Here's my attempt, assuming that any digit D is such that 1 <= D <= 9:
x = 100*A + 10*B + C
y = 100*K + 10*L + M
2A * 3B * 5C = 12 * 2K * 3L * 5M
one possibility: 2A * 3B * 5C = 2(4K) * 3(3L) * 5M
A = 4K, B = 3L, C = M
this limits K to { 1, 2 } and L to { 1, 2, 3 }
therefore: x - y = 300K + 20L will have six different values: 320, 340, 360, 620, 640, 660.
In fact, 12 can be written as follows to comply to 1 <= D <= 9:
set 1) 1 * 2 * 6
set 2) 1 * 3 * 4
set 3) 2 * 2 * 3
If we generate permutations of these, we will have solution sets { x, y, z } such that A = xK, B = yL, C = zM.
Let's take a look at the possible values of the variable on the right for each multiple.
a) if Left = 1 * Right, then 1 <= Right <= 9 and 1 <= Left <= 9; difference digit will be 0
b) if Left = 2 * Right, then 1 <= Right <= 4 and Left belongs to { 2, 4, 6, 8 }; difference digits can be { 1, 2, 3, 4 }
c) if Left = 3 * Right, then 1 <= Right <= 3 and Left belongs to { 3, 6, 9 }; difference digits can be { 2, 4, 6 }
d) if Left = 4 * Right, then 1 <= Right <= 2 and Left belongs to { 4, 8 }; difference digits can be { 3, 6 }
e) if Left = 6 * Right, then Right = 1 and Left is 6, difference digit will be 5.
So for the first solution set { 1, 2, 6 }, the difference x - y will take values which digits are permutations of { 0, X, 5 } , where X = { 1, 2, 3, 4 }.
For the second solution set { 1, 3, 4 }, the difference x - y will take values which digits are permutations of { 0, X, Y }, X = { 2, 4, 6 } and Y = { 3, 6 }
For the thrid solution set { 2, 2, 3 }, the difference x - y will take values which digits are permutations of { X, X, Y }, X = { 1, 2, 3, 4 } and Y = { 2, 4, 6 }.
We will have 4*3! = 24 possible values from set 1, 6*3! - 3 = 33 possible values from set 2 and for solution set 3, 4*3*3 - 2*2 = 32 for when we pick the same element from X twice and 6*3*3! - 8*3 - 2*3! = 72, with a total of 104 possible values.
If I did not calculate anything wrong, we may have 161 possible values for the difference x - y.
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You did not mention whether all the digits should be different, if that's the case then only the solution set 2 * 2 * 3 should be used.
Left = 2*Right => (1,2) (2,4) (3,6) (4,8) - need to select two from these
Left = 3*Right => (1,3) (2,6) (3,9) - need to select one from these
In this case the only solution is (1,2) (4,8) (3,9), which leads to:
982 - 341 = 641, 928 - 314 = 614, 892 - 431 = 461, 829 - 413 = 416, 298 - 134 = 164, 289 - 143 = 146. The differences are numbers which digits are permutations of { 1, 4, 6 }.
If anyone finds something wrong with this reasoning, please let me know.
Namaste.
