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Consider the set $\{a, b, c\}$ with binary operators $+$ and $*$ defined as follows.

 + a b c a b a c b a b c c a c b
 * a b c a a b c b b c a c c c b

For example, $a + c = c, c + a = a, c * b = c$ and $b * c = a$.

Given the following set of equations:
$$(a * x) + (a * y) = c \\ (b * x) + (c * y) = c$$
The number of solution(s) (i.e., pair(s) (x, y) that satisfy the equations) is

1. 0
2. 1
3. 2
4. 3
edited | 1.1k views
+5
In current GO pdf. It seems this question formatting is not proper.
+2
you are right. i got wrong answer by solving from pdf. :D

Consider each pair

1. $(a,a):$ $(a*a) +(a*a) = a+a = b \neq c.$ So, $(a,a)$ is not possible

2. $(a,b):$ $(a*a) + (a*b) = a +b = a \neq c.$  So, $(a,b)$ is not possible

3. $(a,c):$ $(a*a)+(a*c) = a+c = c$

$\quad \quad \quad \quad (b*a) +(c*c) = b + b =b \neq c.$ So, $(a,c)$ is not possible

4. $(b,a):$ $(a*b) +(a*a) = b +a = a \neq c.$ So, $(b,a)$ is not possible

5. $(b,b):$ $(a*b) + (a*b) = b+b =b \neq c.$ So, $(b,b)$ is not possible

6. $(b,c):$ $(a*b) + (a*c) = b + c = c$

$\quad \quad \quad \quad (b*b) + (c*c) = c+b = c.$ So, $(b,c)$ is a solution

7. $(c,a):$ $(a*c) + (a*a) = c+a = a \neq c.$ So, $(c,a)$ is not possible

8.$(c,b):$ $(a*c) + (a*b) = c+b = c$

$\quad \quad \quad \quad (b*c) +(c*b) = a+c = c.$ So, $(c,b)$ is a solution

9. $(c,c):$ $(a*c) +(a*c) =c+c= b \neq c.$ So, $(c,c)$ is not possible

So, no. of possible solutions is $2$.
edited by
+1
sir how can we say that it is asking for a pair which is true for both the set.....(b,c=satisfying both)

it can be a pair (a,c) satisfying first one not the second....

then the ans can be 3 solution
0

I think asu

Do for equation 1 then only You get very less pair then ony those pair check with equation 2.

0
this solution is too long, isn't there a better way to solve this in lesser time ?

@Habibkhan
+17
Hint:
We want $(a * x) + (a * y) = c$ and $(b * x) + (c * y) = c$.
Note that to make $\text{Operand 1 }+\text{ Operand 2}=c$, pairs can be like this

$\left (\text{Operand 1},\text{ Operand 2} \right )=$
\begin{align} \left (\text{a},\text{ c} \right )\\ \left (\text{b},\text{ c} \right )\\ \left (\text{c},\text{ b} \right ) \end{align}

for each pair above find the value of (x,y) for $(a * x) + (a * y) = c$ , and see if this pair also satisfies $(b * x) + (c * y) = c$.

You may ended up doing same computation as brute force, but it seems like we are doing smartly :P
+1

I have developed an approach let me know if it's correct

First things first what all pairs under $+$ make c?

(1)a+c (2)b+c (3)c+b

Now consider the first equation that is given

(I) $(a*x)+(a*y)=c$

If you look at the table of  $*$, a seems to be the identity element.

Now to satisfy an equation a+c=c, what pair do I need?

Simple take x=a and y=c and hence

$(a*a)+(a*c)=a+c=c$ Seems good.

In a similar fashion if I follow,my possible pairs for equations (1),(2) and (3) are

(a)(a,c)

(b)(b,c)

(c)(c,b)

Now I only need to check whether these pairs satisfy the second equation that is

$(b*x)+(c*y)=c$

Now, for this let me draw two graphs which represent the operation $b*x$ and $c*y$

here an edge from b to c->a means b*c=a

Now using these two graphs, check whether any of the pairs (a,c),(b,c) and (c,b) make any one of the equations below

a+c

b+c

c+b

and only pair (a,c) fails to do that.

So we have only 2 solutions

(b,c) and (c,b)

We want $operand_1 + operand_2=c$ ,

so the pairs can be:
$(a, c)$
$(b, c)$
$(c, b)$

Now, note that identity element of $({a,b,c},*)$ is $\color{RED}{a}.$
So $(a∗x)+(a∗y)=c$ will result in the above 3 pair only.

No need to check equation 1, just see if these pairs satisfy $(b∗x)+(c∗y)=c.$
$(b, c)$ and $(c, b)$ satisfy equation 2 also so there are 2 solutions.

$C$ is the answer.

edited by
ans (C).. 2 solutions (b, c) and (c, b)..
+1 vote

Approaching the problem the following way might speed you up : observe that the final sum required is $c$ and also the fact that $a*x = x, x\in a,b,c$. To get $c$ as the sum the only ways are: \begin{align} a+c &=c\\ b+c&=c\\ c+b&=c \end{align}

Now, to construct $a+c$ using the second equation $(b * x) + (c * y) = c$:
It can be: $(b*\color{blue}{c}=a) \color{red}{\wedge} ((c*\color{blue}{a}=c) \color{red}{\vee} (c*\color{blue}{b}=c))$.
Check $(c,a)$ and $(c,b)$ with the first equation: $(a * x) + (a * y) = c$ of which only $\color{red}{(c,b)}$ satisfies.

Similarly for $b+c$ we can have $(a,a)$ and $(a,b)$ which satisfy second equation of which neither satisfies first equation.

For $c+b$  we only have $(b,c)$ satisfying second equation and also $(a*b)+(a*c)=b+c=c$. Therefore $\color{red}{(b,c)}$ also satisfies first equation.

$\text{Hence, we only have two solutions } \color{red}{(c,b) \text{ and } (b,c)}$.