IPMAT Indore 2019 (SA) - You have been asked to select a positive integer N which is less than 1000, such that it is either a multiple of 4, or a multiple of 6, or an odd multiple of 9. The number of such numbers is | PYQs + Solutions

IPMAT Indore 2019

Number System

Divisibility Rules

Hard

You have been asked to select a positive integer N which is less than 1000, such that it is either a multiple of 4, or a multiple of 6, or an odd multiple of 9. The number of such numbers is

Entered answer:

✅ Correct Answer: 388

Alternate Solution

We need to find the number of positive integers less than 1000 that are either:

Multiples of 4, OR

Multiples of 6, OR

Odd multiples of 9

Multiples of 4 less than 1000:

The last multiple of 4 less than 1000 is 996 = 4 × 249

So there are 249 multiples of 4 less than 1000.

Multiples of 6 less than 1000:

The last multiple of 6 less than 1000 is 996 = 6 × 166

So there are 166 multiples of 6 less than 1000.

Odd multiples of 9 less than 1000:

These are of the form

9(2k+1)

where

k = 0, 1, 2, ...

The largest odd multiple of 9 less than 1000 is 999 = 9 × 111

Since 111 is odd, we need to find how many odd numbers are there from 1 to 111.

For odd numbers from 1 to 111, there are

\lceil\dfrac{111}{2}\rceil = 56

such numbers.

So there are 56 odd multiples of 9 less than 1000.

Using the Principle of Inclusion-Exclusion:

|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|

Where:

A = set of multiples of 4 less than 1000

B = set of multiples of 6 less than 1000

C = set of odd multiples of 9 less than 1000

A ∩ B (Multiples of both 4 and 6):

These are multiples of lcm(4, 6) = 12

The last multiple is 996 = 12 × 83

|A \cap B| = 83

A ∩ C (Multiples of both 4 and odd multiples of 9):

For a number to be in both sets, it must be divisible by 4 and be an odd multiple of 9

Any multiple of 4 is even, but odd multiples of 9 are odd

|A \cap C| = 0

B ∩ C (Multiples of both 6 and odd multiples of 9):

Multiples of 6 are always even, but odd multiples of 9 are odd

|B \cap C| = 0

Since

|A \cap C| = 0

, it follows that

|A \cap B \cap C| = 0

as well.

Calculating the final answer:

|A \cup B \cup C| = 249 + 166 + 56 - 83 - 0 - 0 + 0 = 388

Therefore, there are 388 positive integers less than 1000 that satisfy the given conditions.