Question:
Let \( S(n) \) denote the sum of the digits of a positive integer \(n\). Then the value of \( S(1)+S(2)+\cdots+S(99) \) is
Let \( S(n) \) denote the sum of the digits of a positive integer \(n\). Then the value of \( S(1)+S(2)+\cdots+S(99) \) is
Show Hint
For digit sums, split into place values and count repetitions systematically.
Updated On: May 8, 2026
- \(476\)
- \(998\)
- \(782\)
- \(900\)
- \(855\)
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The Correct Option is D
Solution and Explanation
Concept:
Digit sum from 1 to 99 can be split into:
• Units place contribution
• Tens place contribution
Step 1: Units place contribution
Digits 0–9 repeat 10 times: \[ (0+1+2+\cdots+9) = 45 \] Total: \[ 45 \times 10 = 450 \]
Step 2: Tens place contribution
Digits 0–9 appear each 10 times: \[ 45 \times 10 = 450 \]
Step 3: Total sum
\[ 450 + 450 = 900 \]
Step 4: Note special case
From 1 to 9, tens digit is 0 so already included.
Step 5: Final answer
\[ \boxed{900} \]
• Units place contribution
• Tens place contribution
Step 1: Units place contribution
Digits 0–9 repeat 10 times: \[ (0+1+2+\cdots+9) = 45 \] Total: \[ 45 \times 10 = 450 \]
Step 2: Tens place contribution
Digits 0–9 appear each 10 times: \[ 45 \times 10 = 450 \]
Step 3: Total sum
\[ 450 + 450 = 900 \]
Step 4: Note special case
From 1 to 9, tens digit is 0 so already included.
Step 5: Final answer
\[ \boxed{900} \]
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