In the equation
A*B=C :
A is an 8digit number without repeated digits
B is a single digit, not present in
A C is a number containing all nonzero digits.
Sample solution: (A, B, C)= 51249876*3=153749628
Find additional solutions.
I was thinking if there is an algorithm to build such numbers (
A and
C). I've devise something, not completely satisfactory.
I try here to expose a way to build solutions manually.
For easiness I take B = 3. Then, obviously C begins (left) by 1 or 2, so we have to reserve that digit for that position (C = 1abcdefgh or 2abcdefgh).
When A is multiplied by 3, each digit of A receive from the contiguousright digit an addition (0, 1 or 2 units) and transfer to the contiguousleft another addition (=0, 1 or 2 units). In particular: 1,2 when multiplied by 3 do not transfer additions; 4,5,6 transfer one unit; 7,8,9 transfer two units.
So, when a digit of A is multiplied by 3, the corresponding digit in the product C would result in three possibles digits (only two for number 6, as we don't allow 0 in C).
For example: digit 7 of A ends in 1 when multiplied by 3. But its contiguous number on the right, if existent, would transfer to it an amount of 0,1 or 2 units, so the corresponding digit in C will be 1,2 or 3. This flexibility of these additions is what give way to repetitions in C.
To assure that C has each digit just one time we can:
1. compile a table with the three possibles values for each Adigit.
A= 7 4 1 8 5 2 9 6
C
1 x
2 x x
3 x x x
4 x x x
5 x x x
6 x x x
7 x x x
8 x x x
9 x x
2. choose a disposition, taking care that each row and column (except row 1, as C begins by 1) has just one element chosen (see below).
A= 7 4 1 8 5 2 9 6
C
1 x
2 o x
3 x x o
4 o x x
5 x x o
6 o x x
7 x o x
8 x x o
9 o x
3. Now that the nine digits in C (the product) have been chosen, we have to determine the order of the digits in C.
To do this I have to follow a coherent path that begins with a circle and goes circle by circle to the ninth one.
The only possible circles to begin with are under the Adigits 1, 5 or 6: this is necessary as the firstright Adigit multiplied (*) do not receive transfers from other digits, but only can add to the next one). (Also I should should not finish with circle on row two, as it would lead to a 8digits in C).
Let us try:
I choose to begin with the circle under Adigit 1 (underlined). It is not going to transfer (as 1*3=3), so I go to other circle with has no transfer (they are under Adigits 5 or 6), I choose 5. It transfers 1 (as 5*3=15), so I have to go to a circle having addition of 1. They are under Adigits 2 or 7. I choose 2. It does not transfer (as 3*2=06 and it has receive only one, total will be: 07). So have to go to a circle receiving no additions: only remain circle under Adigit 6. 6 transfers 1, so I go to 7, whose circle has the addition of one. 7 transfers 2 (7+3=21), so it send to 8 or to 9. Choose 8, transfers 2; send me to 9; transfers 2; and send me to 4 which is the last circle.
A: 1 => 5 => 2 => 6 => 7 => 8 => 9 => 4
49876251*3=149628753
Edited on February 14, 2018, 10:50 am

Posted by armando
on 20180214 10:28:58 