# How many different solutions of equation?

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• Mar 30th 2013, 09:30 AM
feferon11
How many different solutions of equation?
How many different solutions has this equation:

$\overline{abc}+\overline{def}+\overline{ghi}=963$

Here $\overline{abc}, \overline{def}, \overline{ghi}$ are three-digit numbers

$a,b,c,d,e,f,g,h,i$- different digits; $a,d,g\neq 0$

I could begin with:
$(a+d+g)\cdot100+(b+e+h)\cdot10+(c+f+i)=963$

Now, $c+f+i$ must be 3, 13 or 23...

Not know how to connect with other.

Could you help to me. Thank you!!!
• Mar 30th 2013, 07:06 PM
chiro
Re: How many different solutions of equation?
Hey feferon11.

You have nine independent variables to start off with.

If c + f + i = *3 (* is a number from 0 to 2 since max is 9+9+9 = 27) then you have one constraint. Now tke the * and move it the 10's group. Let * = p.

(b + e + h + p) = *6. Where again * = (0,1,2 for the same reason as above). If * = q we have

(a + d + g + q) = 9

So in total we have the following constraints:

All numbers between 0 and 9 inclusive
a, d, g > 0
c + f + i = 10*p + 3 (p = 0,1,2)
b + e + h + p = 10*q + 6 (q = 0,1,2)
(a + d + g + q) = 9

These constraints though do not make it easy so I will suggest another way.

Consider the two variable case.

Let abc + def = 963. Since a, d != 0 we start abc = 100.

Now def = 963 - def. This means that you have 963 - 100 = 863 different combinations for both numbers.

In the three variable case you do the same thing but twice.

abc + def + ghi = 963. Fix abc = 100 which means (def + ghi) = 963 - 100 = 863. Now figure out how many possibilities for 863 (it will be 863 - 100 = 763). Then do it for abc = 101, 102, ...., 863.

You should find a pattern in this and I suggest you write it all out to see it for yourself. (Hint: Try and find the relationship given that abc = some number for how many combinations between def and ghi and add up all possibilities for each abc).
• Mar 30th 2013, 08:53 PM
Soroban
Re: How many different solutions of equation?
Hello, feferon11!

This is an Alphametic.
It requires some intelligent trial-and-error.
Algebra usually won't help.

Quote:

How many different solutions has this problem?

. . $\begin{array}{cccc} & ^1 & ^2 & ^3 \\ & A&B&C \\ &D&E&F \\ + & G&H&I \\ \hline & 9&6&3 \end{array}$

Each letter represents a different digit and $A,D,G \ne 0.$

In column-3, $C+F+I$ ends in 3.
So, the sum could be 3, 13 or 23.

If the sum is 3, then $C,F,I$ must be $0,1,2$ in some order.
That leaves $3,4,5,6,7,8,9$ for the other digits.

But in column-1, $A+D+G$ must equal 7, 8 or 9.
And that is impossible with the remaining digits.

Using similar reasoning, I eliminated a few more possibilties.

I found a solution.

The general solution looks like this:

. . $\begin{array}{|c|c|c|} ^1 & ^2 & ^3 \\ \hline 1&3&6 \\ 2&4&8\\ 5&7&9 \\ \hline 9&6&3 \\ \hline \end{array}$

The digits in each column can be arranged in $3!$ ways.

Therefore, there are:. $6\cdot6\cdot6 \,=\, 216$ solutions.