Let the number be x. Smallest 10 digit number is 1,000,000,000=1000^3 Further, smallest 10 digit number that starts with 8 (given number starts with 77)is 8,000,000,000=2000^3 So, x lies between 1000 and 2000 and close to 2000 As the last digit is 7, the number x ends with 3. 1993, 1983 are the two numbers need to be tried. 1983 is the required number.
4 comments:
7797729087 cube of 1983
solved through enumeration and reasoning.. any other better way to solve sir??
how to solve this kind of puzzle in simple manner
Let the number be x.
Smallest 10 digit number is 1,000,000,000=1000^3
Further, smallest 10 digit number that starts with 8 (given number starts with 77)is 8,000,000,000=2000^3
So, x lies between 1000 and 2000 and close to 2000
As the last digit is 7, the number x ends with 3. 1993, 1983 are the two numbers need to be tried. 1983 is the required number.
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