This is my riddle for you, to solve it you need to decipher the letters to digits:
j m b k
+ n h s m
---------
n h b m t
the following rules must apply:
which letter represent which digit?
Solution:
9 5 6 7 + 1 0 8 5 --------- 1 0 6 5 2
That is -
$\begin{array}{rclcrcl}\text j&=&9&~~~~~~~&\text m&=&5\\ \text b&=&6 && \text k&=&7\\ \text n&=&1 && \text h&=&0\\ \text s&=&8 && \text t&=&2\\\end{array}$
Explanation:
n=1 is obvious on inspection.
h=0 is required because it can't be 1, and two digits don't allow carry+h+1>11.
b=m+1 and b+s>=9 because m+h=b and h=0 requires a carry (b cannot equal m)
j=9 is now required (it can't be 8 because b=m+1 and h=0 means no carry, so j cannot be 8).
Trial and error to fit the final digits suffices, since m appears 3 times and constrains things well.
We find with b=m+1 and neither b nor m can be 9, 0, or 1, that m=5 and the rest fall in line.
9567 + 1085 --------- = 10652
Explanation:
It is obvious that n is 1 because you get a carry over from the thousands to the tenths of thousands.
This makes j 8 or 9. Either one of these will make h either 1 or 0. But since 1 is taken, h = 0.
Recap:
jmbk + 10sm --------- = 10bmt
Going on:
In this case, j has to be 9 because 8 will not produce a carry over.
So now we have
9mbk + 10sm --------- = 10bmt
and the remaining digits
2,3,4,5,6,7,8.
b+s is at least 10 (I would say at least 11 since m cannot be 1) and b is m+1.
Stabbing blindly.
b = 8, would make m = 7
and this leads to
978k + 10s7 --------- = 1087t
Doesn't work because s would have to be 8 or 9 which are taken.
Trying the same for b=7 leads to a wrong result also.
Now:
Trying b=6.
956k + 10sm --------- = 1065t
This leads to s = 8.
Trying b=6.
956k + 1085 --------- = 1065t
and the remaining digits are 2,3,4,7. Trying the different combinations we get k=7, t = 2.
