Math Problem Statement

Montrer que les nombres définis dans les quatre expressions sont divisibles par certains nombres.

Solution

The problem you've shared is asking to prove divisibility properties of several sequences. Here's a breakdown of the four problems in the image:

  1. Prove that the number an=4n1a_n = 4^n - 1 is divisible by 3 for all nNn \in \mathbb{N}^* (positive integers).
  2. Prove that the number bn=42n+21b_n = 4^{2n+2} - 1 is divisible by 15 for all nNn \in \mathbb{N}^*.
  3. Prove that the number cn=32n+1+2n+2c_n = 3^{2n+1} + 2^{n+2} is divisible by 7 for all nNn \in \mathbb{N}.
  4. Prove that the number dn=4n+6n1d_n = 4n + 6n - 1 is divisible by 9 for all nNn \in \mathbb{N}^*.

Let's outline the methods for solving each problem:

1. an=4n1a_n = 4^n - 1 divisible by 3

We can use modular arithmetic. Notice that 41 (mod 3)4 \equiv 1 \ (\text{mod} \ 3). Therefore, 4n1n=1 (mod 3)4^n \equiv 1^n = 1 \ (\text{mod} \ 3), meaning that 4n10 (mod 3)4^n - 1 \equiv 0 \ (\text{mod} \ 3). Hence, ana_n is divisible by 3 for all nn.

2. bn=42n+21b_n = 4^{2n+2} - 1 divisible by 15

To show divisibility by 15, we need to prove divisibility by both 3 and 5:

  • Divisibility by 3: As in the first part, 41 (mod 3)4 \equiv 1 \ (\text{mod} \ 3), so 42n+21 (mod 3)4^{2n+2} \equiv 1 \ (\text{mod} \ 3). Therefore, 42n+210 (mod 3)4^{2n+2} - 1 \equiv 0 \ (\text{mod} \ 3).
  • Divisibility by 5: Notice that 41 (mod 5)4 \equiv -1 \ (\text{mod} \ 5), so 42n+2=(1)2n+21 (mod 5)4^{2n+2} = (-1)^{2n+2} \equiv 1 \ (\text{mod} \ 5), and 42n+210 (mod 5)4^{2n+2} - 1 \equiv 0 \ (\text{mod} \ 5).

Thus, bnb_n is divisible by both 3 and 5, so it is divisible by 15.

3. cn=32n+1+2n+2c_n = 3^{2n+1} + 2^{n+2} divisible by 7

This can also be tackled using modular arithmetic:

  • 33 (mod 7)3 \equiv 3 \ (\text{mod} \ 7), and powers of 3 repeat in cycles mod 7: 3133^1 \equiv 3, 3223^2 \equiv 2, 3363^3 \equiv 6, etc.
  • 22 (mod 7)2 \equiv 2 \ (\text{mod} \ 7), and powers of 2 also follow a cycle mod 7.

By applying these properties and examining the expression 32n+1+2n+23^{2n+1} + 2^{n+2}, you will see that it is divisible by 7 for all nNn \in \mathbb{N}.

4. dn=4n+6n1d_n = 4n + 6n - 1 divisible by 9

This is a linear expression. Simplifying, dn=10n1d_n = 10n - 1. We need to show that 10n10 (mod 9)10n - 1 \equiv 0 \ (\text{mod} \ 9). This simplifies to n1 (mod 9)n \equiv 1 \ (\text{mod} \ 9). Thus, dnd_n is divisible by 9 for all nNn \in \mathbb{N}^*.


Would you like detailed solutions or explanations for any of these steps?

Here are 5 related questions you can explore:

  1. How does modular arithmetic help in proving divisibility?
  2. Can you extend the divisibility results to larger powers of numbers?
  3. How do different prime factors (like 3, 5, and 7) influence divisibility rules?
  4. What other methods can be used to prove divisibility besides modular arithmetic?
  5. How do patterns in exponents simplify solving divisibility problems?

Tip: Modular arithmetic is a powerful tool for solving divisibility problems, especially when powers of numbers are involved.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Divisibility
Algebraic Expressions

Formulas

a_n = 4^n - 1
b_n = 4^{2n+2} - 1
c_n = 3^{2n+1} + 2^{n+2}
d_n = 4n + 6n - 1

Theorems

Properties of Powers in Modular Arithmetic
Divisibility Rules

Suitable Grade Level

Grades 10-12