How do I reconstruct the following exact long division in which the digits that have been replaced by letters except for the middle digit - the '8' - in the quotient. I've never seen one of these before and I am trying to solve gps coordinates using this problem. GPS coordinates are going to be: N 38 4A.BC(D+E) W 121 0F.GH(C+G) ...so need to solve the problem so I can fill in A,B,C,D,E,F,G,H. Can someone please help. I've been banging my head against the wall for 2 days trying to figure where to start. Hopefully the JPG division problem posts below from an image upload.
I ran a computer program and I was wondering why it didn't find a solution. It's because some numbers presented as 4-digit number are in fact 3-digit numbers.
Edit: WAIT! HOLD ON!
I made a small error. I double checked and there are 124 solutions that correctly result in a long division without requiring there be 3-digit numbers represented by a 4-digit number.
Edit 2: I iterated over the big number (XXBDXXXX), which was stupid. If I iterate over AXC and FX8XX, I find only the above solution (124 * 80809) run it
First I will replace the $X$'s with different lowercase letters, so its easier to refer to them.
The first thing we can notice is, that there are no subtraction steps for $a$ and $b$. This means they are both $0$.
Next look at the first subtraction step: $efBD - kmJ = Gn$. The only possible value for $k$ is $9$. If it was lower, it wouldn't be possible to get a 4 digit number by adding $kmJ$ and $Gn$. If $k$ is $9$, then $ef$ must be $10$. The same applies for the second subtraction step: $Gngh - opH = qr$.
Now look at the product: $AdC * 8 = 9pH$. It's obvious that $A$ must be $1$. Otherwise the product would be too big to fit in 3 digits. Knowing this we can also deduce that in the third product ($1dC * c = stuE$), the letter $s$ is $1$ as well.
Look again at the product $1dC * 8= 9pH$. We can see that $1dC$ must be in the range $113..124$ to get a 3 digit product starting with $9$. Knowing that range and looking at the second product $1dC * F = 9ph$ we can see that $F$ must be $8$ (because $113*9=1017$ and $124*7=868$). This also means that $9mJ$ and $9pH$ are equal.
Knowing the range for $1dC$ and looking at the product $1dC * c = 1tuE$ we can also deduce, that $c$ must be $9$ (because $124*8=992$). As $123 * 80809 = 9939507$ we know that $1dC$ must be bigger than $123$, and there is only $124$ left in the possible range.
Now we can determine the result for the third product $124 * 9 = 1116$. We also know the values for $BDg$ because the dividend must be in the range $10020316$ (assuming remainder $0$) to $10020439$ (assuming remainder 123).
Using $1002 - 9mJ = 10$ we can determine that $mJ$ is $92$.
We know that $h$ is either $3$ or $4$, so $q$ must be $1$ giving the solution at the top of the answer. This is the last digit which I could find without assuming a reminder.
Dividing the puzzle up into sections:
Some reasoning (more to come when I'm back home with a scanner):
Assuming no leading zeroes:
The rest is trivial-ish: