Equation - Algebra

Let $a+b+c+d+e=k$ . Then the five equations reduce to:

$\begin{cases} a(k-a) = 128 & \implies k = \dfrac {128}a+a = \dfrac {2^7}a+a \\ b(k-b) = 155 & \implies k = \dfrac {155}b+b = \dfrac {5\cdot 31}b+b \\ c(k-c) = 203 & \implies k = \dfrac {203}c+c = \dfrac {7\cdot 29}c+c \\ d(k-d) = 243 & \implies k = \dfrac {243}d+d = \dfrac {3^5}d+d \\ e(k-e) = 275 & \implies k = \dfrac {275}e+e = \dfrac {5^2\cdot 11}e+e \end{cases}$

Let us check the possible values of $k$ using the $k=\dfrac {5\cdot 31}b+b$ and $k=\dfrac {7\cdot 29}c+c$ .

$\begin{cases} b = 1, & k = 156 \\ b = 5, & k = 36 \\ b = 31, & k = 36 \\ b = 155, & k = 156 \end{cases} \quad \begin{cases} c = 1, & k = 204 \\ b = 7, & k = 36 \\ b = 29, & k = 36 \\ b = 203, & k = 204 \end{cases}$

This shows that $k=36$ and that $b=5$ and $c=7$ , because the other values are too large. We can find the other values of $a$ , $d$ and $e$ as follows:

$\begin{aligned} \frac {128}a+a & = k = 36 \\ \implies a^2 - 36a + 128 & = 0 \\ (a-4)(a-32) & = 0 \\ \implies a & = 4 \end{aligned}$

Similarly,

$\begin{aligned} d^2 - 36d + 243 & = 0 \\ (d-9)(d-27) & = 0 \\ \implies d = 9 \end{aligned}$

$\begin{aligned} e^2 - 36e + 275 & = 0 \\ (e-11)(e-25) & = 0 \\ \implies e = \boxed{11} \end{aligned}$

Let us check the answer $k = a+b+c+d+e = 4+5+7+9+11 = 36$ .

After simplifying: - 1. a(b+c+d+e) = 128 = 27

• 1. b(a+c+d+e) = 155 = 5 x 31
• 1. c(a+b+d+e) = 203 = 7 x 29
• 1. d(a+b+c+e) = 243
• 1. e(a+b+c+d) = 275

Equation 2 can reached the solution that b = 5, a+c+d+e = 31 or b = 31, a+c+d+e=5

Equations 3 can reach the conclusion that c = 7, a+b+d+e = 29 or c= 29, a+b+d+e = 7

b = 5 and c = 7 are proven true => a+d+e = 24 => d+e = 24 - a

a(5+7+24=-a) = 128

a=32 (false) or a=4

d+e=20

e=20-d

d(47+5+e)=243

(d-27)(d-9) = 0

d=27 (rejected), d=9

e=11

(check the result with the last equation)

• Answer: a=4, b=5, c=7, d=9, e=11