Mistakes give rise to problems- 7

Note that, If any of $b,c=0$ , the equations takes an invalid form:- $\frac{a}{0}=\frac{\frac{a}{b}}{0},$ which implies that $b,c\neq0$

$Now\: We\: divide\: in\: 2\: cases.$

$Case\: 1:\: a = 0$

If $a=0,$ $b$ and $c$ could be any integer from $-10$ to $10$ inclusive other than $0$ . Thus we have $20$ independent choices for $b$ , and $20$ independent choices for $c$ . which implies there are $20\times20=400$ choices for $b$ and $c$ while $a=0$ $.$ $.$ $Case\: 2:\: a \neq 0$

We also know $b,c\neq0$ .

$Now,$ $\frac{a}{\left (\frac{b}{c} \right )} = \frac{\left ( \frac{a}{b} \right )}{c}$ $\Rightarrow a \times c = \left ( \frac{b}{c} \right ) \times \left ( \frac{a}{b} \right )$ $\Rightarrow a \times c = \frac{a}{c}$ $\Rightarrow c^{2} = 1$ $\Rightarrow c =\pm 1$

Now we have $2$ choices for $c$ , $20$ independent choices for $a$ and $20$ independent choices for $b$ . Thus we get $2\times20\times20=800$ ordered triplet of integers for $a\neq0$

Thus we get a total of $400+800=\boxed{1200}$ ordered triplet of integers while the $\color{#D61F06}{Mistake}$ is true.

Note that the problem is asking for number of solutions of $\dfrac{ac}{b} = \dfrac{a}{bc}$ .

In this form, as $b$ is in denominator, it can be any nonzero number, so it has $20$ choices.

Now the problem is just reduced to number of solutions of $\dfrac{a}{c} = ac$ .

Note that here $a=0$ is a solution, giving $20$ pairs $(a,c)$ because $c\neq 0$ .

If $a$ is non-zero, it has 20 choices, then $c$ can have values only $-1$ and $1$ , hence this gives $40$ ordered pairs $(a,c)$ .

Thus we have total $60$ pairs of $(a,c)$ and b has 20 choices, giving $60\times 20 = \boxed{1200}$ ordered triples $\color{#3D99F6}{(a,b,c)}$

$\displaystyle \frac{ac}{b} = \frac{a}{bc}$

$\displaystyle abc^{2} = ab$

$b$ can't be $0$ , so it has $20$ choices.

$\displaystyle ac^{2} = a$

$\displaystyle a(c^{2} - 1) = 0$

If $\displaystyle a = 0; c = -10,-9,...,-2,-1,1,2...,9,10$ (20 pairs)

If $\displaystyle c = 1,-1; a = -10,-9,...,-2,-1,1,2,...,9,10$ (40 pairs not overcount $a = 0$ )

Therefore, the number of solutions $= 20(20+40) = 1200$ . ~

To have the quantities defined, we must have b and c <> 0. Now,

ac/b = a/(bc),

(a/b)(c-1/c) = 0,

a = 0 or c = +/-1.

There are two cases to consider.

(i) b <> 0, c = +/-1.

There are 21-1 = 20 choices for b, 2 choices for c, and 21 choices for a for a total of 20 * 2 * 21 choices.

(ii) b <> 0, c <> 0, +/-1, and a = 0.

There are 21-1 = 20 choices for b, 21-3 = 18 choices for c, and 1 choice for a for a total of 20 * 18 * 1 choices.

In all, there are 20 * 2 * 21+20 * 18 * 1 = 20(42+18) = 1200 choices.

Brute force approach $O(n^3)$ :

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counter = 0
for a in range(-10, 11):
    for b in range(-10, 11):
        for c in range(-10, 11):
            try:
                if a/(b/c) == (a/b)/c:
                    counter += 1
            except ZeroDivisionError:
                continue
print(counter)

IF (b <> 0) AND (c <> 0) AND (a/ (b/ c) = a/ b/ c) THEN INC (count);

Scientific approach tells that there are 1200 of them for -10 to 10 of a, b and c.

Basically, when (c = 1) OR (c = -1) OR (a = 0), it shall count.

Answer: 1200

Venn diagram is the best for these questions