1

Problem:

If $a,b,c,d,e>0$, and $a,b,c,d,e$ are all positive integers, on the condition that $b>a$, solve the following system of equations:

\begin{align} abcde &= 240\ 240\ \ \ (1)\\ abcd+e &= 17\ 174\ \ \ \ \ (2)\\ abc+d+e &= 1\ 347\ \ \ \ \ \ \ (3)\\ ab+c+d+e &= 149\ \ \ \ \ \ \ \ \ \ (4)\\ a+b+c+d+e &= 60\ \ \ \ \ \ \ \ \ \ \ \ (5) \end{align}

Background Information on the Problem:

Funnily enough, my brother made up this problem with random numbers, as he claims, and he absolutely refuses to tell me the solution unless if I agree that I'm stupid. But I'm not.

My Working:

Basically, what I did was look at equation $(1)$ and try to find factors of $240\ 240$, since $a,b,c,d,e$ are all positive integers. I have no idea how to proceed from here. I could look at equations $(1)$ and $(2)$, and use guess and check to find $e$, but that would be extremely time consuming.

Could anyone please help me with this problem? Thank you.

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3 Answers3

1

Let $x_1 = abcd$ & $y_1 = e$ ;
We have $x_1y_1 = 240240$ [[ Eq 1 ]] & $x_1+y_1 = 17174$ [[ Eq 2 ]]
This gives us $x_1 = 17160$ and $y_1 = 14$

Let $x_2 = abc$ & $y_2 = d$ ;
We have $x_2y_2+14 = 17174$ [[ Eq 2 ]] & $x_2+y_2+14 = 1347$ [[ Eq 3 ]]
This gives us $x_2 = 1320$ and $y_2 = 13$

Let $x_3 = ab$ & $y_3 = c$ ;
We have $x_3y_3+13+14 = 1347$ [[ Eq 3 ]] & $x_3+y_3+13+14 = 149$ [[ Eq 4 ]]
This gives us $x_3 = 110$ and $y_3 = 12$

Let $x_4 = a$ & $y_4 = b$ ;
We have $x_4y_4+12+13+14 = 149$ [[ Eq 4 ]] & $x_4+y_4+12+13+14 = 60$ [[ Eq 5 ]]
This gives us $x_4 = 10$ and $y_4 = 11$

$(a,b,c,d,e) = (10,11,12,13,14)$

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Prem
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1

By subtracting from equation $(4)$ equation $(5)$, we have:

$$ab-a-b-89=0\leftrightarrow a=\frac{b+89}{b-1}=1+\frac{90}{b-1}$$

Because $a$ must be an integer and $b>a$, it must be: $$b=11\,\,(a=10)\,\,\lor\,\,b=16\,\,(a=7)\,\,\lor\,\,b=31\,\,(a=4)\,\,\lor\,\,b=46\,\,(a=3)\,\,\lor\,\,b=91\,\,(a=2)$$

Now, observe that $240240\ (\text{mod}\ 31)\neq 0$ and $240240\ (\text{mod}\ 46)\neq 0$, so $b=31$ and $b=46$ can't be accepted. $b=7$, $b=11$ and $b=91$ are possible solutions.

Substracting from the $(3)$ the $(4)$ we have:

$$abc-ab-c=1198$$

Now, you know the value of $ab$, so there are three possibilities: $$112c-112-c=1198\leftrightarrow c=\frac{1310}{111}\not\in \mathbb{N}\,\,\lor\,\,182c-182-c=1198\leftrightarrow c=\frac{1380}{181}\not \in\mathbb{N}\,\,\lor\,\, 110c-110-c=1198\leftrightarrow c=\frac{1308}{109}=12$$

So, by now we have: $$a=10\,\, b=11\,\, c=12$$

Substracting the $(3)$ from the $(2)$ we have:

$$1320d-d-1320=15827\leftrightarrow d=\frac{17147}{1319}=13$$

Finally, from the first equation: $$17160e=240240\leftrightarrow e=14$$

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Matteo
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0

After posting the question for a while, I have found another solution. Similar to Prem's solution, I first multiply equation $(2)$ by $e$ and substitute equation $(1)$. Note: I calculated $\sqrt{293\ 985\ 316}$ by using this strange formula to do any square root from this link: https://www.youtube.com/watch?v=HBdVVFqTrUU.

This gives

\begin{align} abcde+e^2&=17\ 174e\\ e^2-17\ 174e+abcde&=0\\ e^2-17\ 174e+240\ 240&=0\\ e&=\frac{-(-17\ 174)\pm\sqrt{(17\ 174)^2-4\times240\ 240}}{2}\\ e&=\frac{17\ 174\pm\sqrt{294\ 946\ 276-960\ 960}}{2}\\ e&=\frac{17\ 174\pm\sqrt{293\ 985\ 316}}{2}\\ e&=\frac{17\ 174\pm17\ 146}{2}\\ e&=\frac{28}{2}\ \text{or}\ \frac{34\ 320}{2}\\ e&=14\ \text{or}\ 17\ 160 \end{align}

But $e\neq17\ 160$ because of equation $(5)$, and the fact that $a,b,c,d,e$ are all positive. Therefore, $e=14$. Using the quadratic formula a few more times, we get $(a,b,c,d,e)=(10,11,12,13,14)$.

Note: Instead of using the quadratic formula, we could have factorised $240\ 240$ when evaluating $e$, which is much easier. Factorisation works for the entire process in this solution, and becomes increasing easier as the numbers become smaller.

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