Solution:Let the HCF = x and LCM = 10x
According to the question,
x + 10x = 451 ⇒ 11x = 451 ⇒ x = 41
So, HCF = x = 41 and LCM = 10x = 10 × 41 = 410
Sum of numbers = 287
Possible pairs = (41, 246), (82, 205)
But L.C.M. of (41, 246) = 246 ≠ 410
Satisfying pair = (82, 205)
So, one of the numbers = 82