GATE CS 2017 Set 2 QGA-8 — General Aptitude

We are given that $X$ is a $30$-digit number starting with the digit $4$ followed by $7$. We can write the range of $X$ using scientific notation:

To find the number of digits in $X^3$, we cube the entire inequality:

Let us calculate the bounds:

• Lower bound: $$(4.7)^3 \times 10^{87} = 103.823 \times 10^{87} = 1.03823 \times 10^{89}$$
• Upper bound: $$(4.8)^3 \times 10^{87} = 110.592 \times 10^{87} = 1.10592 \times 10^{89}$$

Combining these, we get:

Since $10^{89} \le X^3 < 10^{90}$, the value of $X^3$ lies strictly within the range of numbers that have exactly $90$ digits (as $10^{89}$ is the smallest $90$-digit number and $10^{90}$ is the smallest $91$-digit number).

Thus, $X^3$ has 90 digits.