110 11 10 2 5
11
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.
LCM: 5720 GCD: 11
i think it is 10 times 11 but it said it was wrong for me on hegarty maths
The answer is 10.
11 x 10 = 110
11 x 10 = 110
11 and 9 is a factor pair for 99. 11 and x^9 is a factor pair for 11x^9.
11 x 10 = 110
11 and 12 are a factor pair of 132.
The GREATEST common factor is 11.
110 11 10 2 5
Since 22 is a factor of 110, it is automatically the GCF of this problem.
11 and 110 are not relatively prime because they have 11 as a common factor.
Factor the numbers into primes: 55 = 5 * 11 143 = 13 * 11 110 = 2 * 5 * 11 The ONLY factor they all have in common is 11. 11 is the greatest (and only) common factor.
The greatest common factor of 88 110 and 143 is 11.