Question

1. (a) How many integers from 197 to 603 are divisible by 4?
(b) How many integers from 97 to 995 are divisible by 6?
(c) If the largest of 87 consecutive integers is 255 then what is the smallest?

2. Compute the following:

(a)9!

(b)P(15,8)

(c)8!

(d)P(3,6)

Answer 1

(1)

(a)

From 197 to 603:

Least number divisible by 4 = 200

Maximum number divisible by 4 = 600

Number of integers from 1 to 200 that are divisible by 4 = 200/4 = 50

Number of integers from 1 to 600 that are divisible by 4 = 600/4 = 150

So,

Number of integers from 200 to 600 that are divisible by 4 = 150 - 50 = 100

So,

Answer is:

100

(b)

From 97 to 995:

Least number divisible by 6 = 102

Maximum number divisible by 6 = 990

Number of integers from 1 to 97 that are divisible by 6 = 102/6 = 17

Number of integers from 1 to 995 that are divisible by 6 = 990/6 = 165

So,

Number of integers from 97 to 995 that are divisible by 4 = 165 - 17= 148

So,

Answer is:

148

(c)

Given:

The largest of the 87 consecutive integers = 255

Let the smallest integer be x.

The 87 consecutive integers are:

x, x + 1, x + 2,..., x + 86

So,

The largest of 87 consecutive integers = x + 86           (1)

Given:

The largest of the 87 consecutive integers = 255         (2)

Equating (1) & (2), we get:

x = 86 = 255

So,

x = 255 - 86= 169

So,

Answer is:

169

(2)

(a)

9! = 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 362880

So,

Answer is:

362880

(b)

So,

Answer is:

259459200

(c)

8! = 8 X 7 X 6 X 5X 4 X 3 X 2 X 1 = 40320

So,

Answer is:

40320

(d)

P(3,6) is not valid because in P(n,r), n should be greater than or equal to r.