Anonymous
Feb 7
How many three digit numbers divisible by three can be formed using the digits 2, 3, 4, 5 and 6 if repetition of digits is allowed ?
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ChatGPT
Feb 7
To form a three-digit number divisible by three, the sum of its digits must be divisible by three. The only way to achieve this with the given digits is to use three of the digits 3, 3, 6, and 6.
We can choose the first digit in $\binom{4}{1} = 4$ ways (either 3, 3, 6, or 6).
We can choose the second digit in $\binom{3}{1} = 3$ ways (any of the remaining three digits).
We can choose the third digit in $\binom{2}{1} = 2$ ways (either of the remaining two digits).
Therefore, there are $4 \cdot 3 \cdot 2 = \boxed{24}$ three-digit numbers divisible by three that can be formed using the digits 2, 3, 4, 5, and 6.
We can choose the first digit in $\binom{4}{1} = 4$ ways (either 3, 3, 6, or 6).
We can choose the second digit in $\binom{3}{1} = 3$ ways (any of the remaining three digits).
We can choose the third digit in $\binom{2}{1} = 2$ ways (either of the remaining two digits).
Therefore, there are $4 \cdot 3 \cdot 2 = \boxed{24}$ three-digit numbers divisible by three that can be formed using the digits 2, 3, 4, 5, and 6.
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