Numeration a numeral with 2 variables

Question

could you help me ?

If $a,b,c$ in $\mathbb{N}$ such that

$(\overline{ab})^{3}=\overline{1c8ab}$

then the value of $2b-a-c$ is ?

Thanks!

Answer 1

Well, we can start by noting that the cubic number on the right-hand side has five digits and starts with 1; by inspection $22\le\overline{ab}\le27$. By inspecting these, we find that only 24 has a cube with the 8 in the hundreds place: $24^3=13824$. Hence $a=2,b=4,c=3$ and $2b-a-c=3$.

Answer 2

Assuming that $a, b, c$ are actually digits (rather than any arbitrary natural number), we can proceed as follows:

The last digit of any natural number $n$ matches the last digit of $n^3$ unless that last digit is one of $2, 3, 7, 8$. The fact that the cube has five digits, the first of which is $1$, indicates that $a$ must be $2$, since $20^3 = 8000$ and $30^3 = 27000$. That leaves the possibilities $20, 21, 24, 25, 26, 29$, which can be checked manually.

Answer 3

This solution hopes to minimize cubing and cube squaring:

Let's call $n=\overline{ab}$.

Note that $n^3-n$ ends with two zeros.

That is, $100$ divides $(n-1)n(n+1)$. Then, since there can't be two multiples of $5$ among three consecutive numbers, one of this three numbers is a multiple of $25$

Since $10000<n^3<20000$ we have $20<10\sqrt[3]{10}\le n\le 10\sqrt[3]{20}<30$. Therefore, $n$ is $24$, $25$ or $26$. But $25\cdot26\cdot 27$ is not a multiple of $100$, so $n$ is $24$ or $25$. Since every power of $5$ greater than $25$ ends with $125$ or $625$, it must be $n=24$.

Sadly we must cube $24$ to verify that the hundreds digit of $24^3$ is $8$ and to obtain $c$.