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Let's go through these questions one by one:

### 1. **Montrer que la somme de trois entiers consécutifs est divisible par 3.**
We can represent the three consecutive integers as \( n \), \( n+1 \), and \( n+2 \).
The sum of these integers is:

\[
n + (n+1) + (n+2) = 3n + 3 = 3(n+1)
\]

Since \( 3(n+1) \) is divisible by 3, the sum of three consecutive integers is always divisible by 3.

### 2. **Montrer de même que la somme de cinq entiers consécutifs est divisible par 5 et que la somme de sept entiers consécutifs est divisible par 7.**
#### For five consecutive integers:
Represent the integers as \( n \), \( n+1 \), \( n+2 \), \( n+3 \), and \( n+4 \).
The sum is:

\[
n + (n+1) + (n+2) + (n+3) + (n+4) = 5n + 10 = 5(n+2)
\]

Thus, the sum is divisible by 5.

#### For seven consecutive integers:
Represent the integers as \( n \), \( n+1 \), \( n+2 \), \( n+3 \), \( n+4 \), \( n+5 \), and \( n+6 \).
The sum is:

\[
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6) = 7n + 21 = 7(n+3)
\]

Thus, the sum is divisible by 7.

### 3. **Montrer que la somme de six entiers consécutifs n'est jamais divisible par 6.**
Represent the six consecutive integers as \( n \), \( n+1 \), \( n+2 \), \( n+3 \), \( n+4 \), and \( n+5 \).
The sum is:

\[
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) = 6n + 15
\]

Since \( 6n + 15 = 6n + 3 \times 5 \), the sum will not be divisible by 6 because of the remainder 15, which leaves 3 modulo 6.

### 4. **Plus généralement, on étudie la somme de \( k \) entiers consécutifs en notant \( n \) le plus petit d'entre eux.**
Let the integers be \( n \), \( n+1 \), \( n+2 \), ..., \( n+k-1 \).

#### 4a. **Réduire cette somme à l'aide des suites arithmétiques.**
The sum of the first \( k \) consecutive integers starting at \( n \) is:

\[
S_k = n + (n+1) + (n+2) + \dots + (n+k-1)
\]
\[
S_k = k \cdot n + \left(0 + 1 + 2 + \dots + (k-1)\right)
\]

The sum of the first \( k-1 \) integers \( 0 + 1 + 2 + \dots + (k-1) \) is \( \frac{(k-1)k}{2} \), so:

\[
S_k = k \cdot n + \frac{k(k-1)}{2}
\]

#### 4b. **Montrer que cette somme est divisible par \( k \) si et seulement si \( k \) est impair.**
The sum is:

\[
S_k = k \cdot n + \frac{k(k-1)}{2}
\]

For this to be divisible by \( k \), the second term \( \frac{k(k-1)}{2} \) must also be divisible by \( k \). This happens if \( k \) is odd, as \( k-1 \) is even, making \( \frac{k-1}{2} \) an integer, which ensures the entire expression is divisible by \( k \).

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Would you like a deeper explanation for any part of this solution?

Here are five related questions:
1. How would the formula change for \( k \)-consecutive numbers starting from any non-integer?
2. What if the sequence skips values, e.g., taking only even integers?
3. How would we handle divisibility for a sequence of powers (e.g., squares or cubes)?
4. Is there a general rule for when sums of any sequence of integers are divisible by the count?
5. How can these sums be related to the properties of modular arithmetic?

Tip: When working with consecutive integers, grouping terms can help reveal patterns of divisibility.

Montrer que la somme de trois entiers consécutifs est divisible par 3. Montrer de même que la somme de cinq entiers consécutifs est divisible par 5 et que la somme de sept entiers consécutifs est divisible par 7. Montrer que la somme de six entiers consécutifs n'est jamais divisible par 6. Plus généralement, on étudie la somme de k entiers consécutifs en notant n le plus petit d'entre eux. Réduire cette somme à l'aide des suites arithmétiques. Montrer que cette somme est divisible par k si et seulement si k est impair.

Learn how to prove the divisibility of sums of consecutive integers for specific cases such as 3, 5, 7, and more generally k integers. This solution covers key concepts in divisibility and arithmetic sequences, suitable for high school math students in Grades 10-12.

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