# Cryptarithmetic Problem – Possible ways of getting 6, 7, 8, 9

## Possible ways of getting 6

Unit Digit = 6 e.g. 6, 16, 36, 56

A x B = _6
Explanation
1 x 6 =_6(06)
or
6 x 1 =_6(06)
1 x 6 =_6(06) A = 1 B = 6
6 x 1 =_6(06) A = 6 B = 1
2 x 3 =_6(06)
or
3 x 2 =_6(06)
2 x 3 =_6(06) A = 2 B = 3
3 x 2 =_6(06) A = 3 B = 2
4 x 4 =_6(16)4 x 4 =_6(16) A = 4 B = 4
6 x 6 =_6(36)6 x 6 =_6(36) A = 6 B = 6
8 x 2 =_6(16)
or
2 x 8 =_6(16)
8 x 2 =_6(16) A = 8 B = 2
2 x 8 =_6(16) A = 2 B = 8
9 x 4 =_6(36)
or
4 x 9 =_6(36)
9 x 4 =_6(36) A = 9 B = 4
4 x 9 =_6(36) A = 9 B = 9
8 x 7 =_6(56)
or
7 x 8 =_6(56)
8 x 7 =_6(56) A = 8 B = 7
7 x 8 =_6(56) A = 7 B = 8

## Possible ways of getting 7

Unit Digit = 7 e.g. 7, 27

A x B = _7
Explanation
1 x 7 =_7(07)
or
7 x 1 =_7(07)
1 x 7 =_7(07) A = 1 B = 7
7 x 1 =_7(07) A = 7 B = 1
3 x 7 =_7(27)
or
7 x 3 =_7(27)
3 x 7 =_7(27) A = 3 B = 7
7 x 3 =_7(27) A = 7 B = 3

## Possible ways of getting 8

Unit Digit = 8 e.g. 8, 18, 28, 48

A x B = _8
Explanation
1 x 8 =_8(08)
or
8 x 1 =_8(08)
1 x 8 =_8(08) A = 1 B = 8
5 x 1 =_8(08) A = 8 B = 1
2 x 4 =_8(08)
or
4 x 2 =_8(08)
2 x 4 =_8(08) A = 2 B = 4
4 x 2 =_8(08) A = 4 B = 2
2 x 9 =_8(18)
or
9 x 2 =_8(18)
2 x 9 =_8(18) A = 2 B = 9
9 x 2 =_8(18) A = 9 B = 2
3 x 6 =_8(18)
or
6 x 3 =_8(18)
3 x 6 =_8(18) A = 3 B = 6
6 x 3 =_8(18) A = 6 B = 3
7 x 4 =_8(28)
or
7 x 7 =_8(28)
7 x 4 =_8(28) A = 7 B = 4
6 x 3 =_8(28) A = 4 B = 7
6 x 8 =_8(48)
or
8 x 6 =_8(48)
6 x 8 =_8(48) A = 6 B = 8
8 x 6 =_8(48) A = 8 B = 6

## Possible ways of getting 9

Unit Digit = 9 e.g. 9, 49

A x B = _9
Explanation
1 x 9 =_9(09)
or
9 x 1 =_9(09)
1 x 9 =_9(09) A = 1 B = 9
9 x 1 =_9(09) A = 9 B = 1
7 x 7 =_9(49)7 x 7 =_9(49) A = 7 B = 7
3 x 3 =_9(09)3 x 3 =_9(09) A = 3 B = 3