Five is such a distraction

Here , we can observe that unit digit of $\color{#D61F06}{n^{5}}$ is always $\color{#D61F06}{n}$ itself.

Therefore , the sum to find the unit digit of the question can be written as follows :

$\Rightarrow$ $1+2+3+4+ \ldots +123$

$\Rightarrow$ $Sum$ $=\frac{123×124}{2}$

$\Rightarrow Unit$ $digit =$ $\color{#3D99F6}{\huge \boxed{6}}$

1^5 = 1 2^5= 32 3^5=243 u can notice that units digit of k ^5 will always be k (e.g. 19^5=?,???,??9) and so on. then , for numbers from 1 ^ 5 to 10 ^ 5 unit digit will be : 1, 2,3,4,5,6,7,8,9,0. there summation = 45

for numbers from 1^5 to 120^5 summation will be = 45 *12= 540 units digits will be zero, now we have 121^5 +122^5+123^5 which there unit digits will be 1+2+3 = 6.

That's it

To make it more easier, one can also go with this:

There are two parts- One is explained above i.e. n^5 has its units place as 'n' itself.
The other is the summation of any consecutive 20 numbers shall have its unit's place as '0'.

So, applying these parts: (1) We have 1+2+3+......+123. (2) We will have sum of 1 to 20 ending up with units place as 0. Similarly, sum of 21 to 40 ending up with units place as 0. So this goes on till 120. We are left with 121, 122 and 123, whose units place, when added up gives us 6.

So, 6 is the answer.

The unit digit of abc....k^5 is always k.

Thus the unit digit is: 1 + 2 + 3 + ..... + 123 => 6