Introduction to Modular Arithmetic and Finding the Smallest Number

This article delves into a unique problem in number theory and modular arithmetic. Specifically, we will explore how to find the smallest number that meets a series of modular conditions and an additional divisibility rule. This problem is a fascinating exercise in using the concepts of modular arithmetic, the least common multiple (LCM), and congruences.

Understanding the Problem

The challenge is to find the smallest number x that, when divided by 2, 4, 6, 8, 10, and 12, leaves a remainder of 1 in each case. Additionally, this number x should be divisible by 13 without any remainder. Let's break down the steps to solve this problem using practical methods and mathematical principles.

Step 1: Combining the First Six Conditions

The first step in solving this problem is to recognize that each of the first six conditions can be represented as a single congruence. This is because each condition states that x leaves a remainder of 1 when divided by 2, 4, 6, 8, 10, or 12. Mathematically, this can be expressed as:

x ≡ 1 (mod 2), x ≡ 1 (mod 4), x ≡ 1 (mod 6), x ≡ 1 (mod 8), x ≡ 1 (mod 10), x ≡ 1 (mod 12)

Combining these conditions, we can say:

x - 1 ≡ 0 (mod LCM(2, 4, 6, 8, 10, 12))

Where LCM stands for the least common multiple.

Step 2: Calculating the Least Common Multiple (LCM)

To find the LCM of 2, 4, 6, 8, 10, and 12, we examine their prime factorizations:

2 2^1 4 2^2 6 2^1 × 3^1 8 2^3 10 2^1 × 5^1 12 2^2 × 3^1

The LCM takes the highest power of each prime factor:

Highest power of 2 is 2^3 (from 8) Highest power of 3 is 3^1 (from 6 and 12) Highest power of 5 is 5^1 (from 10)

Therefore, the LCM is:

lcm(2, 4, 6, 8, 10, 12) 2^3 × 3^1 × 5^1 8 × 3 × 5 120

Step 3: Setting Up the Equation

With the LCM found, we can now write the equation for x:

x - 1 ≡ 0 (mod 120) implies x 120k 1 for some integer k

Step 4: Incorporating the Condition with 13

Next, we need to find the smallest number that fits the condition x ≡ 0 (mod 13). Substituting our equation for x, we have:

120k 1 ≡ 0 (mod 13)

This can be rearranged to:

120k ≡ -1 (mod 13)

We then simplify 120 mod 13:

120 9 × 13 3 implies 120 ≡ 3 (mod 13)

Thus, the equation becomes:

3k ≡ -1 (mod 13) or 3k ≡ 12 (mod 13)

Next, we solve for k by finding the multiplicative inverse of 3 modulo 13. Testing values, we find:

3 × 9 27 ≡ 1 (mod 13) implies the inverse of 3 is 9

Multiplying both sides of the equation 3k ≡ 12 by 9:

9 × 3k ≡ 9 × 12 (mod 13)

Simplifying, we get:

k ≡ 108 (mod 13)

Finally, we simplify 108 mod 13:

108 8 × 13 4 implies 108 ≡ 4 (mod 13)

Thus, the smallest non-negative solution for k is 4.

Step 5: Finding the Smallest Number

Substituting k back into the equation for x:

x 120k 1 120 × 4 1 480 1 481

Conclusion

The smallest number that meets all the conditions is:

481