How many 3 digit numbers can be formed from the digits 2,3,5,6,7 and 9 which are divisible by 5 and none of the digits is repeated?
come on, do some thinking.
If a number is divisible by 5, it must end in 5 or 0. In this case, 5 must be the 3rd digit.
That leaves 2,3,6,7,9 for the other two digits.
So, there are 5 ways to pick the 1st digit. Once it has been picked, there are only 4 choices for the 2nd digit.
So, there are 5*4 = 20 ways to form the 3-digit number.
To find the number of 3-digit numbers that can be formed from the digits 2, 3, 5, 6, 7, and 9 which are divisible by 5 and have no repeated digits, we can break down the problem into steps:
Step 1: Count the total number of digits available
In this case, the digits available are 2, 3, 5, 6, 7, and 9. So, we have a total of 6 digits.
Step 2: Determine the last digit
To be divisible by 5, the last digit of the number must be either 5 or 0. However, among the available digits, only 5 is an option. So, the last digit of our 3-digit number will be 5.
Step 3: Count the number of choices for the first digit
Since none of the digits can be repeated, we have 5 available choices for the first digit. We exclude the digit 5, as it has already been used for the last digit.
Step 4: Count the number of choices for the second digit
Again, since none of the digits can be repeated, we have 4 choices remaining for the second digit. We exclude the digit 5 and the digit already chosen for the first digit.
Step 5: Multiply the choices together
To find the total number of 3-digit numbers meeting the given conditions, we multiply the number of choices for each digit:
5 choices for the first digit × 4 choices for the second digit = 20
Therefore, there are 20 different 3-digit numbers that can be formed from the digits 2, 3, 5, 6, 7, and 9 which are divisible by 5 and have no repeated digits.
To find the number of 3-digit numbers that can be formed using the digits 2, 3, 5, 6, 7, and 9, with the condition that the number must be divisible by 5 and none of the digits can be repeated, we can follow these steps:
Step 1: Count the number of digits that are divisible by 5
Out of the given digits, 5 is the only one divisible by 5.
Step 2: Count the number of digits that are not divisible by 5
Out of the given digits, 2, 3, 6, 7, and 9 are not divisible by 5. So there are 5 digits that are not divisible by 5.
Step 3: Count the number of choices for each digit of the 3-digit number
For the first digit, we have 1 choice (only 5 is divisible by 5).
For the second digit, we have 5 choices (any of the 5 digits not divisible by 5).
For the third digit, we have 4 choices (any of the remaining 4 digits that have not been used yet).
Step 4: Multiply the number of choices for each digit together
Multiply the number of choices for each digit together: 1 * 5 * 4 = 20.
Therefore, there are 20 three-digit numbers that can be formed using the digits 2, 3, 5, 6, 7, and 9, which are divisible by 5 and none of the digits are repeated.