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Put 4 distinct 1 digit non-negative numbers (A, B, C, D) to every vertex of a rectangle.
Then put the last digit of the sum of 2 connected vertexes at the sides of the rectangle.
Create smaller rectangle connecting the sides of the bigger rectangle.
Do above procedure once more (using the 4 new numbers which do not have to be distinct).
The last 4 numbers must be the same.
Then sum (A, B, C, D), let’s call it X.

E = last digit of (A+B)           G = last digit of (B+D)
H = last digit of (C+D)           F = last digit of (A+C)
I = last digit of (F+E)           J = last digit of (E+G)
K = last digit of (F+H)           L = last digit of (H+G)
I = J = K = L

Find (A, B, C, D) with the smallest X

Bonus Puzzle : Find (A, B, C, D) with the largest X

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For the last four to all be equal the middle four must be

a rotation of $(n,m,m,n)$ with $m-n \pmod{10}=5$ (with ns and ms on opposite corners).

The only way to do that is to have the first four be

some rotation of $(p,q,r,s)$ with $p+q \pmod{10}=r+s \pmod{10}$
while $p+r=q+s$

The smallest solution in $X$ is thus

$12$ with $A,B,C,D$ some rotation of $(0,1,6,5)$ or $(1,0,5,6)$

The largest would be

$24$ with $A,B,C,D$ some rotation of $(3,4,9,8)$ or $(4,3,8,9)$

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  • $\begingroup$ Oh sory my question was wrong, I have edited it, the second rectangle do not have to be distinct. There are solutions for this question. $\endgroup$ – Jamal Senjaya Oct 25 '16 at 4:31
  • $\begingroup$ @JamalSenjaya - how is it now? $\endgroup$ – Jonathan Allan Oct 25 '16 at 5:01
  • $\begingroup$ you are correct now ! $\endgroup$ – Jamal Senjaya Oct 25 '16 at 5:05
  • $\begingroup$ Only thing I missed: the first condition means there are twice as many solutions as I suggest... fixing $\endgroup$ – Jonathan Allan Oct 25 '16 at 5:06
  • $\begingroup$ pretty sure i had the complete solution before ya, but we both got there. nice job :) $\endgroup$ – Rubio Oct 25 '16 at 6:00
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Minimum sum of A, B, C, D is

12 - using 0,1,5,6 such that each side totals 1, 6, or 11;
e.g. A=0, B=1, C=6, D=5 gives sides AB=1, BD=6, AC=6, CD=11.
this uses values 0 and 1 to minimize the sum, and matches them with the lowest complementary pair that gives usable modulus values after summing with the neighbors.

This gives E, F, G, H of

alternating ones and sixes

which in turn gives I/J/K/L of

7, by adding a one and a six.

There are multiple solutions that fit

8 total, actually.
0,1,6,5; 0,6,1,5; 1,0,5,6; 1,5,0,6;
5,1,6,0; 5,6,1,0; 6,0,5,1; 6,5,0;1.

Bonus -

Largest total sum is 24 by essentially the opposite method -
using 8 and 9 to maximize the digit sum, and matching them with the complementary values 3 and 4 to give pairable modulus sums of 7, 12, and 17.
e.g. A=9, B=8, C=3, D=4 (for sides AB=17, BD=12, AC=12, CD=7)
and then the E,F,G,H values are alternating 2s and 7s,
summing to I/J/K/L of 9.

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Here it is:

1^^^^3^^^^2
^^^1^^^1^^^
8^^^^^^^^^8
^^^1^^^1^^^
7^^^^3^^^^6

The lowest sum should be $1+2+7+6=16$(I don't include zero)

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Here is my solution

X=(1,4,9,6) or (2,3,8,7) or (3,2,7,8) or (4,1,6,9).
Getting a second rectangle of 0-5-0-5 is the only way I found so far and every solutions are = 20 so no lowest or smallest.

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  • $\begingroup$ +1 as the first true answer, but still not the smallest or the largest one. $\endgroup$ – Jamal Senjaya Oct 25 '16 at 5:02

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