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I want to find out the positive integer solutions of variables of the following :

(i) $x_1x_2x_3x_4x_5 =1260$ ( ii ) $2x+3y+4z = 24 $

MY WORK:

For (i), I only know to find out the factors of $1260$, like $1260=2^2.3^2.5.7$ . So total factors are $36$ . Then, I try to take combination $\binom{36}{5}$ , for which I get $376992$ . But the answer is $5625$. I know I'm wrong, but can't advance forward.

For (ii), I am more clueless. I assume that any number can't be greater than $\frac{24}{4}=6$ ... Then I try to select $3$ values out of $6$ ... For which I get $20$ . But the answer is $19$ .

N. B: Sorry if I sound silly... I've faced such questions in simple forms earlier only...

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How many solutions does the equation $x_1x_2x_3x_4x_5 = 1260$ have in the positive integers?

You correctly found that $1260 = 2^2 \cdot 3^2 \cdot 5 \cdot 7$. Let $x_i = 2^{\alpha_i}3^{\beta_i}5^{\gamma_i}7^{\delta_i}$, where $1 \leq i \leq 5$. Then \begin{align*} \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5 & = 2 \tag{1}\\ \beta_1 + \beta_2 + \beta_3 + \beta_4 + \beta_5 & = 2 \tag{2}\\ \gamma_1 + \gamma_2 + \gamma_3 + \gamma_4 + \gamma_5 & = 1 \tag{3}\\ \delta_1 + \delta_2 + \delta_3 + \delta_4 + \delta_5 & = 1 \tag{4} \end{align*} are each equations in the nonnegative integers.

A particular solution of equation 1 corresponds to the placement of four addition signs in a row of two ones. For instance, $$1 + + + 1 +$$ corresponds to the solution $\alpha_1 = 1$, $\alpha_2 = \alpha_3 = 0$, $\alpha_4 = 1$, $\alpha_5 = 1$, while $$+ + + + 1 1$$ corresponds to the solution $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$, $\alpha_5 = 2$. The number of such solutions is the the number of ways we can place four addition signs in a row of two ones, which is $$\binom{2 + 4}{4} = \binom{6}{4}$$ since we must choose which four of the six positions required for two ones and four addition signs will be filled with addition signs.

Equation 2 differs from equation 2 only in the names of the variables, so it too has $$\binom{6}{4}$$ solutions in the nonnegative integers.

Equation 3 has five solutions, depending on which of the five variables is equal to 1.

Equation 4 also has five solutions for the same reason.

Since the distribution of the powers of $2$, $3$, $5$, and $7$ can be made independently, the number of solutions of the equation $x_1x_2x_3x_4x_5 = 1260$ in the positive integers is $$\binom{6}{4}\binom{6}{4}\binom{5}{1}\binom{5}{1} = 5625$$

How many solutions does the equation $2x + 3y + 4z = 24$ have in the positive integers?

As Saaquib Mahmood observed, $$2x + 3y + 4z = 24 \implies 3y = 24 - 2x - 4z$$ so $y$ must be even. Let $y = 2y'$. Then \begin{align*} 3(2y') & = 24 - 2x - 4z\\ 6y' & = 24 - 2x - 4z\\ 3y' & = 12 - x - 2z \tag{1} \end{align*} Since we seek a solution in the positive integers, we require that $1 \leq y' \leq 3$.

If $y' = 1$, then we obtain \begin{align*} 3 & = 12 - x - 2z\\ x & = 9 - 2z \end{align*} Choosing a value for $z$ determines the value of $x$. Since $x$ and $z$ are positive integers, $1 \leq z \leq 4$, so the equation has four solutions in he positive integers.

If $y' = 2$, then we obtain \begin{align*} 6 & = 12 - x - 2z\\ x & = 6 - 2z \end{align*} which has two solutions in the positive integers since $1 \leq z \leq 2$.

If $y ' = 3$, then we obtain \begin{align*} 9 & = 12 - x - 2z\\ x & = 3 - 2z \end{align*} which has one solution in the positive integers since $z = 1$.

Thus, we obtain $1 + 2 + 4 = 7$ solutions in the positive integers.

The given answer is the number of solutions in the nonnegative integers.

How many solutions does the equation $2x + 3y + 4z = 24$ have in the nonnegative integers?

As above, we obtain $$3y' = 12 - x - 2z$$ However, now $y'$ can assume the values $0, 1, 2, 3, 4$.

If $y' = 0$, then $$x = 12 - 2z$$ which has seven solutions in the nonnegative integers since $0 \leq z \leq 6$.

If $y' = 1$, then $$x = 9 - 2z$$ which has five solutions in the nonnegative integers since $0 \leq z \leq 4$.

If $y' = 2$, then $$x = 6 - 2z$$ which has four solutions in the nonnegative integers since $0 \leq z \leq 3$.

If $y' = 3$, then $$x = 3 - 2z$$ which has two solutions in the nonnegative integers since $0 \leq z \leq 1$.

If $y' = 4$, then $$x = 0 - 2z$$ which has one solution in the nonnegative integers since $x = z = 0$.

Thus, there are $7 + 5 + 4 + 2 + 1 = 19$ solutions in the nonnegative integers.

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In Part (ii), you cannot actually put $z= 6$ because in that case you'll have to put $x= y = 0$. Thus $z$ can only be from the set $\{ 1, 2, 3, 4, 5 \}$.

Using the same logic, $y$ can only be from the set $\{ 1, 2, 3, 4, 5, 6, 7 \}$, and $z$ can only be from the set $\{ 1, 2, \ldots, 11 \}$.

Moreover, since $$ 3y = 24 - 4z-2x = 2 ( 12 - 2z-x), $$ so your $y$ must be even. Thus $y$ can only be from the set $\{ 2, 4, 6 \}$.

Now as $$2(x + 2z) = 2x + 4z = 24 - 3y = 3(8 - y),$$ so $x + 2z$ must be a multiple of $3$.

Thus $x \in \{ 1, 2, \ldots, 11 \}$, $z \in \{ 1, 2, 3, 4, 5 \}$, and $x+2z$ is a multiple of $3$.

Thus for $x= 1$, $z \in \{ 1, 4 \}$; for $x= 2$, $z \in \{ 2, 5 \}$; for $x= 3$, $z \in \{ 3 \}$; for $x = 4$, $z \in \{ 1, 4 \}$; for $x= 5$, $z \in \{ 2, 5 \}$; for $x = 6$, $z \in \{ 3 \}$; for $x = 7$, $z \in \{ 1, 4, \}$; for $x= 8$, $z \in \{ 2, 5 \}$; for $x= 9$, $z \in \{ 3 \}$; for $x= 10$, $z \in \{ 1, 4 \}$; and for $x= 11$, $z \in \{ 2, 5 \}$.

This reasoning narrows down your possibilities considerably.

Hope this helps.

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  • $\begingroup$ Can I also do it using combination? If yes, how ? $\endgroup$ – Entrepreneur Jun 24 '18 at 18:08

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