The L.C.M. of two different numbers are 30. Which of the following cannot be their H.C.F.?
12
A fundamental property relating the Highest Common Factor (HCF) and Least Common Multiple (LCM) of two numbers is that the HCF must always be a divisor of the LCM. In mathematical terms, if HCF is the highest common factor and LCM is the least common multiple of two numbers, then LCM must be divisible by HCF without any remainder.
Given that the LCM of two different numbers is 30, we need to identify which of the potential HCF values provided does not divide 30.
We examine each option to see if it divides the LCM (30):
Check divisibility: $ \frac{30}{15} = 2 $. Since 15 divides 30 evenly, 15 can be the HCF.
Check divisibility: $ \frac{30}{6} = 5 $. Since 6 divides 30 evenly, 6 can be the HCF.
Check divisibility: $ \frac{30}{10} = 3 $. Since 10 divides 30 evenly, 10 can be the HCF.
Check divisibility: $ \frac{30}{12} = \frac{5}{2} = 2.5 $. Since 12 does not divide 30 evenly (the result is not an integer), 12 cannot be the HCF.
Based on the divisibility rule, the HCF must divide the LCM. Among the given options, only 12 does not divide 30. Therefore, 12 cannot be the HCF of two numbers whose LCM is 30.
A frog was at the bottom of an 80 m deep well. It attempted to come out of it by jumping. In each jump, it covered 1.15 m but slipped down by 0.75 m. The number of jumps after which it would be out of the well is:
The smallest perfect square number divisible by each of 6 and 12 is:
Which is the least four-digit number which when divided by 5, 6 and 8 leaves remainder 2 in each case?
A frog was at the bottom of an 80 m deep well. It attempted to come out of it by jumping. In each jump, it covered 1.15 m but slipped down by 0.75 m. The number of jumps after which it would be out of the well is:
The smallest perfect square number divisible by each of 6 and 12 is:
Which is the least four-digit number which when divided by 5, 6 and 8 leaves remainder 2 in each case?