Math Problem Statement

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.

Solution

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 nn, n+1n+1, and n+2n+2. The sum of these integers is:

n+(n+1)+(n+2)=3n+3=3(n+1)n + (n+1) + (n+2) = 3n + 3 = 3(n+1)

Since 3(n+1)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 nn, n+1n+1, n+2n+2, n+3n+3, and n+4n+4. The sum is:

n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10=5(n+2)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 nn, n+1n+1, n+2n+2, n+3n+3, n+4n+4, n+5n+5, and n+6n+6. The sum is:

n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)=7n+21=7(n+3)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 nn, n+1n+1, n+2n+2, n+3n+3, n+4n+4, and n+5n+5. The sum is:

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

Since 6n+15=6n+3×56n + 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 kk entiers consécutifs en notant nn le plus petit d'entre eux.

Let the integers be nn, n+1n+1, n+2n+2, ..., n+k1n+k-1.

4a. Réduire cette somme à l'aide des suites arithmétiques.

The sum of the first kk consecutive integers starting at nn is:

Sk=n+(n+1)+(n+2)++(n+k1)S_k = n + (n+1) + (n+2) + \dots + (n+k-1) Sk=kn+(0+1+2++(k1))S_k = k \cdot n + \left(0 + 1 + 2 + \dots + (k-1)\right)

The sum of the first k1k-1 integers 0+1+2++(k1)0 + 1 + 2 + \dots + (k-1) is (k1)k2\frac{(k-1)k}{2}, so:

Sk=kn+k(k1)2S_k = k \cdot n + \frac{k(k-1)}{2}

4b. Montrer que cette somme est divisible par kk si et seulement si kk est impair.

The sum is:

Sk=kn+k(k1)2S_k = k \cdot n + \frac{k(k-1)}{2}

For this to be divisible by kk, the second term k(k1)2\frac{k(k-1)}{2} must also be divisible by kk. This happens if kk is odd, as k1k-1 is even, making k12\frac{k-1}{2} an integer, which ensures the entire expression is divisible by kk.


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

Here are five related questions:

  1. How would the formula change for kk-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.

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Math Problem Analysis

Mathematical Concepts

Divisibility
Consecutive Integers
Arithmetic Sequences

Formulas

Sum of n consecutive integers: S_k = k·n + (sum of first k-1 integers)
Sum of first (k-1) integers: (k-1)k/2

Theorems

Divisibility Rule: A sum is divisible by k if k divides both the sum of the integers and the sequence properties allow divisibility.

Suitable Grade Level

Grades 10-12