Solution:
1000=23×53and 2000=24×53
Since LCM (c. a) and LCM (b, c) is24×53 and LCM (a, b) = 23×53 so the factor 24 must be present in c.
Hence c=24×5x, where x ranges from 0 to 3 Therefore, there are four possible values of C.
Since, HCF of (a, b) = K×53, it means
a=2y×53
b=2z×53
x = 0 to 3, y = 0, then z = 3 → 4 cases,
x = 0 to 3, y = 1, then z = 3 → 4 cases,
x = 0 to 3, y = 2, then z = 3 → 4 cases,
x = 0 to 3, y = 3, then z = 3 → 4 cases,
x = 0 to 3, y = 3, then z = 2 → 4 cases,
x = 0 to 3, y = 3, then z = 1 → 4 cases,
x = 0 to 3, y = 3, then z = 0 → 4 cases.
Hence, total cases = 28.
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