Find the smallest integer a > 2 such that 2|a,3| a + 1 ,4|a+2,5|a+3,6|a+4

We are tasked with finding the smallest integer \( a > 2 \) that satisfies the following conditions:

\[
\begin{aligned}
2 \mid a, \\
3 \mid a + 1, \\
4 \mid a + 2, \\
5 \mid a + 3, \\
6 \mid a + 4.
\end{aligned}
\]

### Step 1: Translate the divisibility conditions

We need to express these divisibility conditions as congruences:

1. \( 2 \mid a \) means \( a \equiv 0 \pmod{2} \).
2. \( 3 \mid a + 1 \) means \( a + 1 \equiv 0 \pmod{3} \), or \( a \equiv 2 \pmod{3} \).
3. \( 4 \mid a + 2 \) means \( a + 2 \equiv 0 \pmod{4} \), or \( a \equiv 2 \pmod{4} \).
4. \( 5 \mid a + 3 \) means \( a + 3 \equiv 0 \pmod{5} \), or \( a \equiv 2 \pmod{5} \).
5. \( 6 \mid a + 4 \) means \( a + 4 \equiv 0 \pmod{6} \), or \( a \equiv 2 \pmod{6} \).

### Step 2: Simplify the system of congruences

We now have the following system of congruences:

\[
\begin{aligned}
a &\equiv 0 \pmod{2}, \\
a &\equiv 2 \pmod{3}, \\
a &\equiv 2 \pmod{4}, \\
a &\equiv 2 \pmod{5}, \\
a &\equiv 2 \pmod{6}.
\end{aligned}
\]

### Step 3: Combine the congruences

Note that the second, third, fourth, and fifth congruences all say \( a \equiv 2 \) modulo some number. This suggests that \( a = 2 \pmod{\text{LCM}(3, 4, 5, 6)} \), where the least common multiple of 3, 4, 5, and 6 is:

\[
\text{LCM}(3, 4, 5, 6) = 60.
\]

Thus, \( a \equiv 2 \pmod{60} \).

### Step 4: Solve the system with the condition \( a \equiv 0 \pmod{2} \)

We also know \( a \equiv 0 \pmod{2} \), which must be true in addition to \( a \equiv 2 \pmod{60} \). Since \( a \equiv 2 \pmod{60} \) satisfies \( a \equiv 0 \pmod{2} \), we do not need any additional modification. Therefore, the solution to the system is:

\[
a = 60k + 2 \quad \text{for some integer } k.
\]

### Step 5: Find the smallest \( a > 2 \)

The smallest \( a > 2 \) is obtained by setting \( k = 1 \), giving:

\[
a = 60(1) + 2 = 62.
\]

Thus, the smallest integer \( a > 2 \) that satisfies all the conditions is \( a = 62 \).

### Final Answer:
\[
\boxed{62}
\]

---

Would you like more details on the solution process or any further explanations? Here are 5 related questions to explore:

1. How can we use the Chinese Remainder Theorem to solve systems of congruences?
2. What is the least common multiple (LCM), and how is it calculated for multiple numbers?
3. How do we check if a solution satisfies all the conditions in a system of congruences?
4. What happens if one of the conditions is modified, for example, changing \( 6 \mid a + 4 \) to \( 6 \mid a + 3 \)?
5. How can we generalize this problem to solve similar problems with different moduli?

**Tip:** When solving systems of congruences with multiple conditions, check if some of the moduli are multiples of others. This can help simplify the system early on.

Find the smallest integer a > 2 such that 2|a, 3|a + 1, 4|a + 2, 5|a + 3, 6|a + 4

Find the smallest integer greater than 2 that satisfies the divisibility conditions 2|a, 3|a+1, 4|a+2, 5|a+3, 6|a+4. This step-by-step solution involves modular arithmetic and the least common multiple (LCM), making it suitable for students studying number theory and modular systems.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation