# Reconstructing long division problems

How do I reconstruct the following exact long division in which the digits that have been replaced by letters except for the middle digit - the '8' - in the quotient.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.
So I ran it again and it found 2350 solutions!
20 solutions have XXBDXXXX mod AXC = 0. (The problem didn't say it had to be 0)
Here's all 20 solutions with mod 0:

124 * 80800 = 10019200 (A,B,C,D,E,F,G,H = 1,0,4,1,0,8,0,2)
124 * 80801 = 10019324 (A,B,C,D,E,F,G,H = 1,0,4,1,4,8,0,2)
124 * 80802 = 10019448 (A,B,C,D,E,F,G,H = 1,0,4,1,8,8,0,2)
124 * 80803 = 10019572 (A,B,C,D,E,F,G,H = 1,0,4,1,2,8,0,2)
124 * 80804 = 10019696 (A,B,C,D,E,F,G,H = 1,0,4,1,6,8,0,2)
124 * 80805 = 10019820 (A,B,C,D,E,F,G,H = 1,0,4,1,0,8,0,2)
124 * 80806 = 10019944 (A,B,C,D,E,F,G,H = 1,0,4,1,4,8,0,2)
124 * 80807 = 10020068 (A,B,C,D,E,F,G,H = 1,0,4,2,8,8,1,2)
124 * 80808 = 10020192 (A,B,C,D,E,F,G,H = 1,0,4,2,2,8,1,2)
124 * 80809 = 10020316 (A,B,C,D,E,F,G,H = 1,0,4,2,6,8,1,2)
111 * 90800 = 10078800 (A,B,C,D,E,F,G,H = 1,0,1,7,0,9,0,8)
111 * 90801 = 10078911 (A,B,C,D,E,F,G,H = 1,0,1,7,1,9,0,8)
111 * 90802 = 10079022 (A,B,C,D,E,F,G,H = 1,0,1,7,2,9,0,8)
111 * 90803 = 10079133 (A,B,C,D,E,F,G,H = 1,0,1,7,3,9,0,8)
111 * 90804 = 10079244 (A,B,C,D,E,F,G,H = 1,0,1,7,4,9,0,8)
111 * 90805 = 10079355 (A,B,C,D,E,F,G,H = 1,0,1,7,5,9,0,8)
111 * 90806 = 10079466 (A,B,C,D,E,F,G,H = 1,0,1,7,6,9,0,8)
111 * 90807 = 10079577 (A,B,C,D,E,F,G,H = 1,0,1,7,7,9,0,8)
111 * 90808 = 10079688 (A,B,C,D,E,F,G,H = 1,0,1,7,8,9,0,8)
111 * 90809 = 10079799 (A,B,C,D,E,F,G,H = 1,0,1,7,9,9,0,8)


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.
Only 1 of those solutions has XXBDXXXX mod AXC = 0:
124 * 80809 = 10020316 (A,B,C,D,E,F,G,H = 1,0,4,2,6,8,1,2)
In all 124 solutions, the A..H are the same.

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

• Yay, my solution is valid! Feb 19, 2016 at 12:45
• @LogicianWithAHat And you did it without a computer. Kudos! I guess the problem is presented wrong. It only has solutions if numbers have leading zeroes and it gives the impression there should only be 1 solution. Feb 19, 2016 at 12:47
• Boo, my solution's less valid :( There's a 'J' value that needs outputting too Feb 19, 2016 at 13:02
• Well, the J isn't used anywhere, so I ignored it. But it's 2 Feb 19, 2016 at 13:03
• Thank you Logician, the only way for me to know for sure if this is correct is for me to plug the coordinates into my GPSr and go to the location and once there I should be able to find a container with a log book in it within +/-30 feet of the location. It's a treasure hunt. Wll let you know what I find Feb 19, 2016 at 19:56

There is a unique solution for the relevant part of the division:

              8 0 8 0 9      +----------------1 2 4 | 1 0 0 2 0 X X X          9 9 2        -------            1 0 0 X              9 9 2            -------                1 X X X                1 1 1 6                -------
The remaining numbers cannot be found, because the remainder of the division is not specified. However, they are not required to find the values for interesting letters:
A=1, B=0, C=4, D=2, E=6, F=8, G=1, H=2, J=2
The final coordinates point to a house in Folsom, CA.

# Explanation

First I will replace the $X$'s with different lowercase letters, so its easier to refer to them.

              F a 8 b c
+----------------
A d C | e f B D g h i j
k m J
-------
G n g h
o p H
-------
q r i j
s t u E
-------


The first thing we can notice is, that there are no subtraction steps for $a$ and $b$. This means they are both $0$.

              F 0 8 0 c
+----------------
A d C | e f B D g h i j
k m J
-------
G n g h
o p H
-------
q r i j
s t u E
-------


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: $Gngh - opH = qr$.

              F 0 8 0 c
+----------------
A d C | 1 0 B D g h i j
9 m J
-------
1 0 g h
9 p H
-------
q r i j
s t u E
-------


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.

              F 0 8 0 c
+----------------
1 d C | 1 0 B D g h i j
9 m J
-------
1 0 g h
9 p H
-------
q r i j
1 t u E
-------


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.

              8 0 8 0 c
+----------------
1 d C | 1 0 B D g h i j
9 m J
-------
1 0 g h
9 m J
-------
q r i j
1 t u E
-------


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.

              8 0 8 0 9
+----------------
1 2 4 | 1 0 B D g h i j
9 m J
-------
1 0 g h
9 m J
-------
q r i j
1 t u E
-------


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).

              8 0 8 0 9
+----------------
1 2 4 | 1 0 0 2 0 h i j
9 m J
-------
1 0 g h
9 m J
-------
q r i j
1 1 1 6
-------


Using $1002 - 9mJ = 10$ we can determine that $mJ$ is $92$.

              8 0 8 0 9
+----------------
1 2 4 | 1 0 0 2 0 h i j
9 9 2
-------
1 0 0 h
9 9 2
-------
q r i j
1 1 1 6
-------


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.

• Thank you Sleafar, I will plug the coordinates into my GPSr and go to the location and once there I should be able to find a container with a log book in it within +/-30 feet of the location. It's a treasure hunt. Will let you know what I find. Appreciate your work. Feb 19, 2016 at 20:10
• @salwc2k That's probably the most elaborate solution check on this site. ;) Feb 19, 2016 at 20:15
• @salwc2k Btw. you can only accept one answer, if you accept another, the previous one in unaccepted. Feb 19, 2016 at 20:19
• You mentioned, “The remaining numbers cannot be found, because the remainder of the division is not specified.” However, the question mentions that it is an exact long division and my understanding is that the phrase, “exact long division” means the remainder is zero. Aug 23, 2022 at 0:28

Solution:

A = 1, B = 0, C = 4, D = 2, E = 6, F = 8, G = 1, H = 2, J = 2

Dividing the puzzle up into sections: Some reasoning (more to come when I'm back home with a scanner):

We know from the fact that the division is done in 3 steps rather than 5 that the solution is of the form F080X.
8 * AXC is a 3 digit number. This means that A = 1
Sum 2, GXXX - XXH has a 2-digit solution. G has a maximum value of 1
Similarly, Sum 1, XXBD - XXJ has a 2-digit solution; assuming no leading zero, this makes the first X equal to 1 (and the second to 0). This makes XXJ into 9XJ, making F either 8 or 9