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  • How many different three-digit numbers can be formed using the first 8 prime numbers without repetition, if the number must be divisible by 3?  Op

How many different three-digit numbers can be formed using the first 8 prime numbers without repetition, if the number must be divisible by 3?

 

Option: 1

182


Option: 2

188


Option: 3

156


Option: 4

186


Answers (1)

best_answer

To determine the number of different three-digit numbers that can be formed using the first 8 prime numbers (2, 3, 5, 7, 11, 13, 17, 19) without repetition, if the number must be divisible by 3, we can use the concept of combinations.

First, let's determine the total number of three-digit numbers that can be formed using these 8 prime numbers without repetition. Since the first digit cannot be 0, we have 7 options for the first digit. For the remaining two digits, we have 6 options for the second digit and 5 options for the third digit.

The total number of three-digit numbers that can be formed without the restriction of being divisible by 3 is:

7 (options for the first digit) \times6 (options for the second digit) \times 5 (options for the third digit) = 210

Now let's consider the restriction that the number must be divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3.

Out of the given prime numbers, the digits that are divisible by 3 are 3 and 7. Thus, to form a three-digit number that is divisible by 3, we need at least one 3 or one 7 in the number.

Now, let's calculate the number of three-digit numbers that are divisible by 3:

Case 1: The number contains one 3 and no 7:

We have 3 positions to place the digit 3 (excluding the first position since it cannot be 0). For the remaining two digits, we have 6 options for the second digit and 5 options for the third digit.

3 (options for the position of digit 3)\times 6 (options for the second digit) \times5 (options for the third digit) = 90
 

Case 2: The number contains one 7 and no 3:

Similar to Case 1, we have 3 positions to place the digit 7. For the remaining two digits, we have 6 options for the second digit and 5 options for the third digit.

3 (options for the position of digit 7)\times 6 (options for the second digit) \times 5 (options for the third digit) = 90


Case 3: The number contains both 3 and 7:

We have 2 positions to place the digit 3 (excluding the first position since it cannot be 0). For the remaining position, we have 4 options for the second digit. The digit 7 will be placed in the remaining position.

2 (options for the position of digit 3) \times 4 (options for the second digit) = 8

The total number of three-digit numbers that can be formed using the first 8 prime numbers without repetition and are divisible by 3 is:

90 + 90 + 8 = 188

Therefore, there are 188 different three-digit numbers that can be formed using the first 8 prime numbers without repetition, if the number must be divisible by 3.

 

Posted by

Divya Prakash Singh

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