GATE EE 2017 Set 2 — Question 63

Let $X$ be the number. The number of digits in any positive integer $N$ is given by $\lfloor \log_{10} N \rfloor + 1$. We need to find the number of digits in $X^3$, which is $\lfloor \log_{10}(X^3) \rfloor + 1 = \lfloor 3\log_{10} X \rfloor + 1$.We are given that $X$ is a 30-digit number. This implies that $10^{29} \le X < 10^{30}$.
Furthermore, we are told that $X$ starts with the digit 4 followed by the digit 7. This provides a more precise range for $X$:$$4.7 \times 10^{29} \le X < 4.8 \times 10^{29}$$Now, let's find the range for $\log_{10} X$:$$\log_{10}(4.7 \times 10^{29}) \le \log_{10} X < \log_{10}(4.8 \times 10^{29})$$$$\log_{10}(4.7) + 29 \le \log_{10} X < \log_{10}(4.8) + 29$$Using a calculator, we find $\log_{10}(4.7) \approx 0.6721$ and $\log_{10}(4.8) \approx 0.6812$.
Substituting these values:$$29 + 0.6721 \le \log_{10} X < 29 + 0.6812$$$$29.6721 \le \log_{10} X < 29.6812$$To find the number of digits in $X^3$, we need to evaluate $\lfloor 3\log_{10} X \rfloor + 1$. Let's find the range for $3\log_{10} X$:$$3 \times 29.6721 \le 3\log_{10} X < 3 \times 29.6812$$$$89.0163 \le 3\log_{10} X < 89.0436$$For any value within this range, the floor (the greatest integer less than or equal to the value) is 89.$$\lfloor 3\log_{10} X \rfloor = 89$$Therefore, the number of digits in $X^3$ is:$$\lfloor 3\log_{10} X \rfloor + 1 = 89 + 1 = 90$$Alternative Method:
We have the range for $X$ as $4.7 \times 10^{29} \le X < 4.8 \times 10^{29}$.
Let's cube this inequality:$$(4.7 \times 10^{29})^3 \le X^3 < (4.8 \times 10^{29})^3$$$$(4.7)^3 \times 10^{87} \le X^3 < (4.8)^3 \times 10^{87}$$Now, we calculate the values of $(4.7)^3$ and $(4.8)^3$:
$(4.7)^3 = 103.823$
$(4.8)^3 = 110.592$
Substituting these back:$$103.823 \times 10^{87} \le X^3 < 110.592 \times 10^{87}$$Let's write these numbers in standard scientific notation:$$1.03823 \times 10^{89} \le X^3 < 1.10592 \times 10^{89}$$A number $N$ has $k$ digits if $10^{k-1} \le N < 10^k$. In our case, $X^3$ is bounded between two numbers that are both greater than or equal to $10^{89}$ and less than $10^{90}$.
Thus, $10^{89} \le X^3 < 10^{90}$.
This shows that $X^3$ is a 90-digit number.