Natural Numbers Divisible by 7: Step-by-Step Count

The question asks us to find the count of natural numbers that lie strictly between 3 and 200 and are also divisible by 7. Understanding these conditions is key to solving the problem efficiently.

Understanding Natural Numbers and Divisibility

• Natural Numbers: These are the positive integers used for counting, starting from 1 (i.e., 1, 2, 3, 4, and so on).
• Divisible by 7: A number is considered divisible by 7 if, when divided by 7, it leaves absolutely no remainder. This means the number is a perfect multiple of 7.

Identifying the Range for Numbers

The phrase "between 3 and 200" implies that the numbers we are looking for must be greater than 3 and, at the same time, less than 200. So, we are searching for natural numbers \(n\) such that \(3 < n < 200\). This means 3 and 200 themselves are not included in our count.

Finding the First Multiple of 7

Our first step is to pinpoint the smallest natural number that is greater than 3 and is also perfectly divisible by 7.

• Let's list the first few multiples of 7: 7, 14, 21, 28, and so on.
• When we look at this list, the very first multiple of 7 that is greater than 3 is 7 itself.
• Therefore, our starting number for this sequence (the first term) is \(a_1 = 7\).

Finding the Last Multiple of 7

Next, we need to find the largest natural number that is less than 200 and is also perfectly divisible by 7.

• To determine this, we can perform a simple division of 200 by 7.
• When we divide \(200 \div 7\), we get 28 with a remainder of 4.
• This calculation tells us that \(7 \times 28 = 196\). If we were to take the next multiple, \(7 \times 29 = 203\), which is already greater than 200.
• Since our numbers must be less than 200, the largest multiple of 7 that fits this condition is 196.
• So, our ending number for this sequence (the last term) is \(a_n = 196\).

Calculating the Total Count of Numbers

The natural numbers we have identified (7, 14, ..., 196) form an arithmetic progression (AP). In this AP, the first term is 7, the common difference between consecutive terms is also 7 (because they are multiples of 7), and the last term is 196. We need to find how many terms are in this sequence.

Method 1: Using the Arithmetic Progression Formula

The standard formula used to find the \(n^{\text{th}}\) term of an arithmetic progression is: \[ a_n = a_1 + (n-1)d \] Let's break down what each variable represents in our problem:

• \(a_n\) is the last term in our sequence, which is 196.
• \(a_1\) is the first term in our sequence, which is 7.
• \(d\) is the common difference between terms, which is 7 (since all numbers are multiples of 7).
• \(n\) is the number of terms in the sequence, which is what we are trying to find.

Now, let's substitute these known values into the formula and solve for \(n\): \[ 196 = 7 + (n-1)7 \] First, subtract 7 from both sides of the equation: \[ 196 - 7 = (n-1)7 \] \[ 189 = (n-1)7 \] Next, divide both sides by 7: \[ \frac{189}{7} = n-1 \] \[ 27 = n-1 \] Finally, add 1 to both sides to find the value of \(n\): \[ n = 27 + 1 \] \[ n = 28 \]

Method 2: Using Simple Division Logic

An alternative and often quicker way to find the count is by understanding the multipliers of 7.

• The numbers we are counting are \(7 \times k\), where \(k\) is a natural number.
• We found that the first number in our range is \(7 \times 1 = 7\). So, the smallest multiplier \(k\) is 1.
• We also found that the last number in our range is \(7 \times 28 = 196\). So, the largest multiplier \(k\) is 28.
• This means the multiples of 7 we are interested in are \(7 \times 1, 7 \times 2, \dots, 7 \times 28\).
• The total number of terms in this sequence is simply the number of multipliers from 1 to 28, which is \(28 - 1 + 1 = 28\).

Conclusion

Both calculation methods consistently show that there are 28 natural numbers lying strictly between 3 and 200 that are perfectly divisible by 7.