The number of positive integral solutions of $abc =$ $30$ is

Question

The number of positive integral solutions of $abc =$ $30$ is

My attempt
Factors of $30$ are $3 \cdot 5 \cdot 2$.
Therefore $a$ will have three choices, similarly $b$ & $c$ will have $2$ choices & $1$ choice.
So,it's answer must be $3 \cdot 2 \cdot 1$

But answer in my textbook is given is $3 \cdot 3 \cdot 3$. .It states that"$2$ can be assigned to either $a$ or $b$ or $c$". Similarly, each of $3$ and $5$ can be assigned in $3$ ways. Thus, the number of solutions is $3 \cdot 3 \cdot 3$

According to me, if $a$ is already assigned to $2$ than how can other letters ($b$ or $c$) have same value as that of $a$.

Please explain me where am I wrong?

Answer 1

$abc=30$

First write $30$ as the prime factors

$30=2\times3\times5$

$a=2^{x_1}\times3^{y_2}\times5^{z_3}$

$b=2^{x_2}\times3^{y_2}\times5^{z_2}$

$c=2^{x_3}\times3^{y_3}\times5^{z_3}$

$$abc=30$$ $$2^{{x_1}+{y_1}+{z_1}+{x_2}+{y_2}+{z_2}+{x_3}+{y_3}+{z_3}}=30$$note that $$x_1+x_2+x_3=1...(1)$$$$y_1+y_2+y_3=1....(2)$$$$z_1+z_2+z_3=1.....(3)$$ Notice that $(1),(2),(3)$ have only $3$ solutions each

So, the total number of solutions are $3\times3\times3=27$

Answer 2

Hint: (Slightly different point of view than the book; but will lead to same answer).

There are many ways to write $30$ as a product of three factors, such as $1\cdot 1\cdot 30$ or $1\cdot 5\cdot 6$.

I think you have neglected some of these cases.

Answer 3

The texbook is right.

Start with $1\cdot1\cdot1$ and assign $2$, say to $b$ and get $1\cdot2\cdot1$.

Now assign $3$, say to $a$ and get $3\cdot2\cdot1$.

Then assign $5$ to $a$ again, giving $15\cdot2\cdot1$.

On every stage, you have three choices.