Cryptarithmetic Problem – Possible ways of getting 0, 1, 2

Possible ways of getting 0

Unit Digit = 0 e.g. 0, 10, 20, 30

A x B = _0
Explanation
5 x 2 =_0(10)
or
2 x 5 =_0(10)
5 x 2 =_0(10) A x B =_0 A = 5 B = 2
2 x 5 =_0(10) A x B =_0 A = 2 B = 5
4 x 5 =_0(20)
or
5 x 4 =_0(20)
4 x 5 =_0(20) A x B =_0 A = 4 B = 5
5 x 4 =_0(20) A x B =_0 A = 5 B = 4
5 x 6 =_0(30)
or
6 x 5 =_0(30)
5 x 6 =_0(30) A x B =_0 A = 5 B = 6
6 x 5 =_0(30) A x B =_0 A = 6 B = 5
8 x 5 =_0(40)
or
5 x 8 =_0(40)
8 x 5 =_0(40) A x B =_0 A = 8 B = 5
6 x 5 =_0(40) A x B =_0 A = 5 B = 8

Possible ways of getting 1

Unit Digit = 1 e.g. 1, 21, 81

A x B = _0
Explanation
1 x 1 = _1(1)A x A = A
7 x 3 = _1(21)
or
3 x 7 = _1(21)
7 x 3 =_1(21) A x B =_1 A = 7 B = 3
3 x 7 =_1(21) A x B =_1 A = 3 B = 7
9 x 9 = _1(81)A x A = _C(C=1)

Possible ways of getting 2

Unit Digit = 2 e.g. 2, 12, 32, 42, 72

A x B = _2
Explanation
1 x 2 = _2(02)
or
2 x 1 = _2(02)
1 x 2 = _2(02) A x B =_2 A = 1 B = 2
2 x 1 = _2(02) A x B =_2 A = 2 B = 1
2 x 6 = _2(12)
or
6 x 2 = _2(12)
2 x 6 = _2(12) A x B =_2 A = 2 B = 6
6 x 2 = _2(12) A x B =_2 A = 6 B = 2
3 x 4 = _2(12)
or
4 x 3 = _2(12)
3 x 4 = _2(12) A x B =_2 A = 3 B = 4
4 x 3 = _2(12) A x B =_2 A = 4 B = 3
8 x 4 = _2(32)
or
4 x 8 = _2(32)
8 x 4 = _2(32) A x B =_2 A = 8 B = 4
4 x 8 = _2(32) A x B =_2 A = 4 B = 8
7 x 6 = _2(42)
or
6 x 7 = _2(42)
7 x 6 = _2(42) A x B =_2 A = 7 B = 6
6 x 7 = _2(42) A x B =_2 A = 6 B = 7
9 x 8 = _2(72)
or
8 x 9 = _2(72)
9 x 8 = _2(72) A x B =_2 A = 9 B = 8
8 x 9 = _2(72) A x B =_2 A = 8 B = 9

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