GATE CSE 2017 Set 2 | Question: GA-8

Question

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

• A. $90$ digits
• B. $91$ digits
• C. $92$ digits
• D. $93$ digits

Comments

Whenever we multiply, m (digits) * n (digits) the resultant will be of m+n digits. So n*n*n will be of maximum 3n digits. Hence answer will be 3*30=90digits

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4{777….upto 29digits}  

4{000….upto 29digits} cube is 64{000….upto 87digits} = 89 digits

5{000….upto 29digits} cube is 125{000….upto 87digits} = 90 digits

It can be 89 or 90 digits.

 

For more precise answer take one more digit in to consideration.

47{000….upto 28digits} cube is 103823{000….upto 84digits} = 90 digits

48{000….upto 28digits} cube is 110592{000….upto 84digits} = 90 digits

Answer = 90 digits.

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Wrong Answer give by manish

Answer 1

$X = 4777\dots $ ( $7$ $29$ times)

It can be written as $X = 4.7777\dots *10^{29}$

$X^3 = (4.777\dots*10^{29})^3 = (4.777\dots)^3* 10^{87}$

Now, even if we round up $4.777\dots$ to $5$, we could represent $5^3 = 125$ in $3$ digits. So, We can say $(4.77\dots)^3$ also has $3$ digits before decimal point.

So, $X^3 $ requires $3 + 87 = 90$ digits.

Correct Answer: $A$

Comments

I count manually and I found 89

can anyone help me which case I missed?

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@mcjoshi
it said
X is a 30 digit number starting with the digit 4 followed by the digit 7
=> 47(--28 digits need not be 7--) right? had they been taken as zeroes, the answer still remains same though
Let 47*10^28 be the number then (47*10^28)^3 is (47^3 * 10*84) = 103823 * 10^84 => 90 digits

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Yes you are right..

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We can check for the max case to get the upper bound. 

X³ will be max when all the remaining 28 digits are 9. 

So, 47(99....9_{28 times}) = 4*10²⁹ + 7*10²⁸+ 9*10²⁷ + 9*10²⁶+.....9*10⁰ =40*10²⁸ +7*10²⁸+9(10²⁷+....+10⁰) = 

$47*10^{28} +9 * \frac{10^{28}-1}{10-1} =47*10^{28}+ 9 * \frac{10^{28}-1}{9} =47*10^{28}+10^{28}-1 = 48*10^{28}-1$

Now X³ = (48*10²⁸)³ -3*(48*10²⁸)² + 3*48*10²⁸-1  [(a-b)³= a³ -3a²b+3ab²-b³]

The a³ is the most dominating one.

48³=103823 and 10^28*3=10⁸⁴

So, 103823*10⁸⁴ has 6+84=90 digits.

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No bro, it contains 90 digits exactly

Check it out:-

https://www.calculator.net/big-number-calculator.html?cx=477777777777777777777777777777&cy=3&cp=20&co=pow

X$^{3}$is109,063,100,137,174,211,248,285,322,358,863,799,725,651,577,503,429,355,281,208,000,137,174,211,248,285,322,359,396,433

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Why you are counting manually dude you are just prone to mistake. Simply write a program in minutes to do it for you. Declare an array and initialize with this result and then find its length simply.

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@King Suleiman best answer 😁

Answer 2

Here's a simple approach purely based on observation:

Note that $47^{3}$ is $103823$ (6 digits)

Start with a smaller number, say $4700$ : It's cube will be $103823$ followed by $000000$.
How about $4799$? It's cube will be $110522$ followed by $894400$. Did number of digits increase? No.

Trying with more number of digits after 47, we can see that the first part $47^{3}$ will always give 6 digits and the remaining number of digits after it are getting tripled in number, eg for two digits after $47$ we got six digits in the cube of $4700$ as shown above.

So extending it for $47$ followed by $28$ more digits, we will get $6$ digits for $47^{3}$ and $3*28$ more digits for the number after $47$.

We get total : $6 + 84 = 90$ digits.

Option A. 

Answer 3

Ans (A) 90

Answer 4

Here’s the exact and quick approach:

Simply we know that in scientific notation say

(a)x=4.777777…. *$10^{29}$

(b)10x= 47.77777…. *$10^{29}$

Substracting b-a:

9x=43 *$10^{29}$

x= $\frac{43*10^{29}}{9}$

$x^{3}$= 109.063*$10^{87}$

So it has 3+87=90 digits