We have integers 1, 2, 3 … 1000
∴ Total number of outcomes, n(S) = 1000
Number of integers which are multiple of 2 are 2, 4, 6, 8, 10, … 1000
Let p be the number of terms
We know that, ap = a + (p – 1) d
Here, a = 2, d = 2 and ap = 1000
Putting the value, we get
2 + (p – 1)2 = 1000
⇒ 2 + 2p – 2 = 1000
p = 1000/2
⇒ p = 500
Total number of integers which are multiple of 2 = 500
Let the number of integers which are multiple of 9 be n.
Number which are multiples of 9 are 9, 18, 27, …999
∴ nth term = 999
We know that, an = a + (n – 1) d
Here, a = 9, d = 9 and an = 999
Putting the value, we get
9 + (n – 1)9 = 999
⇒ 9 + 9n – 9 = 999
n = 999/9
⇒ n = 111
So, the number of multiples of 9 from 1 to 1000 is 111.
The multiple of 2 and 9 both are 18, 36, … 990.
Let m be the number of terms in above series.
∴ mth term = 990
We know that, am = a + (m – 1) d
Here, a = 9 and d = 9
Putting the value, we get
18 + (m – 1)18 = 990
⇒ 18 + 18m – 18 = 990
m = 990/18
⇒ m = 55
Number of multiples of 2 or 9
= No. of multiples of 2 + no. of multiples of 9 – No. of multiples of 2 and 9 both
= 500 + 111 – 55
= 556 = n (E)
