Question:
How many even integers \( n \), where \( 100 \leq n \leq 200 \), are divisible neither by seven nor by nine?
How many even integers \( n \), where \( 100 \leq n \leq 200 \), are divisible neither by seven nor by nine?
Show Hint
Count all the even numbers in the range first, then remove those divisible by 7 or by 9. Be careful not to subtract the common multiples twice.
Updated On: Nov 7, 2026
- 40
- 37
- 39
- 38
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We must count the even integers from \(100\) to \(200\) (both ends included) that are not divisible by \(7\) and also not divisible by \(9\). The easy way is to count all even numbers first and then remove the ones we do not want.
Step 2: Count all even integers:
The even numbers are \(100, 102, 104, \ldots, 200\). Their count is
\[ \dfrac{200 - 100}{2} + 1 = 51 \]
Step 3: Count the even numbers divisible by 7:
An even number divisible by \(7\) must be divisible by \(\mathrm{lcm}(2, 7) = 14\). The multiples of \(14\) between \(100\) and \(200\) are \(14 \times 8 = 112\) up to \(14 \times 14 = 196\).
That gives \(14 - 8 + 1 = 7\) numbers: \(112, 126, 140, 154, 168, 182, 196\).
Step 4: Count the even numbers divisible by 9:
An even number divisible by \(9\) must be divisible by \(\mathrm{lcm}(2, 9) = 18\). The multiples of \(18\) in the range are \(18 \times 6 = 108\) up to \(18 \times 11 = 198\).
That gives \(11 - 6 + 1 = 6\) numbers: \(108, 126, 144, 162, 180, 198\).
Step 5: Remove the overlap:
Numbers divisible by both are counted in both lists above. These are the even multiples of \(\mathrm{lcm}(2, 7, 9) = 126\). In the range only \(126\) itself qualifies, because \(252\) is already above \(200\). So the overlap is \(1\).
Even numbers divisible by \(7\) or by \(9\): \(7 + 6 - 1 = 12\).
Step 6: Subtract:
\[ 51 - 12 = 39 \]
The other options do not match this count. For example, \(38\) comes from forgetting to add the overlap back (\(51 - 7 - 6 = 38\)), and \(40\) comes from adding the overlap back twice (\(51 - 7 - 6 + 2 = 40\)).
Final Answer:
There are \(39\) such even integers, which is option 3. \[ \boxed{39 \text{ (Option 3)}} \]
We must count the even integers from \(100\) to \(200\) (both ends included) that are not divisible by \(7\) and also not divisible by \(9\). The easy way is to count all even numbers first and then remove the ones we do not want.
Step 2: Count all even integers:
The even numbers are \(100, 102, 104, \ldots, 200\). Their count is
\[ \dfrac{200 - 100}{2} + 1 = 51 \]
Step 3: Count the even numbers divisible by 7:
An even number divisible by \(7\) must be divisible by \(\mathrm{lcm}(2, 7) = 14\). The multiples of \(14\) between \(100\) and \(200\) are \(14 \times 8 = 112\) up to \(14 \times 14 = 196\).
That gives \(14 - 8 + 1 = 7\) numbers: \(112, 126, 140, 154, 168, 182, 196\).
Step 4: Count the even numbers divisible by 9:
An even number divisible by \(9\) must be divisible by \(\mathrm{lcm}(2, 9) = 18\). The multiples of \(18\) in the range are \(18 \times 6 = 108\) up to \(18 \times 11 = 198\).
That gives \(11 - 6 + 1 = 6\) numbers: \(108, 126, 144, 162, 180, 198\).
Step 5: Remove the overlap:
Numbers divisible by both are counted in both lists above. These are the even multiples of \(\mathrm{lcm}(2, 7, 9) = 126\). In the range only \(126\) itself qualifies, because \(252\) is already above \(200\). So the overlap is \(1\).
Even numbers divisible by \(7\) or by \(9\): \(7 + 6 - 1 = 12\).
Step 6: Subtract:
\[ 51 - 12 = 39 \]
The other options do not match this count. For example, \(38\) comes from forgetting to add the overlap back (\(51 - 7 - 6 = 38\)), and \(40\) comes from adding the overlap back twice (\(51 - 7 - 6 + 2 = 40\)).
Final Answer:
There are \(39\) such even integers, which is option 3. \[ \boxed{39 \text{ (Option 3)}} \]
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