Perfect Cube

$\text{Using TI-83 calculator, found cub root of 888 to be greater than 9. }\\ \text {Let B cub be the number. Since the unit digit of B cub is 8, the unit }\\ \text{digit of B must be 2 or 8.}\\ \text{2 at unit digit can be obtained from unit digit 8 by adding 4.}\\ \text{8 at unit digit can be obtained from unit digit 2 by adding 6.}\\ \text{This can be easily obtained from the following: }\\ x+1~~ store~~ x ~~~and~~ then~~ 5+(-1)^x. \text{Starting with B=8, and x=0,}\\\color{#3D99F6} {x+1~ STR~ x ~~{\huge :}~~~~~ 5+(-1)^x ~~STR~~A~~{\huge :}~~~~~~B+A~~STR~~B ~~{\huge :}~~~~~~B^3 }\\ \text{Repeat till the result has last three digits as 888.}\\ \text {This is when the number is 7077888.}\\````\\$

$\text{However, solution by logic too is possible.} \\ Let ~~n,~ r, B~\in~{\large Z^+}~~such~ that~~ r^3=B ~~is ~the ~required ~~number. \\ \text {To have 8 at unit digit of B, unit digit of r can only be 8 or 2.}\\ \text {But with 8 as unit digit of r, the tenth digit of B will always be odd} \\ \text {and so can not be 8. So only 2 can be at the unit position of r.} \\$ $With~~~(10n+2)^3~~\text{ the digit of B at the tenth place is even and}\\periodic.~~\{10*(5n-1) +2 \}^3 ~~\text {always ends in 88. }\\ \text{So try for n=1,2,3..that is 42, 92, 142, 192, 242....} \\r=42\implies B=74088,~~~r=92\implies B=778688,~~~\\r=142\implies B=2863288,~~~~r=192\implies B=\color{#D61F06}{\large \boxed{ 7077888} }$

Let such a cube be $k^{3}$

$k=(10m+2)$

$1000|(600m^{2}+120m-880)$

$100|(120m-880)$

$m=(10n+4)$ or $m=(10n+9)$

$m=(10n+9), n=1$ . The least solution.