In mathematics, we often encounter problems where we need to find the smallest number that meets specific conditions. One such problem involves finding the smallest number that leaves a remainder of 5 when divided by 20, 25, and 40. In this article, we will explore a detailed solution to this problem and provide several examples to illustrate the process.

Problem Statement

The problem is to find the smallest number x such that when x is divided by 20, 25, and 40, the remainder is 5. Symbolically, we can write this condition as:

x ≡ 5 (mod 20), x ≡ 5 (mod 25), x ≡ 5 (mod 40)

Solution

To solve this problem, we can use a step-by-step approach based on the concept of the Least Common Multiple (LCM).

Solution 1

Let X be the required number. The problem states that X - 5 is divisible by 18, 20, and 30. Therefore, we can write:

X - 5 LCM{18, 20, 30}

The LCM of 18, 20, and 30 is 180. Substituting this into the equation, we get:

X - 5 180

X 180 5 185

Solution 2

We can also solve this problem by finding the LCM of the divisors and then adding the remainder:

Prime factor of 16: 2 x 2 x 2 x 2Prime factor of 28: 2 x 2 x 7Prime factor of 40: 2 x 2 x 2 x 5LCM of 16, 28, and 40  2 x 2 x 2 x 2 x 5 x 7  560Adding the remainder 5, the smallest number is 560   5  565

Solution 3

Another approach involves finding the LCM of 20, 24, 35, and 45, then adjusting for the remainder 5:

20  2^2 x 524  2^3 x 335  5 x 745  3^2 x 5LCM of 20, 24, 35, and 45  2^3 x 3^2 x 5 x 7  2520The required number must be the smallest 5-digit number. 2520 * 4  1008010080   5  10085Hence, the smallest 5-digit number that when divided by 20, 24, 35, and 45 leaves a remainder of 5 is 10085.

General Approach

In each of the examples provided, we see a common pattern in solving such problems:

1. Subtract the remainder from the condition, then find the LCM of the divisors. 2. Add the remainder back to the result of the LCM calculation.

Conclusion

The smallest number that divides 20, 25, and 40 leaving a remainder of 5 is 185. This can be verified by checking that 185 - 5 180, which is the LCM of 18, 20, and 30.

In summary, understanding the LCM and how to apply it to solve such remainder problems is essential. This method can be generalized to find the smallest number for any set of divisors and remainders.