In how many whole numbers between 100 and 999 is the middle digit equal to the sum of the other two digits?
45
Here's why:
If first digit is k, middle is k+i, last is i.
Max k+i is 9, so i ranges from 0 to 9-k
So, there are
9 numbers starting with 1,
8 starting with 2,
...
1 starting with 9
sum of numbers from 1 to 9 is 9*10/2 = 45
45
45 btw do u like Olivia Rodrigo?
To find the number of whole numbers between 100 and 999 where the middle digit is equal to the sum of the other two digits, we can break down the problem into steps:
Step 1: Determine the range of values for each digit.
- The first digit can be any number from 1 to 9.
- The middle digit can be any number from 0 to 9 since it can be equal to the sum of the other two digits.
- The last digit can be any number from 0 to 9 as well.
Step 2: Calculate the number of possibilities for each digit.
- The first digit has 9 possibilities (1-9) since it cannot be 0.
- The middle digit has 10 possibilities (0-9).
- The last digit also has 10 possibilities (0-9).
Step 3: Calculate the total number of possibilities.
To find the total number of possibilities, we multiply the number of possibilities for each digit.
Total possibilities = Number of possibilities for the first digit * Number of possibilities for the middle digit * Number of possibilities for the last digit
Total possibilities = 9 * 10 * 10 = 900
So, there are 900 whole numbers between 100 and 999 where the middle digit is equal to the sum of the other two digits.