All of the digits of a three-digit integer are distinct and non-zero. Furthermore, the three-digit integer is divisible by 17. Find the largest three-digit integer that has these properties.
To make the division easy, we want to find a multiple of $17$ which is close to $900$. Trying some trials, we find that $17\cdot53=901$, so our number is $901$. However, we need our number to have distinct digits. To maintain divisibility by $17$, note that $34$ is a multiple of $17$, so the number $17\cdot((10\cdot3)+4)=906$ is a multiple of $17$ and uses distinct digits. So we must find multiples of $17$ which are less than $906$. Trying some more multiples of $17$, we find that $17\cdot(10\cdot5+1)=17\cdot51=867$, and so $\boxed{867}$ is the largest number which satisfies the condition.
To find the largest three-digit integer that is divisible by 17 where all the digits are distinct and non-zero, we can start by listing the three-digit numbers in descending order and checking if each number meets the given conditions.
First, let's list the three-digit numbers in descending order:
999, 998, 997, ..., 100.
Now, let's check each number to see if it is divisible by 17.
Starting with 999:
999 divided by 17 has a remainder of 14, so it is not divisible by 17.
Moving on to 998, we find that 998 divided by 17 has a remainder of 14 as well.
Continuing with 997, we find that 997 divided by 17 has a remainder of 7.
Proceeding in the same manner, we find that 996 divided by 17 is not divisible evenly, and the process continues until we eventually find a number that is divisible by 17.
Next, we try 986:
986 divided by 17 has a remainder of 15.
Trying 985 next:
985 divided by 17 also has a remainder of 15.
Next, trying 984:
984 divided by 17 leaves a remainder of 15.
We continue this process until we come across a number that is divisible by 17.
Trying the number 952 next:
952 divided by 17 results in a remainder of 0! This means that 952 is divisible by 17.
However, we need to check if all the digits are distinct and non-zero. In this case, the digits are not distinct, as there is a repeated digit "5".
So, we try the next number, 951:
951 divided by 17 has a remainder of 14.
We continue this process until we find a number that meets all the conditions.
Trying 935 next:
935 divided by 17 results in a remainder of 14.
Moving on to 934:
934 divided by 17 has a remainder of 15.
Trying the number 920 next:
920 divided by 17 leaves a remainder of 4.
Next, trying 918:
918 divided by 17 has a remainder of 7.
Finally, we try the number 901:
901 divided by 17 leaves a remainder of 12.
Based on our calculations, there are no three-digit integers that meet all the given conditions. Hence, there is no largest three-digit integer that is divisible by 17 where all the digits are distinct and non-zero. The answer is "None".