Math Problem Statement
What is the sum of all digits which appear in all the integers from 10 to 100?
Solution
To find the sum of all digits that appear in the integers from 10 to 100, we can break it down by looking at the tens digits and the ones digits separately.
Step 1: Analyze the tens digits
The numbers from 10 to 99 consist of two digits. For the tens digit, it cycles through 1 to 9. Each digit from 1 to 9 will appear 10 times because every set of 10 consecutive numbers (10-19, 20-29, etc.) shares the same tens digit.
So, the total sum of the tens digits is:
Step 2: Analyze the ones digits
For the ones digit, the numbers repeat the sequence 0, 1, 2, ..., 9 across each set of 10 numbers (10-19, 20-29, ..., 90-99). Therefore, each digit from 0 to 9 appears exactly 9 times as a ones digit.
So, the total sum of the ones digits is:
Step 3: Total sum
Now, we add the sum of the tens digits and the sum of the ones digits:
Thus, the sum of all digits which appear in the integers from 10 to 99 is 855.
Let me know if you'd like more details or have any questions!
Here are five related questions:
- How would the sum change if we considered numbers from 100 to 199?
- What is the sum of all digits appearing in the numbers from 1 to 50?
- How would the method change if we included numbers from 0 to 100?
- How can this problem be generalized to any range of numbers?
- What is the sum of all digits in the numbers from 200 to 300?
Tip: For problems like this, breaking the digits down by place value simplifies the calculations.
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Math Problem Analysis
Mathematical Concepts
Arithmetic
Place Value
Summation
Formulas
Sum of tens digits: (1 + 2 + 3 + ... + 9) × 10 = 45 × 10 = 450
Sum of ones digits: (0 + 1 + 2 + ... + 9) × 9 = 45 × 9 = 405
Theorems
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Suitable Grade Level
Grades 5-7