Find the greatest ten digit positive multiple of 12 using each digit once and only once .
To find the greatest ten-digit positive multiple of 12 using each digit once and only once, we can start by recognizing that any number divisible by 12 must be divisible by both 3 and 4.
Since the sum of the digits of any number divisible by 3 is also divisible by 3, we can quickly determine that the sum of the digits 0 to 9 is 45 (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45).
Therefore, the sum of the remaining two digits in our ten-digit number must be divisible by 3. To maximize the number, we want the remaining two digits to be as large as possible.
To find the largest possible two-digit number divisible by 3, we can consider the multiples of 3 from 99 downwards. We find that 99 is the largest two-digit number divisible by 3.
Now, let's arrange the remaining eight digits (0 to 9 excluding 9 and 9) in descending order, which gives us:
8, 7, 6, 5, 4, 3, 2, 1, 0
To form the largest possible ten-digit number, we can place 99 at the beginning of the number, followed by the remaining eight digits in descending order.
So, the greatest ten-digit positive multiple of 12 using each digit once and only once is:
99, 8, 7, 6, 5, 4, 3, 2, 1, 0