added = ( 60 - 40 ) = 20 . answer c
subtract(multiply(multiply(multiply(const_2, const_2), 5), 3), divide(2400, multiply(multiply(multiply(const_2, const_2), 5), 3)))