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How many three digit numbers divisible by three can be formed using the digits 2, 3, 4, 5 and 6 without repetition?
Anonymous
Feb 7
How many three digit numbers divisible by three can be formed using the digits 2, 3, 4, 5 and 6 without repetition?
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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 3, 4, and 5.
We can arrange these three digits in $3! = 6$ ways. However, we must exclude the arrangement 345, since it is not a three-digit number.
Therefore, there are $6-1=\boxed{5}$ three-digit numbers divisible by three that can be formed using the digits 2, 3, 4, 5, and 6 without repetition.
We can arrange these three digits in $3! = 6$ ways. However, we must exclude the arrangement 345, since it is not a three-digit number.
Therefore, there are $6-1=\boxed{5}$ three-digit numbers divisible by three that can be formed using the digits 2, 3, 4, 5, and 6 without repetition.
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