GATE 2017 EE (EE2) – Question 8
X is a 30-digit number beginning with digits 47. How many digits does $X^3$ have?
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Correct answer: (A) 90 digits
Explanation
The leading digits imply $4.7\times10^{29}\le X<4.8\times10^{29}$. Cubing gives a lower bound $103.823\times10^{87}$ and upper bound $110.592\times10^{87}$.
Both lie between $10^{89}$ and $10^{90}$. Every integer in that interval has 90 digits. Therefore **X³ has 90 digits, A**.
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