Any factor between 0 and 1.
No, for 0 to be a factor, the numbers would have to be divisible by 0. You can't divide any number by 0, so it can't be a factor.1 is a factor of every positive number.
It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form. So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers). Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1. Example: n = 72 = 2*2*2*3*3 = (2^3)*(3^2) so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2. When (a, b) = (0, 0) the factor is 1. (a, b) = (1, 0) the factor is 2. (a, b) = (2, 0) the factor is 4. (a, b) = (3, 0) the factor is 8. (a, b) = (0, 1) the factor is 3. (a, b) = (1, 1) the factor is 6. (a, b) = (2, 1) the factor is 12. (a, b) = (3, 1) the factor is 24. (a, b) = (0, 2) the factor is 9. (a, b) = (1, 2) the factor is 18. (a, b) = (2, 2) the factor is 36. (a, b) = (3, 2) the factor is 72.
No. Apart from -1 and 1, no other whole number is a factor of 1.
Since zero has infinitely many factors, it can be said that any pair of numbers (0,n) - where n is an integer - are a factor pair for zero.1 = 1 * 1
Any factor between 0 and 1.
No, for 0 to be a factor, the numbers would have to be divisible by 0. You can't divide any number by 0, so it can't be a factor.1 is a factor of every positive number.
0 has no factors and 1 only has 1 factor which is 1.
It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form. So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers). Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1. Example: n = 72 = 2*2*2*3*3 = (2^3)*(3^2) so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2. When (a, b) = (0, 0) the factor is 1. (a, b) = (1, 0) the factor is 2. (a, b) = (2, 0) the factor is 4. (a, b) = (3, 0) the factor is 8. (a, b) = (0, 1) the factor is 3. (a, b) = (1, 1) the factor is 6. (a, b) = (2, 1) the factor is 12. (a, b) = (3, 1) the factor is 24. (a, b) = (0, 2) the factor is 9. (a, b) = (1, 2) the factor is 18. (a, b) = (2, 2) the factor is 36. (a, b) = (3, 2) the factor is 72.
No. Apart from -1 and 1, no other whole number is a factor of 1.
Since zero has infinitely many factors, it can be said that any pair of numbers (0,n) - where n is an integer - are a factor pair for zero.1 = 1 * 1
0 is not a factor of any number other than itself. If N is any number, than there is no other number P such that N = 0 * P and that means that 0 is not a factor of N. The only number that is a factor of all numbers is 1.
4(x + 1) = 0.
The factor theorem states that for any polynomial function f(x), if f(a) = 0, then (x-a) is a factor of f(x). Let f(x) = x3-2x2-8x-5. If (x+1) is a factor, then f(-1) = 0. (x+1 = x - (-1)) Input x = -1 into f: (-1)3-2(-1)2-8(-1)-5 f(-1) = -1 -2 + 8 - 5 f(-1) = 0. Since f(-1) = 0, (x+1) is a factor of x3-2x2-8x-5. Q.E.D.
for 1 its 0 and 1 for 2 its o,1,2
the greatest common factor is 1. 34 and 5 have only two common factors 0 and 1. 1 is the greatest factor
(2m + 1) and (6m - 1)