Problem 19 · 2018 Math Kangaroo
Hard
Number Theory
place-value
How many digits does the result of the calculation \(\frac{1}{9}\cdot 10^{2018}\cdot(10^{2018}-1)\) have?
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Answer: D — 4036
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Hint 1 of 2
\(\tfrac{1}{9}(10^{2018}-1)\) is the repunit made of 2018 ones.
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Hint 2 of 2
Multiplying by \(10^{2018}\) just appends 2018 zeros.
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Approach: recognize the repunit, then append zeros
- \(10^{2018}-1\) is 2018 nines, so \(\tfrac{1}{9}(10^{2018}-1)\) is the number made of 2018 ones (2018 digits).
- Multiplying by \(10^{2018}\) appends 2018 zeros, adding 2018 more digits.
- Total digits = \(2018 + 2018 = \) 4036.
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