A three-digit number has all digits nonzero. These digits are multiplied together and then this product is subtracted from the original number.What is the largest possible value of this difference?

To find the largest possible value of the difference, we need to maximize both the original number and minimize the product of its digits.

Step 1: Maximizing the original number The largest three-digit number is 999.

Step 2: Minimizing the product of its digits The smallest product of three non-zero digits is 1, which occurs when all the digits are 1. However, 111 is not the largest three-digit number.

To get the largest number with the smallest product, we need to have a 1 in the hundreds place and 0 in the tens and units place. But the problem states that all digits are non-zero. So, the next smallest product would be 2 (112), and the largest such number is 912.

Step 3: Subtract the product of the digits from the original number 912 - (192) = 912 - 18 = 894

So, the largest possible value of the difference is 894.

To solve this problem, we need to find a three-digit number that maximizes the difference between the number itself and the product of its digits.

Step 1: Understand the problem We are looking for a three-digit number where all digits are non-zero. This means our number is between 100 and 999. We need to find the number that gives the maximum difference when we subtract the product of its digits from the number itself.

Step 2: Formulate a strategy Since we want to maximize the difference, we want to maximize the number itself and minimize the product of its digits. The maximum three-digit number is 999, but this doesn't minimize the product of its digits (999=729).

Step 3: Implement the strategy We can start by trying to minimize the product of the digits. The smallest non-zero digit is 1, so we want as many 1s as possible. However, we also want to maximize the number itself, so we want the 1s to be in the least significant places. This suggests that our number should be in the form of 9x1 or 8x1, where x is a digit between 1 and 9.

Step 4: Test the strategy Let's test the numbers 911, 811, 821, 831, ..., 891. We find that 911 gives a difference of 911 - (911) = 902, 821 gives a difference of 821 - (821) = 805, and so on.

Step 5: Find the maximum difference After testing all the numbers, we find that 911 gives the maximum difference of 902. So, the largest possible value of the difference is 902.

To solve this problem, we need to find a three-digit number that maximizes the difference between the number itself and the product of its digits.

Step 1: Understand that the largest three-digit number is 999. However, the product of its digits is 999=729, which is not the largest possible product of three digits (which is 987=504).

Step 2: Realize that to maximize the difference, we need to maximize the number itself while also minimizing the product of its digits.

Step 3: The largest three-digit number with the smallest possible product of its digits is 500 (500=0).

Step 4: Subtract the product of the digits from the number itself to find the difference: 500 - 0 = 500.

Therefore, the largest possible value of this difference is 500.

To find the largest possible value of the difference, we need to maximize both the original number and minimize the product of its digits.

Step 1: Maximizing the original number The largest three-digit number is 999.

Step 2: Minimizing the product of its digits The smallest product of three non-zero digits is 111 = 1. However, 111 is not larger than 999.

So, we need to find the smallest product of three non-zero digits that can be part of a number that is less than 999. This happens with 121 = 2, in the number 912.

Step 3: Subtract the product from the original number 912 - 2 = 910

So, the largest possible value of this difference is 910.