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Algebra Level 3

If a , b , c a,b,c are natural numbers and if
a + b + c + a b + b c + a c + a b c = 1000 a + b+ c + ab + bc+ac + abc = 1000
Find the value of a + b + c 4 \dfrac{ a+b+c}4 .


The answer is 7.

Raven Herd by Raven Herd , 21, India

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1 solution

a + b + c + a b + b c + c a + a b c + 1 = 1000 + 1 ⟹ ( a + 1 ) ( b + 1 ) ( c + 1 ) = 7.11.13 \begin{aligned} a + b+ c+ab+bc+ca+abc+\color{#3D99F6}{1} = 1000 + \color{#3D99F6}{1}\implies (a+1)(b+1)(c+1)=7.11.13 \end{aligned}

It is symmetric in a,b,c ∈ N ⟹ a + 1 = 7 , b + 1 = 11 , c + 1 = 13 \text{It is symmetric in a,b,c } \in \mathbb N\implies a+1=7,b+1=11,c+1=13

Solving we get a = 6 , b = 10 , c = 12 a=\color{#D61F06}{6},b=\color{#D61F06}{10},c=\color{#D61F06}{12}

a + b + c 4 = 6 + 10 + 12 4 = 7 \frac{a+b+c}{4} = \frac{6+10+12}{4} = \boxed{7}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice observation! But there is another way too.

Raven Herd - 3 years, 9 months ago

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Show us the way, master!

Adhiraj Dutta - 1 year, 6 months ago

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