Let A=1,4,7 and B=2,3,8.

Question

Let $A={1,4,7}$ and $B={2,3,8}$. Then the number of elements, in the relation $R={(({a}_{1},{b}_{1}),({a}_{2},{b}_{2}))\in ((A	imes B)	imes (A	imes B)):{a}_{1}+{b}_{2}$ divides ${a}_{2}+{b}_{1}}$ is $____$ .

Answer: 18

Accepted Answer

We need to count the number of ordered pairs

$(({a}_{1},{b}_{1}),({a}_{2},{b}_{2}))\in (A	imes B)	imes (A	imes B)$

such that

${a}_{1}+{b}_{2}{a}_{2}+{b}_{1}.$

Given

$A={1,4,7},\, B={2,3,8}$

first note that

$A	imes B={(1,2),(1,3),(1,8),(4,2),(4,3),(4,8),(7,2),(7,3),(7,8)}$

So there are $9$ elements in $A	imes B$, and hence $9	imes 9=81$ ordered pairs in $(A	imes B)	imes (A	imes B)$.

We now count only those for which

${a}_{1}+{b}_{2}|{a}_{2}+{b}_{1}.$

Let

$x={a}_{1}+{b}_{2},\, y={a}_{2}+{b}_{1}.$

Since ${a}_{1},{a}_{2}\in {1,4,7}$ and ${b}_{1},{b}_{2}\in {2,3,8}$, possible values of such sums are:

$1+2=3,\, 1+3=4,\, 1+8=9,$

$4+2=6,\, 4+3=7,\, 4+8=12,$

$7+2=9,\, 7+3=10,\, 7+8=15.$

So possible values are

$3,4,6,7,9,10,12,15.$

Now we count how many times each value occurs as a sum of one element from $A$ and one from $B$:

$3$ occurs once: $(1,2)$

$4$ occurs once: $(1,3)$

$6$ occurs once: $(4,2)$

$7$ occurs once: $(4,3)$

$9$ occurs twice: $(1,8),(7,2)$

$10$ occurs once: $(7,3)$

$12$ occurs once: $(4,8)$

$15$ occurs once: $(7,8)$

Let this frequency be $f(s)$.

Now for each possible value of $x={a}_{1}+{b}_{2}$, we count possible $y={a}_{2}+{b}_{1}$ such that $x|y$.

Also, number of ways to get a given $x$ is $f(x)$, and number of ways to get a given $y$ is $f(y)$.

Hence total number required is

$\sum _{x|y}f(x)f(y)$

where $x,y$ run over the above sums.

Now check each $x$:

1. For $x=3$
Multiples among the possible sums:

$3,6,9,12,15$

So contribution is

$f(3)(f(3)+f(6)+f(9)+f(12)+f(15))$

$=1(1+1+2+1+1)=6$

2. For $x=4$
Multiples among possible sums:

$4,12$

Contribution:

$1(1+1)=2$

3. For $x=6$
Multiples among possible sums:

$6,12$

Contribution:

$1(1+1)=2$

4. For $x=7$
Multiples among possible sums:

$7$

Contribution:

$1\cdot 1=1$

5. For $x=9$
Multiples among possible sums:

$9$

Contribution:

$2\cdot 2=4$

6. For $x=10$
Multiples among possible sums:

$10$

Contribution:

$1\cdot 1=1$

7. For $x=12$
Multiples among possible sums:

$12$

Contribution:

$1\cdot 1=1$

8. For $x=15$
Multiples among possible sums:

$15$

Contribution:

$1\cdot 1=1$

Now add all contributions:

$6+2+2+1+4+1+1+1=18$

Therefore, the number of elements in the relation $R$ is

$18$

Solutions · 1