Question

N(n+5)-(n-3)(n+2) делится на 6

Answer 1

Решение в приложении

Answer 2

To determine if the expression N(n+5) - (n-3)(n+2) is divisible by 6, we can simplify the expression and check if it satisfies the divisibility rule for 6.

Let's expand and simplify the expression: N(n+5) - (n-3)(n+2) = Nn + 5N - (n^2 - 3n + 2n - 6) = Nn + 5N - (n^2 - n - 6) = Nn + 5N - n^2 + n + 6

Now, we need to check if this expression is divisible by 6. For a number to be divisible by 6, it must be divisible by both 2 and 3.

Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8). In our expression, we don't have any restrictions on N, so we can ignore it. Let's focus on the term -n^2 + n since it affects the divisibility by 2.

If we factor out -n from this term, we get: -n^2 + n = -n(n - 1)

For a number to be divisible by 2, at least one of the factors must be divisible by 2. Since n(n - 1) represents a product of two consecutive integers, one of them will always be even. Therefore, -n^2 + n is divisible by 2.

Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. In our expression, we don't have any restrictions on N, so we can ignore it. Let's focus on the term -n^2 + n + 6 since it affects the divisibility by 3.

To check the divisibility by 3, we need to consider the sum of the coefficients of n: 1 + (-1) = 0

Since the sum of the coefficients is 0, -n^2 + n + 6 is divisible by 3.

Combining both divisibility tests, we conclude that the expression N(n+5) - (n-3)(n+2) is divisible by 6 for any value of N and n.