GATE CS 2017 Set 2 Q.63 (GA Q.8): Quantitative Aptitude

$X$ is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number $X^3$ will have

90 digits

91 digits

92 digits

93 digits

Official answer and solution

Official answer: A
From: Official GATE 2017 answer key

Solution

Determine the lower and upper bounds of the 30-digit number $X$ in scientific notation and compute the bounds for $X^3$ .

A number $N$ with $d$ digits satisfies $10^{d-1} \le N < 10^d$ .
We are given that $X$ is a 30-digit number whose first two digits are 4 and 7.
Therefore, $X$ satisfies the tight bounds:

4.7 \times 10^{29} \le X < 4.8 \times 10^{29}

Cube both bounds:

X^3 \ge (4.7)^3 \times (10^{29})^3 = (4.7)^3 \times 10^{87}

X^3 < (4.8)^3 \times (10^{29})^3 = (4.8)^3 \times 10^{87}

Compute the cubes of the leading digits:

(4.7)^3 = 4.7 \times 4.7 \times 4.7 = 22.09 \times 4.7 = 103.823

(4.8)^3 = 4.8 \times 4.8 \times 4.8 = 23.04 \times 4.8 = 110.592

Express in standard scientific notation:

1.03823 \times 10^2 \times 10^{87} \le X^3 < 1.10592 \times 10^2 \times 10^{87}

1.03823 \times 10^{89} \le X^3 < 1.10592 \times 10^{89}

Since $10^{89} \le X^3 < 10^{90}$ , the number $X^3$ has exactly $89 + 1 = 90$ digits.

Final answer

Common trap

Estimating $X \approx 10^{29} \implies X^3 \approx 10^{87}$ (88 digits), forgetting that $4.7^3 > 100 = 10^2$ , which adds 2 additional digits.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 63 of 65, printed as GA Q.8
Section: General Aptitude

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