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