I would divide 110 by 1, 2, 5 and 10.
1, 2, 5, 10, 11, 22, 55, 110
Find the prime factorisation of 110.Find all possible combinations of these. Done!
1, 2, 5, 10, 11, 22, 55, 110 are all I can find at the moment.
The prime factors of 110 are 2, 5 and 11.
110 has eight factors: 1 2 5 10 11 22 55 110.
The factors of 110 are: 1, 2, 5, 10, 11, 22, 55, 110.
The prime factorisation of 110 is 2*5*11.Therefore all factors of 110 are 2p*5q*11r where each of p, q and r can be 0 or 1.
For any number x, you can multiply any of it's factors by x/factor to get x. For 110, 11 is a factor, and 110/11 is 10. So 10 * 11 = 110. In the case of 110, all of the factors are as follows: 1, 2, 5, 22, 55, 110 There is a pattern here; the first and last numbers multiplied together will give you the value 110, and then the 2nd numbers from the front and back will multiply to get 110, and so on. This is true for all integers. If you find the prime factorization of the number, you can determine all the possible combinations of multiplying numbers to get the original number, that is to say, where you can multiply 3, 4 or more numbers together to get the original number. For example, the prime factorization of 110 is as follows: 2, 5, 11. So you can see that 2*5*11=110 as well.
The prime factorisation of 110 is 2*5*11.Therefore all factors of 110 are 2p*5q*11r where each of p, q and r can be 0 or 1.
2, 5, 11
1, 2, 5, 10, 11, 22, 55, 110
All Factors of 110:1, 2, 5, 10, 11, 22, 55, 110
Once all the prime factors of a number have been found, the number of factors the number has and what they are can be found. I'd be finding the prime factors first before finding all the factors of a number, so I'd rather find all the prime factors as it means I can stop before I have to do more work in finding all the factors.