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### 4(ABCD) = DCBA

```
Date: 06/15/97 at 17:36:44
From: Ziva
Subject: Math problem

Find a four digit number such that:

ABCD
*    4
------
DCBA
```

```
Date: 06/17/97 at 16:27:04
From: Doctor Rob
Subject: Re: Math problem

This is one of those digital logic problems.  You must use the facts
that the digits are integers from 0 to 9 and what you know about
arithmetic to figure this one out.  These problems can be tricky
and/or tedious, but this one isn't too bad.

1 <= A <= 9 (since ABCD is a four-digit number, A can't be zero)
0 <= B <= 9
0 <= C <= 9
4 <= D <= 9 (D must be at least 4*A which is at least 4)

Then we have some equations:

4*D = 10*z + A (z is the carry to the tens column)
4*C + z = 10*y + B (y is the carry to the hundreds column)
4*B + y = 10*x + C (x is the carry to the thousands column)
4*A + x = D

There are more inequalities:

0 <= z <= 3
0 <= y <= 3
0 <= x <= 3 (because the multiplier is 4, the largest a carry can be
is 3)

Now we combine the first and last equations by multiplying the last by
4 and substituting for 4*D in the first equation:

4*(4*A + x) = 10*z + A   or   4*x = 10*z - 15*A = 5*(2*z - 3*A)

Since 5 divides the number on the right, 5 must divide 4*x, and
so 5 divides x.  But x is small, from 0 to 3, so x = 0 is forced. Thus
2*z - 3*A = 0, so 2*z = 3*A.

Similarly, 3 divides z, so put z = 3*w, so A = 2*w.
Then D = 8*w, and 4 <= D <= 9.

Clearly D = 8, so w = 1, A = 2, and z = 3.

Similarly, using the middle two equations, we can determine B, C, and
y. This gives you the answer you seek.  You can finish the rest.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
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