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Magazine Chapter 4: Problem 6
Find the smallest integer \(a>2\) such that $$ 2|a, 3| a+1,4|a+2,5| a+3,6 \mid a+4 $$
Short Answer
Expert verified
The smallest integer \(a\) is 62.
Step by step solution
01
Understand the Problem Statement
We need to find the smallest integer \(a > 2\) such that \(a\) satisfies five different divisibility conditions. These are \(2|a, 3|a+1, 4|a+2, 5|a+3,\) and \(6|a+4\).
02
Translate Conditions into Congruences
Convert each divisibility condition into a congruence. These are:\1. \(a \equiv 0 \pmod{2}\) implies \(a\) is even. 2. \(a + 1 \equiv 0 \pmod{3}\) means \(a \equiv 2 \pmod{3}\).3. \(a + 2 \equiv 0 \pmod{4}\) indicates \(a \equiv 2 \pmod{4}\).4. \(a + 3 \equiv 0 \pmod{5}\) suggests \(a \equiv 2 \pmod{5}\).5. \(a + 4 \equiv 0 \pmod{6}\) tells \(a \equiv 2 \pmod{6}\).
03
Combine Congruences
Notice that all conditions are in the form \(a \equiv 2 \pmod{n}\) for some \(n\). Hence, we need \(a \equiv 2\) where \(n = \text{lcm}(3, 4, 5, 6)\).
04
Calculate Least Common Multiple
Find the least common multiple of \(3, 4, 5, \) and \(6\). These numbers factorize as:- \(3 = 3\)- \(4 = 2^2\)- \(5 = 5\)- \(6 = 2 \times 3\)Thus, the least common multiple is \(\text{lcm}(3, 4, 5, 6) = 60\).
05
Solve for Least Integer \(a\)
Since we want \(a > 2\) and satisfy \(a \equiv 2 \pmod{60}\), write \(a = 60k + 2\). The smallest integer solution is achieved by taking \(k = 0\), which gives \(a = 2\). Thus, taking the next smallest integer, we set \(k = 1\), gives \(a = 62\).
06
Verify Solution
Check the divisibility conditions for \(a = 62\):- \(2 \mid 62\) - \(3 \mid 63\) - \(4 \mid 64\) - \(5 \mid 65\) - \(6 \mid 66\) All conditions are satisfied, so \(a = 62\) is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Congruences
Understanding congruences is key in modular arithmetic. Essentially, a congruence is a statement about divisibility. When we say that two numbers are congruent modulo a number, we mean that they leave the same remainder when divided by that number.
For example, if we have the expression \(a \equiv b \pmod{n}\), it means that \(a - b\) is divisible by \(n\), or that both \(a\) and \(b\) leave the same remainder when divided by \(n\).
For example, if we have the expression \(a \equiv b \pmod{n}\), it means that \(a - b\) is divisible by \(n\), or that both \(a\) and \(b\) leave the same remainder when divided by \(n\).
- In our exercise, several congruences are created based on different divisibility conditions. For instance, if \(a + 1 \equiv 0 \pmod{3}\), this is equivalent to saying \(a \equiv 2 \pmod{3}\).
- This translates to \(a\) having a remainder of 2 when divided by 3.
- Congruences simplify the process of determining whether an integer satisfies several divisibility rules simultaneously.
Divisibility Rules
Divisibility rules provide a quick way to find out if one integer is divisible by another, making arithmetic problems easier to solve.
- These rules involve simple checks you can perform mentally, such as checking whether an integer leaves no remainder when divided by a smaller integer.
- In the exercise, each divisibility condition is converted into a congruence. For instance, \(3|a+1\) implies that when \(a+1\) is divided by 3, there is no remainder, i.e., \(a + 1 \equiv 0 \pmod{3}\).
- This translation from a divisibility rule to a congruence allows us to work with modular arithmetic easily.
Least Common Multiple
The least common multiple (LCM) of a set of numbers is the smallest number that is a multiple of each of the given numbers.
- Calculating the LCM is crucial in problems involving synchronization of cycles, like aligning different divisibility conditions in our exercise.
- To find the LCM of \(3, 4, 5, \) and \(6\), we consider the highest power of each prime that appears in the factorization of the numbers.
- For example: \(3=3\), \(4=2^2\), \(5=5\), \(6=2\times3\). Therefore, \(\text{lcm}(3, 4, 5, 6) = 2^2 \times 3 \times 5 = 60\).
Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is a powerful tool in number theory. It provides a way to solve systems of congruences with different moduli. This theorem essentially guarantees the existence and uniqueness of solutions within a prescribed range.
- In the exercise, each condition on \(a\) can be thought of as a congruence, and they are combined using the information linked to their prime factors.
- While we did not directly apply the formal statement of CRT here, the idea behind solving \(a \equiv 2 \pmod{n}\) where \(n = \text{lcm}(3, 4, 5, 6) = 60\) is inspired by it.
- The CRT helps simplify the solving process by enabling us to deal with each congruence separately, further supporting modular arithmetic techniques.
