Introduction

In mathematics, a common problem involves finding the greatest number with a specific remainder when divided by a set of given divisors. This article delves into the process of determining the greatest 5-digit number that, when divided by 3, 5, 7, 9, 8, and 12, leaves a remainder of 2. We'll explore the steps to achieve this and understand the underlying mathematical concepts.

Understanding LCM

To solve such problems effectively, it's essential to understand the concept of the Least Common Multiple (LCM). The LCM of a set of numbers is the smallest positive integer that is divisible by each number in the set. Understanding LCM is pivotal to solving these types of problems efficiently.

The Problem at Hand

We aim to find the greatest 5-digit number that leaves a remainder of 2 when divided by 3, 5, 7, 9, 8, and 12. Let's begin by determining the LCM of these numbers.

Step 1: Find the LCM of 3, 5, 7, 9, 8, and 12

First, we need to find the LCM of the given divisors: 3, 5, 7, 9 (which is 32), 8 (which is 23), and 12 (which is 22 × 3). To find the LCM, we'll take the highest power of each prime number involved:

For 2, the highest power is 3 (from 8 and 12). For 3, the highest power is 2 (from 9). For 7, the highest power is 1 (since it only appears in 7).

Therefore, the LCM of 3, 5, 7, 9, 8, and 12 is:

LCM 23 × 32 × 5 × 7 8 × 9 × 5 × 7 2520.

Step 2: Determine the Greatest 5-Digit Divisible Number

The greatest 5-digit number is 99999. We need to find the greatest 5-digit number that is divisible by 2520. To do this, we divide 99999 by 2520 and round down to the nearest whole number:

99999 ÷ 2520 ≈ 39.7015873

The greatest whole number less than or equal to 39.7015873 is 39. Therefore, the highest number that is a multiple of 2520 is:

2520 × 39 98280.

Step 3: Adjust for the Remainder

We've found that 98280 is the greatest 5-digit number that is exactly divisible by 2520. However, we need a remainder of 2. To achieve this, we can add 2 to the number that is just less than 99999 and divisible by 2520:

98280 2 98282.

Therefore, the greatest 5-digit number that leaves a remainder of 2 when divided by 3, 5, 7, 9, 8, and 12 is 98282.

Alternative Approaches

Let's explore a second approach using a different set of divisors to illustrate another method. Consider finding the greatest 5-digit number when divided by 37, 81, and 12, leaving a remainder of 2. The LCM of 37, 81 (3?), and 12 (22 × 3) is:

For 2, the highest power is 2 (from 12). For 3, the highest power is 4 (from 81). For 37, the highest power is 1 (since it only appears in 37).

Therefore, the LCM of 37, 81, and 12 is:

LCM 22 × 3? × 37 4 × 81 × 37 11904.

The greatest 5-digit number, 99999, is divided by 11904:

99999 ÷ 11904 ≈ 8.432, so the highest multiple of 11904 is 11904 × 8 95232.

To achieve the remainder of 2, we add 2 to 95232:

95232 2 95234.

So, another correct solution is 95234.

Conclusion

The greatest 5-digit number leaving a remainder of 2 when divided by 3, 5, 7, 9, 8, and 12 is 98282. We found this by first determining the LCM of the divisors, then finding the greatest 5-digit number divisible by the LCM, and finally adjusting for the remainder. This method can be applied to solve similar problems involving any set of divisors.

For more such problems or detailed solutions, consider exploring online resources or mathematical forums. If you need to solve similar problems, follow these steps and throw in some practice.