Given the digits 1,2,5,7,8,9,0 and if repetition are not allowed, answer each of the following:

A) how many 3 digit numbers canbve formed that are divisible by 5?
•i got 6x6x1
B) how many numbers would be less then 800(must be 3 digit and divisible by 5)?

A

The last digit must be 0 or 5, and the 1st digit is not 0, so
ending in 0: 5*4*1
ending in 5: 4*4*1

B
toss out all those starting with 8 or 9.

To find the number of 3-digit numbers that are divisible by 5 using the given digits (1, 2, 5, 7, 8, 9, 0) without repetition, follow these steps:

Step 1: Determine the possible choices for the hundreds place.
Since the number must be divisible by 5, the only options for the hundreds place are 5 and 0. Therefore, we have 2 choices.

Step 2: Determine the possible choices for the tens place.
Since repetition is not allowed, there are only 6 remaining digits to choose from (1, 2, 7, 8, 9, 0). However, if the hundreds place is 0, the tens place cannot be 0 because it would create a 2-digit number. Therefore, there are 5 choices for the tens place. If the hundreds place is 5, all 6 remaining digits can be used, so there are 6 choices for the tens place.

Step 3: Determine the possible choices for the units place.
Since repetition is not allowed, there are only 4 remaining digits to choose from (1, 2, 7, 8). Regardless of the choices for the hundreds and tens places, there are 4 choices for the units place.

Step 4: Calculate the total number of possibilities.
To find the total number of 3-digit numbers, we multiply the number of choices at each place value:
Number of choices for hundreds place × Number of choices for tens place × Number of choices for units place
= (2 choices for hundreds place) × (5 choices for tens place) × (4 choices for units place) = 2 × 5 × 4 = 40

Therefore, there are 40 different 3-digit numbers that can be formed using the given digits that are divisible by 5.

Now, to answer the second question about how many of these numbers would be less than 800:

Step 1: Determine the possible choices for the hundreds place.
Since the numbers must be less than 800, the only option for the hundreds place is 5. Therefore, we have 1 choice.

Step 2: Determine the possible choices for the tens place.
To ensure the numbers are less than 800, the tens place can be any of the remaining digits (1, 2, 7, 8, 9). Therefore, there are 5 choices for the tens place.

Step 3: Determine the possible choices for the units place.
Since repetition is not allowed, there are only 4 remaining digits to choose from (1, 2, 7, 8). Regardless of the choices for the hundreds and tens places, there are 4 choices for the units place.

Step 4: Calculate the total number of possibilities.
Following the same calculation as before:
Number of choices for hundreds place × Number of choices for tens place × Number of choices for units place
= 1 (choice for hundreds place) × 5 (choices for tens place) × 4 (choices for units place) = 20

Therefore, there are 20 different 3-digit numbers that can be formed using the given digits, which are divisible by 5 and also less than 800.