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For 2: All the prime factors of an odd number N will be odd. The prime factors to 2N will be all the factors of N as well as the factors of N multiplied by 2.

For example: 21 = 1,3,7,21
42 = 1,3,7,21,
2,6,14,42 (above factors multiplied by 2)

For an even number N, 2N can't have all distinct prime factors by multiplying N's factors by 2
10 - 1,2,5,10
20 - 1,2,5,10
4, ,20

Yurik79, I did not understand how your example satisfies the 2nd option.
8 is divisible by 4 positive numbers and 4 is divisible by 3 numbers. 8 should have 6 positive factors to meet the requirement stated in 2. Am I missing something?

Let O be any odd number. O is prime factorized into a, b and c.

O = a^m x b^n x c^l
Therefore number of factors of O = (m+1)(n+1)(l+1)
2xO = 2^1 x a^m x b^n x c^l
Therefore number of factors of 2O = (1+1)(m+1)(n+1)(l+1); this is twice the number of factors of O.

Now let E be any even number.
E= 2^m x d^n x y^l
Therefore number of factors of E = (m+1)(n+1)(l+1)
// At the minimum m+1 = 2
2xE = 2^(m+1) x d^n x y^l
Therefore number of factors of 2E = (m+2)(n+1)(l+1)

The number of prime factors of 2E will be twice of those of E only if (m+2) = 2(m+1)
This is possible only if m=0. Since E is an even number, m cannot be 0.

Yurik79, I did not understand how your example satisfies the 2nd option. 8 is divisible by 4 positive numbers and 4 is divisible by 3 numbers. 8 should have 6 positive factors to meet the requirement stated in 2. Am I missing something?

my fault))you are absolutely correct thanks for pointing my mistake
yezz I am sorry agree B
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