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

1. $90$ digits
2. $91$ digits
3. $92$ digits
4. $93$ digits

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?
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)
Reduce the problem to small size say 3 digits number being 479 ....

take its cube ..u will find 9 digits in the answer hence it came out that we should just multiply the number of digits with 3.
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

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

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$X = 4777\dots$ ( $7$ $29$ 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: $A$
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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....928 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 1028*3=1084

So, 103823*1084 has 6+84=90 digits.

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

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.

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.

Ans (A) 90

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Here’s another take

47^3 = 6 digits (3*2)

477^3 = 9 digits (3*3)

4777^3 = 12 digits (3*4)

The pattern is 3 multiplied by the number of digits in the base

=> for 30 digits it will be 3 * 30=90

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The product of numbers with m and n digits has either m+n−1 or m+n digits.

https://math.stackexchange.com/questions/1295186/my-proof-that-an-n-digit-number-times-an-n-digit-number-can-be-expressed-as-a-2

So, X^3 can have atmost 90 digits, and there is no option less than 90 digits.

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