If you make a mistake, it could be anything.
But assuming you do it right, the answer is 0.
The last digit of a product depends only on the
last digits of the factors.
Consequently, with two factors, there are 100
possibilities:
0x0, 0x1, 0x2, ... 9x9.
How many of these give 9?
Just four: 9x1, 7x7,
3x3, and 1x9.
How many give 8?
There are twelve 8's. You can readily see this in
a multiplication table:
| x
|
0
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
| 0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
| 1
|
0
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
| 2
|
0
|
2
|
4
|
6
|
8
|
0
|
2
|
4
|
6
|
8
|
| 3
|
0
|
3
|
6
|
9
|
2
|
5
|
8
|
1
|
4
|
7
|
| 4
|
0
|
4
|
8
|
2
|
6
|
0
|
4
|
8
|
2
|
6
|
| 5
|
0
|
5
|
0
|
5
|
0
|
5
|
0
|
5
|
0
|
5
|
| 6
|
0
|
6
|
2
|
8
|
4
|
0
|
6
|
2
|
8
|
4
|
| 7
|
0
|
7
|
4
|
1
|
8
|
5
|
2
|
9
|
6
|
3
|
| 8
|
0
|
8
|
6
|
4
|
2
|
0
|
8
|
6
|
4
|
2
|
| 9
|
0
|
9
|
8
|
7
|
6
|
5
|
4
|
3
|
2
|
1
|
|
