Three consecutive positive integers, the middle one is a complete square number, the product of such three consecutive positive integers is called "wonderful number". What is the greatest common divisor of all "wonderful numbers" less than 2008?

Three consecutive positive integers, the middle one is a complete square number, the product of such three consecutive positive integers is called "wonderful number". What is the greatest common divisor of all "wonderful numbers" less than 2008?

① Any three consecutive positive integers must have a divisor of 3. Therefore, any "wonderful number" must have a factor of 3. ② if the number in the middle of three consecutive positive integers is even and it is a perfect square, it must be divisible by 4. If the number in the middle is odd, the first and third numbers are even, so any "wonderful number" must have a factor of 4. ③ the bits of a perfect square can only be 4 1. 4, 5, 6, 9 and 0. If the bits are 5 and 0, then the middle number will be divisible by 5. If the bits are 1 and 6, then the first number will be divisible by 5. If the bits are 4 and 9, then the third number will be divisible by 5. Therefore, any "wonderful number" must have a factor 5. ④ the above description of "beautiful number" has a factor 3, 4, and 5, that is, a factor 60, that is, the maximum of all wonderful numbers The common divisor is at least 60. On the other hand, 60 = 3 × 4 × 5, 60 is also a "wonderful number", and the maximum convention of wonderful numbers is at most 60. A: the maximum common divisor of all wonderful numbers can only be 60

Given that 1 / 2 n is a perfect square number and 1 / 3 N is a cubic number, what is the minimum value of N

Let n / 2 = P ^ 2
n/3=q^3
be
2p^2=3q^3
p. Let P = 2 ^ x * 3 ^ y, q = 2 ^ s * 3 ^ t
There are
2^(2x+1)3^(2y)=2^(3s)3^(3t+1)
2x+1=3s
2y=3t+1
Let n be the minimum, let x = 1, y = 2
Then s = 1, t = 1
p=2*3^2=18
q=2*3=6
n=6^3*3=648
The minimum positive value of n is 648
648/2=18^2
648/3=6^3

If n is a non-zero natural number, half n is a square number and one third n is a cubic number, what is the minimum value of N?

1 / 2n is a square number. It can be seen that after n prime factor decomposition, the exponent of prime factor 2 is odd, and the exponents of other prime factors are even
Another 1 / 3N is a cubic number. It can be seen that after the prime factor n is decomposed, the exponent of prime factor 3 is divisible by 3, and the other prime factor exponents are multiples of 3
Therefore, in the smallest number n, the exponent of prime factor 2 is an odd number and a multiple of 3, and the minimum is 3; while the exponent of prime factor 3 is an even number and divisible by 3, and the minimum is 4
So the minimum number n is 2 ^ 3 × 3 ^ 4 = 648