Question

what the smallest no which when divided by 35, 56, 91,leaves the remainder 7 in each case?

Answer 1

35 = 7 x 5
56 = 7 x 2 x 2 x 2 
91 = 7 x 13

thus LCM = 7x5x2x2x2x13

now 7 has been added that is 3647 is the required number