January 22, 2001
Problem of the Month

The Multiplying Digits Problem
by David Book
in Problems for Puzzlebusters

If you multiply two whole numbers together, what is the last digit of the answer most likely to be?

Solution to the Problem:

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

There are also twelve each of 2's, 4's, and 6's.

But as you can see from the table, nothing is close to 0.
There are 27 zeroes, so the probability of getting a zero is 27/100. The next closest probability is 12/100.

Correctly solved by:

1. Erin McGinnis	Winchester, VA
2. David Powell	Winchester, VA
3. Matthias	Switzerland
4. Kirstine Wynn	Winchester, VA

Send any comments or questions to: David Pleacher