To solve this problem, we need to find the remainder when the product

$21 \times 2^2 \times 3^2 \times 7^4 \times 5^2$

is divided by 20.

Step-by-Step Solution

1. Prime Factorize 20: The number 20 can be factorized as: $20 = 2^2 \times 5$

2. Calculate the Product Modulo 20: We’ll use modular arithmetic to simplify the calculation, by finding the remainder of each factor modulo 20 and then multiplying.

   • Factor 21: $21 \mod 20 = 1$.
   • Factor $2^2$: $2^2 = 4$, and $4 \mod 20 = 4$.
   • Factor $3^2$: $3^2 = 9$, and $9 \mod 20 = 9$.
   • Factor $7^4$: Calculate $7^4 \mod 20$: $7^2 = 49 \Rightarrow 49 \mod 20 = 9$ Then, $7^4 = (7^2)^2 = 9^2 = 81 \Rightarrow 81 \mod 20 = 1$ So, $7^4 \mod 20 = 1$.
   • Factor $5^2$: $5^2 = 25$, and $25 \mod 20 = 5$.

3. Combine the Results: Now, we multiply the results of each factor modulo 20: $1 \times 4 \times 9 \times 1 \times 5 = 180$ Calculate $180 \mod 20$: $180 \div 20 = 9 \text{ remainder } 0$ So, the remainder is $0$.

Answer

The remainder when $21 \times 2^2 \times 3^2 \times 7^4 \times 5^2$ is divided by 20 is:

$\boxed{0}$

Would you like further details or have any questions?

Related Questions:

1. What is the remainder when $21 \times 2^3 \times 3 \times 7^2$ is divided by 20?
2. How would the solution change if we divided by 25 instead of 20?
3. Can modular arithmetic be used to simplify calculations for large exponents?
4. Why is it useful to break down large products using modular arithmetic?
5. How does the Chinese Remainder Theorem apply to problems like this?

Tip:

Using modular arithmetic with factorization can greatly simplify division remainder problems, especially with large exponents.