Problem 29 · 2020 Math Kangaroo
Stretch
Number Theory
place-valuedigit-sum
The number \(K = 9999\ldots9\) is formed by n digits 9. What is the sum of the digits of the number \(K^3\)?
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Answer: D — \(18n\)
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Hint 1 of 2
Write K as 10^n − 1 and cube it.
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Hint 2 of 2
Expand and check the digit sum for small n to spot the pattern.
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Approach: expand (10^n - 1)^3 and read the digit sum
- K = 10^n − 1, so K³ = 10^(3n) − 3·10^(2n) + 3·10^n − 1.
- For n = 1, K³ = 729 (digit sum 18); for n = 2, K³ = 970299 (digit sum 36).
- The digit sum is 18n.
- So the answer is 18n.
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