Question

The number of positive integral solution of the equation $x_1.x_2.x_3.x_4.x_5 = 1050$ is

• A. 1800
• B. 1400
• C. 1600
• D. 1875

Answer 1

Answer: (D)

1875

Answer 2

The prime factorization of 1050 is $2^1 \times 3^1 \times 5^2 \times 7^1$. We need to find the number of positive integral solutions to $x_1.x_2.x_3.x_4.x_5 = 1050$. This is equivalent to distributing the prime factors among the 5 variables. For a prime factor $p^a$ and $k$ variables, the number of ways to distribute it is $\binom{a+k-1}{k-1}$.

For $2^1$ and $k=5$: $\binom{1+5-1}{5-1} = \binom{5}{4} = 5$. For $3^1$ and $k=5$: $\binom{1+5-1}{5-1} = \binom{5}{4} = 5$. For $5^2$ and $k=5$: $\binom{2+5-1}{5-1} = \binom{6}{4} = 15$. For $7^1$ and $k=5$: $\binom{1+5-1}{5-1} = \binom{5}{4} = 5$.

The total number of solutions is the product of these values: $5 \times 5 \times 15 \times 5 = 1875$.