Question

Доказать, что 641^3 - 541^3 делится на 50.

Answer 1

A^3 - b^3 = (a - b)(a^2 + ab + b^2)

641^3 - 541^3 = (641 - 541)(641^2 + 641*541 + 541^2) = 
= 100*(641^2 + 641*541 + 541^2)

Так как 100 делится на 50, то и все произведение делится на 50, ч т д 

Answer 2

641³ - 541³ = (641 - 541)(641² + 641 * 541 + 541²) =100 * (641² + 641*541 + 541²)
Результат второй скобки можно не подсчитывать, так как первый множитель равен 100, который делится на 50, а если один из множителей делится на 50, то и всё произведение делится на 50.

Answer 3

To prove that 641^3 - 541^3 is divisible by 50, we need to show that the difference (641^3 - 541^3) is divisible by both 2 and 5, which are the prime factors of 50.

Let's start by expressing the expression (641^3 - 541^3) in a factored form:

(641^3 - 541^3) = (641 - 541)(641^2 + 641 * 541 + 541^2)

Now, we need to check if both factors are divisible by 2 and 5.

1. Divisibility by 2: Both 641 and 541 are odd numbers. Therefore, (641 - 541) is an even number since the difference of two odd numbers is even. As a result, the first factor (641 - 541) is divisible by 2.

2. Divisibility by 5: To check if (641^2 + 641 * 541 + 541^2) is divisible by 5, we need to check the remainder when this expression is divided by 5.

Now, calculate the value of the expression (641^2 + 641 * 541 + 541^2):

(641^2 + 641 * 541 + 541^2) = (411,681 + 346,681 + 292,681) = 1,051,043

Since 1,051,043 is not divisible by 5 (the remainder is 3), the second factor (641^2 + 641 * 541 + 541^2) is not divisible by 5.

Now, since the first factor (641 - 541) is divisible by 2, and the second factor (641^2 + 641 * 541 + 541^2) is divisible by neither 2 nor 5, the entire expression (641^3 - 541^3) is divisible by 2 but not by 5.

Since 50 = 2 * 5, and the expression is divisible by 2, we can conclude that (641^3 - 541^3) is divisible by 50.