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Algebra Level 2

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


The answer is 504.

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Raihan Fauzan by Raihan Fauzan , 20, Indonesia

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1 solution

Raihan Fauzan
May 4, 2016

Sum all the equations. You will get

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

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

a + b + c + d + e = 84 4 = 21 a + b + c + d + e = \frac{84}{4} = 21

Use the equation above to find the value of each variables.

You will get

a = 1 , b = 3 , c = 4 , d = 6 , e = 7 a = 1, b = 3, c = 4, d = 6, e = 7

So,

a a b b c c d d e e = 1 × 3 × 4 × 6 × 7 1 \times 3 \times 4 \times 6 \times 7 = 504 \boxed{504}

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