Let the two numbers be 21a and 21b(since hcf = 21 and gcd =1)
21a×21b = 5292
441ab = 5292
ab=4415292
ab = 12 and gcd(a,b) = 1
factor pairs of 12:
(1,12),(2,6),(3,4)
valid pairs are (1,12),(3,4)
each gives a pair of numbers (21,252)(63,84)
so total number of such pairs = 2