How many whole numbers from 99 to 131 contain the digit 1 exactly once?
I'm sure you can count from 99 to 131.
I wouldn't want to do that. It wouldn't work out. Are there fast ways to do the problem? Or do I just have to count?
By the time you posted this you could be through the counting.
by the time you typed out your reply you could have easily counted from 99 to 131 and back again.
um no
To find the number of whole numbers from 99 to 131 that contain the digit 1 exactly once, we need to check each number in this range and count the numbers that meet the given condition.
We start by iterating through the range from 99 to 131, inclusive. For each number, we need to check if it contains the digit 1 exactly once. To do this, we can convert each number to a string and count the occurrences of the digit 1. If the count is equal to 1, we increment a counter variable.
Here's how we can solve it step by step:
1. Initialize a counter variable to zero, let's call it `count`.
2. Iterate through the range from 99 to 131 (inclusive).
3. For each number, convert it to a string.
4. Count the occurrences of the digit 1 in the string representation of the number.
5. If the count is equal to 1, increment the `count` variable.
6. After the loop finishes, `count` will contain the number of whole numbers from 99 to 131 that contain the digit 1 exactly once.
Let's implement this in Python:
```python
count = 0
for num in range(99, 132):
if str(num).count("1") == 1:
count += 1
print(count)
```
By running this code, you will get the answer to your question.