$a$ and $a^3$

Number Theory Level 3

Does there exist a natural number $a$ , such that the sum of the number of digits of $a$ and $a^3$ is equal to $2017$ ?

Yes No

1 solution

Ahmed Abdelaal
Nov 9, 2017

1^3=1 , 9^3=729

10^3=1000 , 99^3=970299

100^3=1000000 , 999^3 = 997002999

let the number of digits of a = n then the number of digits of a^3 = 3n-2 or 3n-1 or 3n

number of digits of a + number of digits of a^3 =4n-2 or 4n-1 or 4n

therefore n=2017/4 or n=2018/4 or n=2019/4 none of which is valid because n must be an integer

So... I don't really get how this proves its impossible to have digits sum op to 2017?

Peter van der Linden - 3 years, 7 months ago

not digits. it's number of digits

Ahmed Abdelaal - 3 years, 7 months ago

aaah ok, then I didn't read the question right.

Peter van der Linden - 3 years, 7 months ago