How many whole numbers, each divisible by 7, lie between 200 and 500?

Answer

To find how many whole numbers divisible by 7 lie between 200 and 500, we can follow these steps:

### Step 1: Identify the first number divisible by 7
We need to find the smallest whole number greater than 200 that is divisible by 7. We can do this by dividing 200 by 7 and rounding up to the nearest whole number.

\[ \text{First number} = 7 \times \lceil \frac{200}{7} \rceil \]

Calculating this gives:

\[ \frac{200}{7} \approx 28.57 \quad \Rightarrow \quad \lceil 28.57 \rceil = 29 \]

Thus,

\[ \text{First number} = 7 \times 29 = 203 \]

### Step 2: Identify the last number divisible by 7
Next, we need to find the largest whole number less than 500 that is divisible by 7. We can do this by dividing 500 by 7 and rounding down to the nearest whole number.

\[ \text{Last number} = 7 \times \lfloor \frac{500}{7} \rfloor \]

Calculating this gives:

\[ \frac{500}{7} \approx 71.43 \quad \Rightarrow \quad \lfloor 71.43 \rfloor = 71 \]

Thus,

\[ \text{Last number} = 7 \times 71 = 497 \]

### Step 3: Form the arithmetic progression (AP)
Now we have the first term \(a = 203\) and the last term \(l = 497\) with a common difference \(d = 7\). The sequence of numbers divisible by 7 between 200 and 500 is:

\[ 203, 210, 217, \ldots, 497 \]

### Step 4: Find the number of terms in the AP
To find the number of terms \(n\) in this arithmetic progression, we can use the formula for the \(n\)-th term of an AP:

\[ l = a + (n-1) \cdot d \]

Substituting the known values:

\[ 497 = 203 + (n-1) \cdot 7 \]

### Step 5: Solve for \(n\)
Rearranging the equation:

\[ 497 - 203 = (n-1) \cdot 7 \]

Calculating the left side:

\[ 294 = (n-1) \cdot 7 \]

Dividing both sides by 7:

\[ n-1 = \frac{294}{7} = 42 \]

Adding 1 to both sides gives:

\[ n = 42 + 1 = 43 \]

### Conclusion
Thus, the total number of whole numbers divisible by 7 that lie between 200 and 500 is **43**.

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