Number Theory or Combinatorics

Number Theory Level 3

Find the number of integral solutions of the equation x 1 x 2 x 3 = 108 x_{1} x_{2} x_{3} = 108 such that the solution triplet ( x 1 , x 2 , x 3 ) (x_1,x_2, x_3) must contain two negative and one positive integers.


The answer is 180.

Shreyansh Mukhopadhyay by Shreyansh Mukhopadhyay , 18, India

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1 solution

As, 108 = 2 2 × 3 3 108=2^{2}\times3^{3}

x 1 × x 2 × x 3 = 108 = 2 2 × 3 3 x_{1}\times x_{2}\times x_{3} =108=2^{2}\times3^{3}

So the number of positive integral solutions are

( 2 + 3 − 1 C 3 − 1 ) × ( 3 + 3 − 1 C 3 − 1 ) = ( 4 C 2 ) × ( 5 C 2 ) = 60 ( ^{2+3-1}C_{3-1})\times ( ^{3+3-1}C_{3-1})=( ^{4}C_{2})\times( ^{5}C_{2})=60

Now, number of possible combinations of three numbers such that two of them are negative is 3

Answer is 60 × 3 = 180 60\times3=\boxed{180}

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