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2#0 = 10
2#2 = 2
2#16 = 8
3#3 = 3
4#9 = 3
5#5 = 5
5#11 = 7
5#17 = 9
6#13 = 7
9#12 = 12
9#16 = 16
10#3 = 7

8#8=2
13#13=19

9#4=2

3#0=10
4#0=10

0#6=2
0#7=3

1000#1000=3000

0#5=5
0#8=2
0#12=4

5#23=3
5#31=3

0#0=2

52#3=3
53#1=3

41#43 = 3
43#41 = 3

1#0 = 10
5#0 = 10

20#0 = 20

4#1 = 3
100#1 = 3

This is a mathematical puzzle, i.e., all numbers actually represent numbers and not letters of the alphabet. Also for the input the numerical value of the inputs and their representation (i.e., the base in which they are writen in) are relevant, not the number of holes, its English name or anything like that. Programming and similar technical help can help, but is not required.

a#b is always valid and unambiguous in every base larger than 2. It is impossible to get 0 as a result. There is a certain edge case that can only happen in base-2 that one might argue could lead to a#b being undefined. But if you choose to define this edge case as 1 (which is mathematically reasonable) you get, e.g. 0b1#0b1=1 and 0b100#0b1=1. In fact, if the twin prime conjecture holds, there are infinitely many such cases.

2#0 = 10
2#2 = 2
2#16 = 8
3#3 = 3
4#9 = 3
5#5 = 5
5#11 = 7
5#17 = 9
6#13 = 7
9#12 = 12
9#16 = 16
10#3 = 7

8#8=2
13#13=19

9#4=2

3#0=10
4#0=10

0#6=2
0#7=3

1000#1000=3000

0#5=5
0#8=2
0#12=4

5#23=3
5#31=3

0#0=2

52#3=3
53#1=3

41#43 = 3
43#41 = 3

1#0 = 10
5#0 = 10

20#0 = 20

4#1 = 3
100#1 = 3 

1#8 = 2
1#08 = 4

1#234 = 2
12#34 = 2
123#4 = 4

This is a mathematical puzzle, i.e., all numbers actually represent numbers and not letters of the alphabet. Also for the input the numerical value of the inputs and their representation are relevant, not the number of holes, its English name or anything like that. Programming and similar technical help can help, but is not required.

a#b is always valid and unambiguous in every base larger than 2. It is impossible to get 0 as a result. There is a certain edge case that can only happen in base-2 that one might argue could lead to a#b being undefined. But if you choose to define this edge case as 1 (which is mathematically reasonable) you get, e.g. 0b1#0b1=1 and 0b100#0b1=1. In fact, if the twin prime conjecture holds, there are infinitely many such cases.

Hint #20

Leading zeros do not matter for a but they do matter for b (see 1#8 and 1#08 example)

2#0 = 10
2#2 = 2
2#16 = 8
3#3 = 3
4#9 = 3
5#5 = 5
5#11 = 7
5#17 = 9
6#13 = 7
9#12 = 12
9#16 = 16
10#3 = 7

8#8=2
13#13=19

9#4=2

3#0=10
4#0=10

0#6=2
0#7=3

1000#1000=3000

0#5=5
0#8=2
0#12=4

5#23=3
5#31=3

0#0=2

52#3=3
53#1=3

41#43 = 3
43#41 = 3

1#0 = 10
5#0 = 10

20#0 = 20

2#0 = 10
2#2 = 2
2#16 = 8
3#3 = 3
4#9 = 3
5#5 = 5
5#11 = 7
5#17 = 9
6#13 = 7
9#12 = 12
9#16 = 16
10#3 = 7

8#8=2
13#13=19

9#4=2

3#0=10
4#0=10

0#6=2
0#7=3

1000#1000=3000

0#5=5
0#8=2
0#12=4

5#23=3
5#31=3

0#0=2

52#3=3
53#1=3

41#43 = 3
43#41 = 3

1#0 = 10
5#0 = 10

20#0 = 20 

4#1 = 3
100#1 = 3

2#0 = 10
2#2 = 2
2#16 = 8
3#3 = 3
4#9 = 3
5#5 = 5
5#11 = 7
5#17 = 9
6#13 = 7
9#12 = 12
9#16 = 16
10#3 = 7

8#8=2
13#13=19

9#4=2

3#0=10
4#0=10

0#6=2
0#7=3

1000#1000=3000

0#5=5
0#8=2
0#12=4

5#23=3
5#31=3

0#0=2

52#3=3
53#1=3

41#43 = 3
43#41 = 3

1#0 = 10
5#0 = 10

2#0 = 10
2#2 = 2
2#16 = 8
3#3 = 3
4#9 = 3
5#5 = 5
5#11 = 7
5#17 = 9
6#13 = 7
9#12 = 12
9#16 = 16
10#3 = 7

8#8=2
13#13=19

9#4=2

3#0=10
4#0=10

0#6=2
0#7=3

1000#1000=3000

0#5=5
0#8=2
0#12=4

5#23=3
5#31=3

0#0=2

52#3=3
53#1=3

41#43 = 3
43#41 = 3

1#0 = 10
5#0 = 10 

20#0 = 20