it is not mentioned that 7 is repeating upto end, the number could be 4799999....(28 times) or 470000.....(28 times) which may give different answer.Am i right?

doubt 2:

can we solve this problem using scientific calculator in exam?

23 votes

$X$ 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

0

doubt 1:

it is not mentioned that 7 is repeating upto end, the number could be 4799999....(28 times) or 470000.....(28 times) which may give different answer.Am i right?

doubt 2:

can we solve this problem using scientific calculator in exam?

it is not mentioned that 7 is repeating upto end, the number could be 4799999....(28 times) or 470000.....(28 times) which may give different answer.Am i right?

doubt 2:

can we solve this problem using scientific calculator in exam?

3

Yes, I agree with your first point that it could be anything, but the answer in any case will be same (you can try it out)

77 votes

Best answer

$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: $A$

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: $A$

10

@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

4

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

X^{3} will be max when all the remaining 28 digits are 9.

So, 47(99....9_{28 times}) = 4*10^{29} + 7*10^{28}+ 9*10^{27} + 9*10^{26}+.....9*10^{0} =40*10^{28} +7*10^{28}+9(10^{27}+....+10^{0}) =

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

Now X^{3} = (48*10^{28})^{3} -3*(48*10^{28})^{2} + 3*48*10^{28}-1 [(a-b)^{3}= a^{3} -3a^{2}b+3ab^{2}-b^{3}]

The a^{3} is the most dominating one.

48^{3}=103823 and 10^{28*3}=10^{84}

So, 103823*10^{84} has 6+84=90 digits.

6 votes

**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.**