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### Multiplying Groups of Numbers

```
Date: 5/9/96 at 12:8:46
From: Anonymous
Subject: Whole Numbers and Multiplication

Arrange the nine digits 1,2,3,4,5,6,7,8 and 9 into three groups. Make
one group of two digits, one group of three digits, and one group of
four digits, so that when you multiply the first group of two times
the second group of three you will get the third group of four.

Example:  39 x 186 = 7,254

There are six other possible combinations.  My 10 year old must find
one. I am his mom, and to be politically correct, I am mathmatically
challenged.

Thanks
```

```
Date: 12/11/96 at 01:02:12
From: Doctor Rob
Subject: Re: Whole Numbers and Multiplication

I would start by looking at the units digit of the two-digit and
three-digit numbers. Let's write the multiplication problem as
abc * de = fghi

Then we'll consider cases, depending on the values of c and e.
Neither one can be a zero, obviously. Neither one can be a 1, either,
since if c = 1, then i = e, which is excluded, and if e = 1, then
i = c, which is also excluded. Similarly, neither one can be a 5,
because 5 times an even number ends in 0, and 5 times an odd number
ends in 5, both of which are impossible. This leaves 42 possible pairs
of values of c and e. Certain others can also be eliminated, since 6
times an even number ends in that same even number. This reduces the
list to 36 possibilities:

c e i   c e i   c e i   c e i   c e i   c e i   c e i   c e i   c e i
-----   -----   -----   -----   -----   -----   -----   -----   -----
2 3 6   2 4 8   2 7 4   2 8 6   2 9 8   3 2 6   3 4 2   3 6 8   3 7 1
3 8 4   3 9 7   4 2 8   4 3 2   4 7 8   4 8 2   4 9 6   6 3 8   6 7 2
6 9 4   7 2 4   7 3 1   7 4 8   7 6 2   7 8 6   7 9 3   8 2 6   8 3 4
8 4 2   8 7 6   8 9 2   9 2 8   9 3 7   9 4 6   9 6 4   9 7 3   9 8 2

The next step is to see which b and d can possibly go with each of
these 36 possibilities. Let us try the first one, c = 2, e = 3, i = 6.
Clearly b and d cannot be any of these or zero, and they must be
unequal, so there are 30 possible pairs.  We know that the digit h
must equal the last digit of the sum of c*d, plus b*e, plus the carry
from the product c*e.  In this case, that is:

2*d + 3*b + 0 = 2*d + 3*b

This last digit h also cannot be 0, 2, 3, 6, b, or h.  We try all
thirty:

1 4 1   1 5 3   1 7 7   1 8 9   1 9 1   4 1 4
4 5 2   4 7 6   4 8 8   4 9 0   5 1 7   5 4 3
5 7 9   5 8 1   5 9 3   7 1 3   7 4 9   7 5 1
7 8 7   7 9 9   8 1 6   8 4 2   8 5 4   8 7 8
8 9 2   9 1 9   9 4 5   9 5 7   9 7 1   9 8 3

Any of these with h = b or h = d or h = 0, 2, 3, or 6 is impossible.
This leaves:

b d h   b d h   b d h   b d h   b d h
-----   -----   -----   -----   -----
1 8 9   5 1 7   5 7 9   5 8 1   7 4 9
7 5 1   8 5 4   9 4 5   9 5 7   9 7 1

or only ten possibilities for this first case.

The next step is to test the remaining values of a for each of these
ten subcases. a cannot be zero, b, c, d, e, h, or i, so there are only
three possibilities in each subcase. Let us try the first one, b = 1,
d = 8, h = 9 (and recall c = 2, e = 3, i = 6).  The three values of a
are 4, 5, or 7.  412*83 = 34196, which is too big, and the other
values of a will all give even bigger answers, so none of them works.

This eliminates the first subcase.

Next we have to try a second subcase, then a third, and so on, until
we either find a solution, or else eliminate all ten of the subcases.
If we eliminate all ten of the subcases, that means that the first of
the 36 cases is eliminated, and we go on to the next.

I know that this is a long and complicated process, but it is sure way
to find all the solutions, and to prove that you have them all.

I wish I could think of some useful shortcuts, but this is all that I
could figure out to solve your problem.

If you need more help, write back again, and we'll try to offer more.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations

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