Question

Product of all the even divisors of N = 1000, is

Answer 1

64 × 10^18

Answer 2

To find the product of all even divisors of N = 1000, we follow these steps:

1. Prime Factorization of N: First, express N in its prime factorized form: $N = 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3$.

2. Form of Divisors: Any divisor of N is of the form $d = 2^a \times 5^b$, where $0 \le a \le 3$ and $0 \le b \le 3$.

3. Identify Even Divisors: For a divisor to be even, its prime factor 2 must have a power of at least 1. So, for even divisors, the exponent 'a' must satisfy $1 \le a \le 3$. The exponent 'b' can still range from $0 \le b \le 3$. Thus, the even divisors are of the form $2^a \times 5^b$ where $a \in \{1, 2, 3\}$ and $b \in \{0, 1, 2, 3\}$.

4. Count of Even Divisors: The number of choices for 'a' is 3 (1, 2, 3). The number of choices for 'b' is 4 (0, 1, 2, 3). The total number of even divisors is $3 \times 4 = 12$.

5. Calculate the Product of Even Divisors (P): Let P be the product of all even divisors. $P = \prod_{a=1}^{3} \prod_{b=0}^{3} (2^a \times 5^b)$

   To calculate this product, we sum the powers of each prime factor.

   • Contribution from prime factor 2: For each value of 'a' (1, 2, 3), there are 4 corresponding values of 'b'. This means $2^1$ appears 4 times, $2^2$ appears 4 times, and $2^3$ appears 4 times in the product. The total power of 2 in P is $4 \times (1 + 2 + 3) = 4 \times 6 = 24$. So, the factor involving 2 is $2^{24}$.

   • Contribution from prime factor 5: For each value of 'b' (0, 1, 2, 3), there are 3 corresponding values of 'a'. This means $5^0$ appears 3 times, $5^1$ appears 3 times, $5^2$ appears 3 times, and $5^3$ appears 3 times in the product. The total power of 5 in P is $3 \times (0 + 1 + 2 + 3) = 3 \times 6 = 18$. So, the factor involving 5 is $5^{18}$.

   Therefore, the product P is $P = 2^{24} \times 5^{18}$.

6. Simplify the Result: We can rewrite the expression to involve powers of 10: $P = 2^{18} \times 2^6 \times 5^{18}$ $P = (2 \times 5)^{18} \times 2^6$ $P = 10^{18} \times 64$ $P = 64 \times 10^{18}$