# puzzle

• Dec 3rd 2012, 09:35 AM
prasum
puzzle
Logic+logic=prolog what is the value of alphabets involved in it.

I was trying the rebues puzzle but couldnt get anything
• Dec 3rd 2012, 05:06 PM
Soroban
Re: puzzle
Hello, prasum!

If you've had no experience with "alphametics",
. . no wonder you're having difficulty.

Quote:

$\begin{array}{cccccc} ^1 &^2 & ^3 & ^4 & ^5 & ^6\\ &L&\Theta&G&I&C \\ +&L&\Theta&G&I&C \\ \hline P&R&\Theta&L&\Theta&G \end{array} \quad \text{where each letter represents a different digit.}$

$\text{I use }\Theta\text{ because the letter }"O"\text{ can be confused with the digit }"0".$

You should be able to see that $P = 1.$

(If you can't, these problems are not for you.)

In column-3, we see that $\Theta + \Theta$ ends in $\Theta.$

Then there are two possible cases:
. . (1) $\Theta = 0$
. . (2) $\Theta = 9$ with a "carry" from column-4.

(1) Suppose $\Theta = 0.$

We have: . $\begin{array}{cccccc} ^1 &^2 & ^3 & ^4 & ^5 & ^6\\ &L& 0&G&I&C \\ +&L& 0&G&I&C \\ \hline 1&R&0 &L& 0&G \end{array}$

In column-6, $C+C \,=\,G$ with no carry.

In column-5: . $I+I$ ends in $0$ with a carry.
. . Hence: $I = 5.$

In column-4: . $G+G+1 \,=\,L$ with no carry.

In column-2: . $L+L \,=\,10+R$

In column-6, let $C = 2.$
. . Then $G = 4.$

In column-4: $L = 9.$

In column-2: $R = 8.$

We have found one solution: . $\begin{array}{cccccc} &9&0&4&5&2 \\ +&9&0&4&5&2 \\ \hline 1&8&0&9&0&4 \end{array}$

There may be other solutions.