
Let the two numbers be x and y such that HCF(x,y) = h => x = ha, y = hb and LCM(x,y) = hab => h(a + b + ab) = 89 => h = 1 and a + b + ab = 89 => a = (89  b)/(1 + b) = 90/(1 + b)  1. Number of positive divisors of 90 (i.e. 2*5*3^{2}) = 2*2*3 = 12. So, 1 + b = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 => (a,b) = (89,0), (44,1), (29,2), (17,4), (14,5), (9,8), (8,9), (5,14), (4,17), (2,29), (1,44) or (0,89) Now a and b are non zero b'coz they are natural numbers. Also HCF(a,b) = 1. So final values for (a,b) = (1,44) , (2,29) , (4,17) , (5,14) , (8,9) i.e. 5 pairs. 