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

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

Answer 2

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.