Question
The sum of 6 natural numbers (not necessarily distinct) is 23. Let M denote the positive difference of the maximum and minimum of the LCM of these 6 numbers. Then total number of divisors that divide M2 but doesn't divide M are:
Answer: Option B
:
B
Minimum occurs when the numbers are 6, 6, 6, 2, 2, 1 (LCM = 6). Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2 (LCM = 420).
To achieve minimum LCM, we have to search for maximum number of GCDs among the 6C2 pairs. To look for maximum LCM, we need the numbers as co-prime.
Thus, M = 414. Now, 414 can be written as 2×32×23⟹ M has 2×3×2 = 12 divisors.
M2 = 22×34×232⟹ M2 has 3×5×3 = 45 divisors.
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:
B
Minimum occurs when the numbers are 6, 6, 6, 2, 2, 1 (LCM = 6). Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2 (LCM = 420).
To achieve minimum LCM, we have to search for maximum number of GCDs among the 6C2 pairs. To look for maximum LCM, we need the numbers as co-prime.
Thus, M = 414. Now, 414 can be written as 2×32×23⟹ M has 2×3×2 = 12 divisors.
M2 = 22×34×232⟹ M2 has 3×5×3 = 45 divisors.
Hence, option (b) (45-12 = 33) is the correct answer.
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