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What is the total number of positive integer solution to the equation

$(x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=15$

a) 20 $\qquad$ $\qquad$ $\qquad$ $\qquad$ b) 18

c) 10 $\qquad$ $\qquad$ $\qquad$ $\qquad$ d) 4

Sol. $(1,1,1,)(2,1,1,1)$, $(1,1,1,)(1,2,1,1)$

$(1,1,1,)(1,1,2,1)$ , $(1,1,1,)(1,1,1,2)$

Hence there are only 4 positive integer solution to the equation.

Is there any other way to approach this question or this is only one which i have done?

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    $\begingroup$ We have $15=3\cdot 5$ as the only possible factorization, because $x_1+x_2+x_3\ge 3$, and $y_1+\cdots +y_4\ge 4$. So we are done, and there are $4$ possibilities. $\endgroup$ Jun 30, 2016 at 15:40

3 Answers 3

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Since $15 = 1\cdot15 = 3 \cdot 5$, the only possibilities are

$$(x_1 + x_2 + x_3, y_1 + y_2 + y_3 + y_4) = (1, 15), (15, 1), (3, 5), (5, 3)$$

Since the variables are all at least $1$, we can rule out the case where $x_1 + x_2 + x_3 = 1$ or where $y_1 + y_2 + y_3 + y_4 = 1, 3$ for the minimum values of each sum exceeds the respective assignments.

This leaves us with

$$(x_1 + x_2 + x_3, y_1 + y_2 + y_3 + y_4) = (3, 5)$$

Here we see that the only possible assignment for $x_1, x_2, x_3$ is $x_1 = x_2 = x_3 = 1$. So we only need to count the number of solutions to

$$y_1 + y_2 + y_3 + y_4 = 5$$

which is possible via the Stars and Bars method. In fact, you don't even need this at all if you consider the fact that $y_1, y_2, y_3, y_4 \ge 1$.

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  • $\begingroup$ I never get when and where to apply Stars and Bars method while solving problem. $\endgroup$ Jun 30, 2016 at 16:13
  • $\begingroup$ @slowbutkeenlearner It is the go-to technique when solving things like counting the number of non-negative solutions to $x_1 + x_2 + x_3 = 100$. $\endgroup$
    – Yiyuan Lee
    Jul 1, 2016 at 11:29
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$15 = 5 * 3$, So $(x_1 + x_2 + x_3) = 3$ or $5$ same for the other one. $(y_1 + y_2 + y_3 + y_4) = 3$ has no solution. $\therefore$ $(x_1 + x_2 + x_3) = 3$ and $(y_1 + y_2 + y_3 + y_4) = 5$. Now you know $x_1, x_2, x_3$ can't be greater than $1$. $\therefore$ the only solution for it is $(1 + 1 + 1)$. In $(y_1 + y_2 + y_3 + y_4) = 5$, only one variable can be $2$, rest have to $1$. Now we can use permutation here as we have to arrange $2$ in $4$ possible position. $$P = {n!\over (n-r)!}$$ $$P = {4!\over (4-1)!} = 4$$

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Since all the integers are positive, we can only have $$x_1+x_2+x_3=3$$ $$y_1+y_2+y_3+y_4=5$$ In the first equation, clearly $x_1=x_2=x_3=1$ In the second,only one is equal to $2$ and everything else is equal to 1 Hence, the solutions: $$(1,1,1)(2,1,1,1)$$ $$ (1,1,1)(1,2,1,1)$$ $$(1,1,1)(1,1,2,1)$$ $$(1,1,1)(1,1,1,2)$$

4 possibilities.

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