How many 3-digit numbers have the units digit larger than the tens digit?
To find the number of 3-digit numbers with the units digit larger than the tens digit, we need to consider the possible values for each digit.
Let's break it down step by step:
Step 1: Consider the possible values for the hundreds digit.
Since we are looking for 3-digit numbers, the hundreds digit cannot be zero, so it can take values from 1 to 9.
Step 2: Consider the possible values for the tens digit.
For the tens digit to be smaller than the units digit, it can only take values between 0 and the selected value for the hundreds digit. For example, if the hundreds digit is 3, then the tens digit can take values from 0 to 3.
Step 3: Consider the possible values for the units digit.
Similarly, the units digit can take any value between 0 and 9, as there are no restrictions on its value.
Step 4: Calculate the total number of numbers.
To find the total number of possibilities, we need to consider all the combinations of the digits we obtained in steps 1, 2, and 3. We can multiply the number of possibilities for each digit together to get the total number of 3-digit numbers.
Now, let's calculate it:
Step 1: There are 9 possible values for the hundreds digit (1 to 9).
Step 2: For each chosen hundreds digit, we have that many possible values for the tens digit.
So, for the tens digit, we have 0, 1, 2, ..., (value of hundreds digit).
Step 3: There are 10 possible values for the units digit (0 to 9).
Step 4: Multiply the possibilities for each digit together:
Total number of 3-digit numbers = Possible values for hundreds digit × Possible values for tens digit × Possible values for units digit
Total number of 3-digit numbers = 9 × (0 + 1 + 2 + ... + 9) × 10
To simplify this, we can use the sum formula for consecutive integers:
Sum of consecutive integers from 0 to n = (n × (n + 1)) / 2
So the calculation becomes:
Total number of 3-digit numbers = 9 × [(9 × (9 + 1)) / 2] × 10
Total number of 3-digit numbers = 9 × (9 × 10) × 10 / 2
Total number of 3-digit numbers = 9 × 9 × 10 × 10 / 2
Now we can solve this expression:
Total number of 3-digit numbers = 9 × 9 × 10 × 10 / 2
Total number of 3-digit numbers = 81 × 100 / 2
Total number of 3-digit numbers = 8100 / 2
Total number of 3-digit numbers = 4050
Therefore, there are 4050 3-digit numbers with the units digit larger than the tens digit.
Take a look in each hundred numbers starting with the digit x:
x10 x20 ... x90 - 9 of them
x21 x31 ... x91 - 8 of them
...
x98 - 1 of them
9+8+...+1 = 45
So, in each hundred numbers there are 45 that have the tens greater than the ones.