Arithmetic Progression Problem Solving Divisibility And Remainders
In this article, we will explore an arithmetic progression problem that involves finding the number of terms within a given sequence that satisfy specific divisibility conditions. The problem at hand asks us to determine the number of terms in the arithmetic progression 201, 204, 207, ..., 300 that are divisible by 3 and leave a remainder of 1 when divided by 5. This requires us to delve into the properties of arithmetic progressions, divisibility rules, and modular arithmetic. By combining these mathematical concepts, we can systematically solve this problem and gain a deeper understanding of number theory.
To begin, let's define what an arithmetic progression is. An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference, often denoted by 'd'. The general form of an arithmetic progression is:
a, a + d, a + 2d, a + 3d, ...
where 'a' is the first term of the sequence.
In our given problem, the arithmetic progression is 201, 204, 207, ..., 300. We can identify the first term (a) as 201 and the common difference (d) as 204 - 201 = 3. Thus, the nth term of this AP can be expressed as:
a_n = a + (n - 1)d = 201 + (n - 1)3
This formula will be crucial in identifying the terms that fall within the given range and satisfy our divisibility conditions. Understanding this fundamental concept of arithmetic progressions is key to solving problems involving sequences of numbers with a constant difference. To further illustrate, let's consider a few examples. In the sequence 2, 5, 8, 11, ..., the first term is 2 and the common difference is 3. Similarly, in the sequence 10, 7, 4, 1, ..., the first term is 10 and the common difference is -3. These simple examples demonstrate how the formula a_n = a + (n - 1)d allows us to determine any term in an arithmetic progression, provided we know the first term and the common difference.
Next, we need to understand the divisibility rules and the concept of remainders. A number is said to be divisible by another number if the remainder is 0 when divided. For example, 15 is divisible by 3 because 15 ÷ 3 = 5 with a remainder of 0. However, if we divide 16 by 3, we get 5 with a remainder of 1. In our problem, we are looking for terms that are divisible by 3, meaning they leave a remainder of 0 when divided by 3, and terms that leave a remainder of 1 when divided by 5. To put it simply, divisibility rules and remainders help us classify numbers based on their relationship with a divisor. Understanding these concepts is crucial for identifying numbers that satisfy specific conditions, such as leaving a particular remainder when divided by another number. Let's delve deeper into this concept with an example. Consider the numbers 11, 12, 13, 14, and 15. When these numbers are divided by 5, the remainders are 1, 2, 3, 4, and 0 respectively. We can see that 11 leaves a remainder of 1, 12 leaves a remainder of 2, and so on. This illustrates how each number can be categorized based on its remainder when divided by a specific divisor. This principle is fundamental to solving our problem, where we need to find numbers that leave a remainder of 1 when divided by 5.
We need to find terms in the arithmetic progression that are divisible by 3. This means that the term can be written in the form 3k, where k is an integer. Simultaneously, these terms must leave a remainder of 1 when divided by 5, which means they can be written in the form 5m + 1, where m is an integer. Thus, we are looking for terms that satisfy both conditions: 3k = 5m + 1. This application of the conditions sets the stage for finding the specific terms within the arithmetic progression that meet our criteria. By combining the divisibility rules with the properties of arithmetic progressions, we can narrow down the possible solutions and efficiently identify the terms that satisfy both conditions. To further clarify this concept, let's consider a few examples. The number 6 is divisible by 3, as 6 = 3 * 2. However, when 6 is divided by 5, it leaves a remainder of 1, so 6 = 5 * 1 + 1. Therefore, 6 satisfies both conditions. On the other hand, the number 9 is divisible by 3 (9 = 3 * 3), but when divided by 5, it leaves a remainder of 4 (9 = 5 * 1 + 4), so it does not satisfy the second condition. This example illustrates how we need to find numbers that can be expressed in both forms, 3k and 5m + 1.
To solve the equation 3k = 5m + 1, we can rearrange it to find a relationship between k and m. Rearranging the equation, we get 3k - 5m = 1. This is a linear Diophantine equation, which can be solved using various methods, such as the Euclidean algorithm or by inspection. In this case, we can observe that one solution is k = 2 and m = 1, since 3(2) - 5(1) = 6 - 5 = 1. This solving the equation provides us with a critical piece of information that will help us identify the terms in the arithmetic progression that meet our criteria. By finding the solutions to this equation, we can determine the specific values of k and m that satisfy both conditions, allowing us to pinpoint the terms that are divisible by 3 and leave a remainder of 1 when divided by 5. To elaborate on this, let's consider why finding solutions to this equation is so important. Each solution (k, m) represents a number that can be expressed in both forms, 3k and 5m + 1. This means that this number is divisible by 3 and leaves a remainder of 1 when divided by 5. Our goal is to find all such numbers within the given arithmetic progression, so finding the solutions to the equation 3k - 5m = 1 is a crucial step in the problem-solving process.
