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$$\large \begin{cases} a + b + c + d = 14 \\ a + b + c + e = 15 \\ a + b + d + e = 17 \\ a + c + d + e = 18 \\ b + c + d + e = 20\end{cases}$$

If a,b,c,d and e satisfy the system of equations above, find the product abcde .

Jul 9, 2020

#1
+25600
+2

help
$$\large \begin{cases} a + b + c + d = 14 \\ a + b + c + e = 15 \\ a + b + d + e = 17 \\ a + c + d + e = 18 \\ b + c + d + e = 20\end{cases}$$
If a,b,c,d and e satisfy the system of equations above, find the product $$abcde$$ .

$$\begin{array}{|lrcll|} \hline (1) & a + b + c + d &=& 14 \\ (2) & a + b + c + e &=& 15 \\ (3) & a + b + d + e &=& 17 \\ (4) & a + c + d + e &=& 18 \\ (5) & b + c + d + e &=& 20 \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (2)-(1) & e-d &=& 15-14 \\ & e-d &=& 1 \\ & \mathbf{ d} &=& \mathbf{e-1} \\\\ (3)-(1) & e-c &=& 17-14 \\ & e-c &=&3 \\\\ & \mathbf{ c} &=& \mathbf{e-3} \\\\ (4)-(1) & e-b &=& 18-14 \\ & e-b &=& 4 \\\\ & \mathbf{ b} &=& \mathbf{e-4} \\\\ (5)-(1) & e-a &=& 20-14 \\ & e-a &=& 6 \\ & \mathbf{ a} &=& \mathbf{e-6} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (3)-(2) & d-c &=& 17-15 \\ & d-c &=& 2 \\\\ & \mathbf{ c} &=& \mathbf{d-2} \\\\ (4)-(2) & d-b &=& 18-15 \\ & d-b &=& 3 \\\\ & \mathbf{ b} &=& \mathbf{d-3} \\\\ (5)-(2) & d-a &=& 20-15 \\ & d-a &=& 5 \\ & \mathbf{ a} &=& \mathbf{d-5} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (4)-(3) & c-b &=& 18-17 \\ & c-b &=& 1 \\\\ & \mathbf{ b} &=& \mathbf{c-1} \\\\ (5)-(3) & c-a &=& 20-17 \\ & c-a &=& 3 \\ & \mathbf{ a} &=& \mathbf{c-3} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{ d=e-1} \\ & \mathbf{ c=e-3} \\ & \mathbf{ b=e-4} \\ & \mathbf{ a=e-6} \\ \hline (1) & a + b + c + d &=& 14 \\ & e-6+e-4+e-3+e-1 &=& 14 \\ & 4e-14 &=& 14 \\ & 4e &=& 28 \quad| \quad : 4 \\ & \mathbf{e} &=& \mathbf{7} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{ c=d-2} \\ & \mathbf{ b=d-3} \\ & \mathbf{ a=d-5} \\ \hline (2) & a + b + c + e &=& 15 \quad | \quad e=7 \\ & a + b + c + 7 &=& 15 \\ & a + b + c &=& 15-7 \\ & a + b + c &=& 8 \\ & d-5+d-3+d-2 &=& 8 \\ & 3d-10 &=& 8 \\ & 3d &=& 18 \quad| \quad : 3 \\ & \mathbf{d} &=& \mathbf{6} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{ b=c-1} \\ & \mathbf{ a=c-3} \\ \hline (3) & a + b + d + e &=& 17 \quad | \quad e=7, \ d=6 \\ & a + b + 6 + 7 &=& 17 \\ & a + b + 13 &=& 17 \\ & a + b &=& 17-13 \\ & c-3 + c-1 &=& 4 \\ & 2c-4 &=& 4 \\ & 2c &=& 8 \quad| \quad : 2 \\ & \mathbf{c} &=& \mathbf{4} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (4) & a + c + d + e &=& 18 \quad | \quad e=7, \ d=6,\ c=4 \\ & a + 4 + 6 + 7 &=& 18 \\ & a + 17 &=& 18 \\ & a &=& 18-17 \\ & \mathbf{a} &=& \mathbf{1} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (5) & b + c + d + e &=& 20 \quad | \quad e=7, \ d=6,\ c=4 \\ & b + 4 + 6 + 7 &=& 20 \\ & b + 17 &=& 120 \\ & b &=& 20-17 \\ & \mathbf{b} &=& \mathbf{3} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline abcde &=& 1*3*4*6*7 \\ \mathbf{abcde} &=& \mathbf{504} \\ \hline \end{array}$$

Jul 9, 2020
#2
+8352
0

4(a + b + c + d + e) = 84

a + b + c + d + e = 21 --- (*)

Subtracting each equation from (*),

e = 21 - 14 = 7

d = 21 - 15 = 6

c = 21 - 17 = 4

b = 21 - 18 = 3

a = 21 - 20 = 1

abcde = 1(3)(4)(6)(7) = 504

Jul 9, 2020