- AHCF(a, b) is always strictly less than LCM(a, b)
- BLCM(a, b) is always equal to a b
- CHCF(a, b) + LCM(a, b) is always equal to a + b
- DHCF(a, b) always divides LCM(a, b)
Step-by-Step Solution
Since HCF(a, b) LCM(a, b) = and the HCF divides both a and , the HCF always divides the LCM. The others fail: if then HCF = LCM, so the HCF is not always strictly smaller; LCM = only when a and are co-prime; and for we get HCF + LCM = 2 + 12 = 14 while .
Answer: D) HCF(a, b) always divides LCM(a, b)