answer:For a number to be divisible by 9, the sum of its digits must be divisible by 9. If the number is of the above form, then the sum of its digits is a+a+a+a+a+a+a+a+a=9a. Thus, whatever digit a is, the sum of its digits will always be divisible by 9. Thus, to minimize the 9 digit number, the number is 111, ! 111, ! 111.

answer:Since 16 and 20 have a common factor of 4, we can simplify [ frac{16}{20}=frac{4cdot 4}{5cdot 4}=frac{4cdot cancel{4}}{5cdot cancel{4}}=frac{4}{5}. ]

answer:I know that 2000 is equal to 2^4 times 5^3, so its prime factors are 2 and 5. The greatest prime factor is 5 and the least prime factor is 2. The positive difference between them is 5 - 2, which is 3.

answer:I need to find a number that is both a multiple of 6 and a multiple of 8. That means it has to be a common multiple of both numbers. I can use the least common multiple (LCM) method to find the smallest common multiple. To find the LCM of 6 and 8, I can factor both numbers into their prime factors. 6 = 2 times 3 and 8 = 2 times 2 times 2. The LCM is the product of the highest power of each prime factor that appears in either factorization. That is, the LCM is 2^3 times 3 = 8 times 3 = 24. Therefore, the smallest positive multiple of 6 that is also a multiple of 8 is 24.