Question 6

Mathematics Sets Medium

The number of all even integers between 99 and 999 which are not multiple of 3 and 5 is

(A) 250
(B) 240
(C) 235
(D) 245
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We want the number of all even integers between \(99\) and \(999\) which are not multiples of \(3\) or \(5\). Step 1: Total even numbers. The first even number after \(99\) is \(100\), and the last even number before \(999\) is \(998\). This is an arithmetic progression: \[ 100, 102, 104, \dots, 998 \] with common difference \(2\). The number of terms is \[ n = \frac{998-100}{2} + 1 = \frac{898}{2} + 1 = 449 + 1 = 450. \] So there are \(450\) even numbers between \(99\) and \(999\). Step 2: Exclude multiples of 3. Even numbers that are multiples of \(3\) are multiples of \(6\). Between \(100\) and \(998\), the first multiple of \(6\) is \(102\), and the last is \(996\). Number of terms: \[ \frac{996 - 102}{6} + 1 = \frac{894}{6} + 1 = 149 + 1 = 150. \] Step 3: Exclude multiples of 5. Even numbers that are multiples of \(5\) are multiples of \(10\). Between \(100\) and \(998\), the first multiple of \(10\) is \(100\), and the last is \(990\). Number of terms: \[ \frac{990 - 100}{10} + 1 = \frac{890}{10} + 1 = 89 + 1 = 90. \] Step 4: Double counting (multiples of both 3 and 5). Even numbers that are multiples of both \(3\) and \(5\) are multiples of \(30\). Between \(100\) and \(998\), the first multiple of \(30\) is \(120\), and the last is \(990\). Number of terms: \[ \frac{990 - 120}{30} + 1 = \frac{870}{30} + 1 = 29 + 1 = 30. \] Step 5: Inclusion--Exclusion. The count of even numbers divisible by \(3\) or \(5\) is \[ 150 + 90 - 30 = 210. \] Step 6: Required count. Thus, the number of even integers between \(99\) and \(999\) not divisible by \(3\) or \(5\) is \[ 450 - 210 = 240. \] \[ \boxed{240} \]

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