Last digit of $1234 \cdot 5678$

Question

How to find the last digit of this multiplication $1234\times5678$. It may seem like homework but its not. I can do simple multiplication to get the result but I see some patter here, the digits are continuous is this some hint?

Answer 1

If you think of $1234$ as $1230 + 4$ and $5678$ as $5670 + 8$, then you'll see that $$\begin{align} 1234\cdot5678 &= (1230 + 4) \cdot (5670 + 8) \\ &= \underbrace{1230 \cdot 5670 + 1230 \cdot 8 + 5670 \cdot 4}_{\text{all multiples of } 10} + 4 \cdot 8 \end{align}.$$ The last digit isn't affected by multiples of $10$. So the last digit is completely determined by the last digit of the product of $8$ and $4$, which is $2$.

From a different perspective, if you're familiar with modular arithmetic, then last-digit questions almost beg to work mod 10. Working mod 10, we see that the question immediately boils down to $8 \cdot 4 \bmod 10$, which is $2$.

The underlying reasoning behind these two answers is the same. $\diamondsuit$