Final answer:
To find the three-digit numbers divisible by 3 formed with the digits 0, 1, and 5, we ensure that 0 is not in the hundreds place and the sum of the digits is divisible by 3. There are only two such numbers possible, which are 150 and 510.
Step-by-step explanation:
The question asks how many three-digit numbers that are divisible by 3 can be formed using the digits 0, 1, and 5 exactly once in each number. A number is divisible by 3 if the sum of its digits is divisible by 3. With the digits 0, 1, and 5, the sum is 0+1+5=6, which is divisible by 3. When forming three-digit numbers, the digit 0 cannot be in the hundreds place as that would not make a valid three-digit number.
Hence, we have two options for the hundreds place (1 or 5), three options for the tens place, and one option for the units place after the hundreds and tens places have been filled. So in essence, we can form 2 valid permutations since the digit 0 restricts the number of permutations to avoid leading zeros. Thus, the two numbers possible are 150 and 510.