Since we have one solution (k = 2, m = 1), we can find the general solutions for k and m. The general solution for linear Diophantine equations of the form ax + by = c is given by:
k = k_0 + (b/gcd(a, b))n m = m_0 + (a/gcd(a, b))n
where (k_0, m_0) is a particular solution, and n is an integer. In our case, a = 3, b = -5, c = 1, gcd(3, -5) = 1, k_0 = 2, and m_0 = 1. Thus, the general solutions are:
k = 2 + 5n m = 1 + 3n
This finding general solutions is a crucial step in determining all the possible terms that satisfy our conditions. By expressing k and m in terms of a general integer n, we can generate an infinite number of solutions to the equation 3k = 5m + 1. However, we are only interested in the solutions that correspond to terms within our arithmetic progression, so we need to find the values of n that give us terms between 201 and 300. To further illustrate the importance of general solutions, consider that each value of n will generate a unique pair of (k, m) that satisfies the equation 3k = 5m + 1. This allows us to systematically explore all possible terms that meet our criteria, ensuring that we do not miss any solutions. The challenge now is to filter these solutions and identify the ones that fall within the specified range of our arithmetic progression.
Now we know that the terms we are looking for are of the form 3k, where k = 2 + 5n. So, the terms are 3(2 + 5n) = 6 + 15n. We need to find the values of n for which these terms are also part of the given arithmetic progression. The nth term of the arithmetic progression is 201 + 3(n' - 1), where n' is the term number in the original sequence. We need to find n such that:
201 ≤ 6 + 15n ≤ 300
This terms in the arithmetic progression need to be within the specified range. By setting up this inequality, we can determine the possible values of n that correspond to terms within our original sequence. This step is crucial for narrowing down the solutions and identifying the specific terms that meet all the conditions. To elaborate, let's consider why it is important to ensure that the terms we find are within the given range. The arithmetic progression has a finite number of terms, and we are only interested in the terms that fall between 201 and 300. Therefore, we need to filter out any solutions that would result in terms outside this range. This inequality allows us to do just that, ensuring that we only consider the relevant solutions.
Let's solve the inequality 201 ≤ 6 + 15n ≤ 300. First, subtract 6 from all parts:
195 ≤ 15n ≤ 294
Now, divide by 15:
13 ≤ n ≤ 19.6
Since n must be an integer, the possible values for n are 13, 14, 15, 16, 17, 18, and 19. This solving the inequality gives us the range of n values that correspond to terms within our arithmetic progression. By determining these possible values of n, we can now identify the specific terms that satisfy both our divisibility conditions and fall within the given range. To further illustrate the significance of this step, consider that each integer value of n within this range represents a unique term in our sequence that is divisible by 3 and leaves a remainder of 1 when divided by 5. Therefore, by finding all the possible integer values of n, we can systematically identify all the terms that meet our criteria.
There are 7 integer values for n (13, 14, 15, 16, 17, 18, and 19), which means there are 7 terms in the arithmetic progression that are divisible by 3 and leave a remainder of 1 when divided by 5. Therefore, the answer is (B) 7. This counting the terms is the final step in solving our problem. By counting the number of integer values of n that satisfy our conditions, we can determine the total number of terms that meet our criteria. This provides us with the answer to the problem and demonstrates the power of combining arithmetic progressions, divisibility rules, and modular arithmetic to solve complex number theory problems. To further elaborate on this, consider that each of the 7 values of n corresponds to a unique term in the arithmetic progression that satisfies both our divisibility conditions. Therefore, by counting these values, we are essentially counting the number of terms that meet our criteria, giving us the final answer to the problem.
In conclusion, we have successfully found the number of terms in the arithmetic progression 201, 204, 207, ..., 300 that are divisible by 3 and leave a remainder of 1 when divided by 5. By understanding arithmetic progressions, divisibility rules, and solving Diophantine equations, we determined that there are 7 such terms. This problem highlights the importance of combining different mathematical concepts to solve complex problems in number theory. In conclusion, the systematic approach we have taken, from understanding the basic concepts to applying them step-by-step, has allowed us to arrive at the correct solution. This problem serves as a valuable exercise in problem-solving and reinforces the importance of a solid foundation in mathematical principles. Furthermore, the techniques used in this problem can be applied to a wide range of similar problems in number theory, making it a valuable learning experience.
