GATE CSE 2017 Set 2 | Question: GA-8
in Quantitative Aptitude recategorized by
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26 votes
26 votes

$X$ 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
in Quantitative Aptitude recategorized by
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5 Comments

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?
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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)
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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.
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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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5 Answers

84 votes
84 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: $A$
edited by

5 Comments

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.

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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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@ best answer 😁

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7 votes
7 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. 

edited by
4 votes
4 votes

Ans (A) 90

0 votes
0 votes

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

0 votes
0 votes

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.

Answer:

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