Smallest Perfect Cube ending in 888

Find the Smallest Perfect cube ending in 888

My Try: First of all its cube root should have Last digit as $2$.

Now if $x888$ is a Perfect Cube Then $$x888=888+1000x=8(111+125x)$$ which means $111+125x$ should be a Perfect cube.

Need a small Hint to Proceed Further

Solution 1:

You've already figured out the ones digit of the root is $2$.

So, the number $(10x + 2)^3 = 1000x^3 + 600x^2 + 120x + 8$ ends with $...888$.

This number modulo $10$ is obviously $8$. So, you remove the $8$ and divide by $10$. Now you got $100x^3 + 60x^2 + 12x$. This number has to end in $...88$.

Now, $100x^3 + 60x^2 + 12x \equiv 12x \,\, (\textrm{mod}\ 10)$

So, $12x$ has to end in an $8$ and thus, $x$ either ends in $4$ or $9$. So, now we've narrowed down possibilities to $x = 4, 9, 14, 19, 24 ...$ If you check these numbers, you find that $x = 19$ is the answer.

Thus, the smallest perfect cube to end in 888 is 192. (Remember the root is $10x + 2$)

EDIT: If you want to do a bit more work, you can figure out the following:

$60x^2$ will always end in $...60$ for $x$ that ends in $4$ or $9$. (If it isn't immediately obvious, work at it a bit more).

Which means that, $12x$ must end in $...28$ to get the desired $...88$ - this narrows down to ALL numbers whose cubes end in $...888$, which are $19,44,69,94$.

Here, an interesting pattern emerges.

All numbers of the form $250x + 192$ (where $x$ is a non-negative integer) are roots of perfect cubes ending in $...888$.

Solution 2:

From where you left it: $$a^3\equiv 111\equiv -14\pmod{125}$$

This implies: $$a^3\equiv 11\pmod{25}$$

You have $4^3\equiv14\equiv-11\pmod{25}$, so $a\equiv -4\equiv 21\pmod{25}$.

Trying possible values of $a$ (21,46,71,96,141) with a calculator, you can find $96^3\equiv 111\pmod{125}$ and $192^3=7077888$.