A three-digit integer has been divided by 9. As a result, the sum of the digits decrease by 9.
How many three-digit numbers possess this property?
To find the numbers that meet this property, we need to consider all three-digit integers, which range from 100 to 999.
Here's one way to approach the problem:
1. Start by considering the first three-digit number, which is 100.
2. Divide 100 by 9 to get 11.11. Since this is not an integer value, we move on.
3. Go to the next three-digit number, which is 101.
4. Divide 101 by 9 to get 11.22. Again, this is not an integer value, so we continue.
5. Repeat this process for all three-digit numbers, incrementing by 1 each time, until we reach 999.
By checking each number individually, we will be able to identify the numbers that meet the given property.
Alternatively, we can also use a mathematical approach to solve this problem:
1. Let the three-digit number be represented by "abc", where a, b, and c are the hundreds, tens, and units digits, respectively.
2. The sum of the digits of the original number is a + b + c.
3. After dividing the number by 9, we have (100a + 10b + c)/9. Since this is an integer, it implies that (100a + 10b + c) is a multiple of 9.
4. According to divisibility rules, a number is divisible by 9 if its digits sum up to a multiple of 9.
5. Therefore, we can express the property as: a + b + c = 9k, where k is a positive integer.
6. We need to find all possible combinations of a, b, and c (ranging from 1 to 9) that satisfy the above equation.
By systematically checking all possible combinations, we can determine the three-digit numbers that meet the given property.