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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

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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?
+2
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)
0
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.

$X = 4777\dots$ ( $7$ $29$ 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: $A$
by Boss (28.7k points)
edited
0
I count manually and I found 89

can anyone help me which case I missed?
+8

@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

0
Yes you are right..
+2

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.

0

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

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.

by Active (3.9k points)
edited

Ans (A) 90

by Active (1.6k points)
ans is A.
by Active (2.6k points)