Japanese Cryptogram

$\large{\begin{array}{c}&& K&I&M&O&N&O \\ + && O&S&A&K&A\\ + \ \ S&A&S&H&I&M&I \\ \hline =K&A&R&A&O&K&E& \end{array}}$

Terminology : I read the cryptogram from left to right. So the first column refers to the leftmost column.

From the second column, seeing that $K+A$ gives $A$ , an impossibility with no carry over, we can deduce that there must be a carry over of 1 from the third column. Therefore, $K=9$ . The carry over of 1 from the second column to the first column implies that $S=8$ .

Now, take a look at the fifth and seventh column. Both column features $O+A+I$ . We deduce that $O+A+I\geq10$ . With $O+A+I <10$ , $O+A+I$ cannot gives $O$ . Even if we consider a carry over of 1 from the sixth column causing a carry over from fifth column to fourth column, only possible if $E=9$ and $O=0$ , this creates contradictions since $9$ is already represented by $K$ while $O$ cannot be $0$ since $OSAKA$ starts with $O$ . We then observe that $E+1=O$ due to the carry over from the sixth column. There must be a carry over. Or else, $O+A+I$ will just give $E$ , as expected from first column. Possible candidates for $E,O$ (in ordered pairs) are $(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)$ .

A consequence of $E+1=O$ is that $A+I=9$ . This is apparent if we look at the final column.

We now take a look at the fifth column, where $N+9+M$ gives $9$ . Knowing that there is a carry over from the first column, we deduce that $N+M=9$ . Possible candidates for $N,M$ and $A,I$ (in unordered pairs) are $(2,7),(3,6),(4,5)$ .

Observe that it is only possible to choose $(2,7),(3,6)$ or $(3,6),(4,5)$ to represent $(N,M)$ and $(A,I)$ , with the two options giving $(4,5)$ and $(1,2)$ for $(E,O)$ respectively. We cannot choose $(2,7),(4,5)$ since no options will be left for $(E,O)$ .

We now prove that $(3,6),(4,5)$ is tbe correct choice for $(N,M)$ and $(A,I)$ . Suppose that we choose $(2,7),(3,6)$ instead. Note that all numbers are exhausted except $0$ and $1$ . Now, take a look at the third column where $I+O+S$ gives $R$ . Note that the choice of $(0,1)$ for $(I,R)$ (unordered) won't fit (with carry over or not), bearing in mind that $O=5$ and $S=8$ .

After all this hard work, it is pretty easy to deduce that $N,M,A.I$ represents $5,4,3,6$ respectively just by some trial and error (plugging in numbers to see if they fit). After all the hard work we did, this small bit of trivial checking cannot be perceived as hand-wavy. Subsequent work also reveals that $R=7$ and $H=0$ .

Sorry this is so awkward in the future but this is how I found it:

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for k,i,m,o,n,s,a,h,r,e in permutations('0123456789',10):
  if '0' in (k,o,s):
    continue
  if int(k+i+m+o+n+o)+int(o+s+a+k+a)+int(s+a+s+h+i+m+i)==int(k+a+r+a+o+k+e):
    print(o+s+a+k+a)