Divisibility and Remainders: Exploring the Patterns with 342 and 18

In the field of mathematics, understanding the divisibility and remainders of numbers is crucial for solving complex problems. This article provides a detailed walkthrough of a specific problem, where a number divisible by 342 leaves a remainder of 47, and we need to find the remainder when this number is divided by 18.

Understanding the Given Information

Let's start by defining the problem. If a number x is divisible by 342 and leaves a remainder of 47 upon division, this can be expressed as:

x 342k - 47, where k is an integer.

Step 1: Simplifying the Problem

The first step is to simplify our expression by dividing both 342 and 47 by 18, the divisor we are ultimately interested in. This involves calculating the remainders of these numbers when divided by 18.

342 mod 18 0, because 342 is exactly divisible by 18 (18 * 19 342).

47 mod 18 11, because when 47 is divided by 18, the quotient is 2, and the remainder is 11 (47 - 36 11).

Step 2: Substitution and Simplification

Substituting these values back into our original equation gives us:

x ≡ 342k - 47 mod 18

x ≡ 0 - 47 mod 18

x ≡ -47 mod 18

Since -47 can be converted to a positive remainder by adding 18:

x ≡ -47 54 mod 18 (since 54 is the next multiple of 18 after -47)

x ≡ 7 mod 18

However, a positive remainder less than 18 is obtained by subtracting 18 from 7:

x ≡ 11 mod 18

Conclusion

Thus, the remainder when the number x (which is divisible by 342 and leaves a remainder of 47) is divided by 18 is 11.

Shortest Method: A Simplified Approach

A much more concise method involves directly constructing the smallest possible number that fits the given criteria:

x 342 - 47 295

Now, dividing 295 by 18:

295 mod 18 11

Generalizing the Solution

This method can be generalized to any number that is divisible by 342 and leaves a remainder of 47. By using the formula x 342k - 47 and simplifying it as shown, the remainder upon division by 18 will always be 11.

Conclusion

Understanding divisibility and the concept of remainders is key to solving such mathematical problems efficiently. The solution presented here not only finds the correct remainder but also provides a clear and concise method to approach similar problems in the future.