So perfect

$X$ must be divisible by $2, 3$ and $7$ to satisfy the condition given.

We can write $X$ as $X = 2^a\times3^b\times7^c$

We can find the smallest that $a$ must be by writing it as

$\frac{X}{2} = 2^{a-1}\times3^b\times7^c$

$a$ must be divisible by $3$ and $7$ to be a perfect cube and perfect seventh power while $a-1$ must also be divisible by 2 to be a perfect square

Therefore, the smallest that $a$ can be is 21

We had solved for $a$

Now let's find $b$

$\frac{X}{3} = 2^a\times3^{b-1}\times7^c$

In the same way, we find $b$ by knowing that it should be divisible by $2$ and $7$ to be a perfect square and perfect seventh power while $b-1$ should be divisible by $3$ to be a perfect cube

Thus, the smallest that $b$ can be is $28$

We had solved for $a$ and $b$ , now let's solve for $c$

$\frac{X}{7} = 2^a\times3^b\times7^{c-1}$

We already knew how to solve it, so we can easily know that the smallest of $c$ can be is $36$

Now we know that the smallest integer $X$ can be written as $X = 2^{21}\times3^{28}\times7^{36}$

The last digit of $2^{21}=2$ , $3^{28}=1$ , $7^{36}=1$

So, $2\times1\times1 = \boxed{2}$

In order to make $X$ satisfy the condition, we let $X=2^a\times 3^b\times 7^c$ . So, this really simplifies our work because it's time for Chinese Remainder Theorem.

Now, we know that $a$ must be in the form of $2k+1$ , where $k$ is any number, so this implies to

$a\equiv 1\pmod 2$

$a\equiv 0\pmod 3$

$a\equiv 0\pmod 7$

Luckily, $2,3,7$ are pairwise prime. So there must exist a number $a$ that satisfy the congruence relation above. We do it in this way:

$a=\overline{\Big(\frac{2\times3\times7}{2}\Big)}\times {\frac{2\times3\times7}{2}}\times 1+\overline{\Big(\frac{2\times3\times7}{3}\Big)}\times {\frac{2\times3\times7}{3}} \times 0+\overline{\Big(\frac{2\times3\times7}{7}\Big)}\times {\frac{2\times3\times7}{7}}\times 0$

where $\overline{p}$ denotes the multiplicative inverse in mod n. (Multiplicative inverse: Let two numbers be $a,b$ . These numbers are said to be the multiplicative inverse of each other if and only if $ab \equiv 1 \pmod p$ and $a,b > p$ )

Therefore, $a=\overline{21}\times21= 1\times 21= 21$ , because $21 \equiv 1 \pmod2$ and 1 is the multiplicative inverse of 1 in mod 2, since $1\times 1=1\equiv\pmod2$ . Similarly for $b$ , we have

$b=\overline{\Big(\frac{2\times3\times7}{2}\Big)}\times {\frac{2\times3\times7}{2}}\times 0+\overline{\Big(\frac{2\times3\times7}{3}\Big)}\times {\frac{2\times3\times7}{3}} \times 1+\overline{\Big(\frac{2\times3\times7}{7}\Big)}\times {\frac{2\times3\times7}{7}}\times 0$

$\implies b=14\times\overline{14}=14\times 2=28$

since $14\equiv 2\pmod 3$ and also, 2 is the multiplicative inverse of 2 in mod 3, beacuse $2\times 2=4 \equiv 1\pmod 3$ . Also for $c$ ,

$c=\overline{\Big(\frac{2\times3\times7}{2}\Big)}\times {\frac{2\times3\times7}{2}}\times 0+\overline{\Big(\frac{2\times3\times7}{3}\Big)}\times {\frac{2\times3\times7}{3}} \times 0+\overline{\Big(\frac{2\times3\times7}{7}\Big)}\times {\frac{2\times3\times7}{7}}\times 1$

$\implies 6\times \overline 6 = 6\times 6 = 36$

Therefore, the smallest integer that satisfy $X$ is $2^{21}\times 3^{28}\times 7^{36}$ . Then, we reduce it to mod 10, we have our desired answer.

$2^{21}\times 3^{28}\times 7^{36} \equiv 2\times 1\times 1 \equiv \boxed{2} \pmod {10}$

Wow. Such problem. So perfect. Much solution. Here is. Wow. (I'm sorry)

The minimum $X$ is of the form $2^A3^B7^C.$ We need to find conditions for $A,$ $B,$ and $C.$ We can use mods while looking at the three fractions.

Take a look at $\frac{X}{2}.$ $\frac{X}{2}=2^{A-1}3^B7^C=k_1^2.$ We can see that $A-1\equiv0\text{ mod }2\Rightarrow A\equiv1\text{ mod }2.$ Also, $B,C\equiv0\text{ mod }2.$

Now look at $\frac{X}{3}.$ $\frac{X}{3}=2^A3^{B-1}7^C=k_2^3.$ We can see that $B-1\equiv0\text{ mod }3\Rightarrow B\equiv1\text{ mod }3.$ Also, $A,C\equiv0\text{ mod }3.$

Lastly, look at $\frac{X}{7}.$ $\frac{X}{7}=2^A3^B7^{C-1}=k_3^7.$ We can see that $C-1\equiv0\text{ mod }7\Rightarrow C\equiv1\text{ mod }7.$ Also, $A,B\equiv0\text{ mod }7.$

Now we can start solving for values of $A,$ $B,$ and $C.$

Start with $A.$ Since $A\equiv0\text{ mod }3$ and $A\equiv0\text{ mod }7,$ we know that $A\equiv0\text{ mod }21.$ Combining this with the knowledge that $A\equiv1\text{ mod }2,$ we can find the smallest possible $A$ is $21,$ and to generalize, $A\equiv21\text{ mod }42.$

Now go to $B.$ Since $B\equiv0\text{ mod }2$ and $B\equiv0\text{ mod }7,$ we know that $B\equiv0\text{ mod }14.$ Combining this with the knowledge that $B\equiv1\text{ mod }3,$ we can find the smallest possible $B$ is $28,$ and to generalize, $B\equiv28\text{ mod }42.$

Finally, solve for $C.$ Since $C\equiv0\text{ mod }2$ and $B\equiv0\text{ mod }3,$ we know that $C\equiv0\text{ mod }6.$ Combining this with the knowledge that $C\equiv1\text{ mod }7,$ we can fid the smallest possible $C$ is $36,$ and to generalize, $C\equiv36\text{ mod }42.$

So we have found that the smallest possible $X$ is $2^{21}3^{28}7^{36}.$ Now we need to find the last digit of this.

Note that $X=2^{1\text{ mod }4}3^{0\text{ mod }4}7^{0\text{ mod }4}.$ Using the recurrence patterns of the last digits of a number raised to the power, we can find that $X\text{ mod }10\equiv2^13^07^0\text{ mod }10\equiv2\text{ mod }10.$ Therefore, the last digit of $X$ is $\boxed{2}$