Problem 15 · AMC 8 Stretch
Core
Number Theory
Counting & ProbabilityArithmetic & Operations
symmetryorganizing-datalogical-reasoning
Find the digit-sum of every number from 1 to 999, then add all those digit-sums together. (The digit-sum of 254 is \(2 + 5 + 4 = 11\).) What is the grand total?
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Answer: 13,500
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Hint 1 of 4
Don't add number by number. Count how many times each digit 1, 2, ..., 9 appears in total across 1 to 999. (Zeros add nothing.)
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Hint 2 of 4
By symmetry, every nonzero digit appears the exact same number of times. So just count how often, say, the digit 3 shows up.
Still stuck? Show hint 3 →
Hint 3 of 4
Count the digit 3 in the ones place, the tens place, and the hundreds place separately. In each place it appears 100 times.
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Approach: Count digit appearances by symmetry
- Adding digit-sums is the same as counting how often each digit appears, weighted by its value. Zeros add nothing, so only digits 1 through 9 matter, and by symmetry each appears equally often.
- Count the digit 3 across 1 to 999 (think 000 to 999, three places): ones place 100 times, tens place 100 times, hundreds place 100 times — so 300 times total.
- Every nonzero digit likewise appears 300 times, so total \(= 300 (1 + 2 + \cdots + 9) = 300 \times 45 = 13{,}500\).
- The grand total is 13,500.
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