If the LCM of three numbers is 120 and their HCF is 2, what is the maximum possi

If the LCM of three numbers is 120 and their HCF is 2, what is the maximum possible product of the three numbers? (2023)

If the LCM of three numbers is 120 and their HCF is 2, what is the maximum possible product of the three numbers? (2023)

Step 1: Understand that LCM (Least Common Multiple) is the smallest number that is a multiple of all three numbers.
Step 2: Understand that HCF (Highest Common Factor) is the largest number that divides all three numbers.
Step 3: Given that the LCM is 120 and the HCF is 2, we can express the three numbers as 2a, 2b, and 2c, where a, b, and c are integers.
Step 4: Since the LCM of the three numbers is 120, we can write it as LCM(2a, 2b, 2c) = 2 * LCM(a, b, c).
Step 5: Therefore, LCM(a, b, c) must equal 120 / 2 = 60.
Step 6: Now, we need to find three numbers a, b, and c such that their LCM is 60 and their product is maximized.
Step 7: The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
Step 8: We can try combinations of these factors to find a, b, and c that give the maximum product while keeping LCM(a, b, c) = 60.
Step 9: After testing combinations, we find that a = 5, b = 3, and c = 4 gives LCM(5, 3, 4) = 60.
Step 10: Now, calculate the original numbers: 2a = 10, 2b = 6, 2c = 8.
Step 11: The product of the three numbers is 10 * 6 * 8 = 480.
Step 12: However, we can also try a = 6, b = 10, c = 1, which gives LCM(6, 10, 1) = 30, not valid.
Step 13: The maximum product occurs when the numbers are 2, 10, and 6, giving a product of 120.

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