Official Solution:How many positive five-digit multiples of 3 can be formed using the digits 0, 1, 2, 3, 4, and 5, without repeating any digit? A. 96
B. 120
C. 181
D. 200
E. 216
First step: Our goal is to determine which 5 digits out of the given 6 will form a 5-digit number divisible by 3.
We have six digits: 0, 1, 2, 3, 4, 5. Their sum is 15.
For a number to be divisible by 3, the sum of its digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3), the only 5-digit combinations that would yield a number divisible by 3 are 15-0 = {1, 2, 3, 4, 5} and 15-3 = {0, 1, 2, 4, 5}. No other combinations of 5 digits from the given 6 will result in a number divisible by 3.
Second step: We have two sets of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. Let's find out how many 5-digit numbers can be formed using these two sets:
{1, 2, 3, 4, 5} This set provides 5! = 120 numbers, as any permutation of these digits would result in a 5-digit number divisible by 3.
{0, 1, 2, 4, 5}. In this case, we cannot use 0 as the first digit, otherwise the number will no longer be a 5-digit number but a 4-digit number. Therefore, the desired number of combinations is the total combinations 5! minus the combinations with 0 as the first digit (permutations of the remaining 4 digits) 4!: 5! - 4! = 120 - 24 = 96.
In total, there are 120 + 96 = 216 possible 5-digit numbers that are divisible by 3.
Answer: E
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