From the 2012 Australian Mathematical Competition (Junior Level):
How many four-digit numbers containing no zeros have the property that whenever any its four digits is removed, the resulting three-digit number is divisible by $3$?
To solve this problem, I listed 4 cases
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If the 1st digit is deleted then the remaining number would have to be a multiple of $3$. That means that there are $10 \cdot 10 \cdot 3 = 300$ cases.
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If the 2nd digit is deleted then the remaining number would have to be a multiple of $3$. That means that there are $10 \cdot 10 \cdot 3 = 300$ cases.
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If the 3rd digit is deleted then the remaining number would have to be a multiple of $3$. That means that there are $10 \cdot 10 \cdot 3 = 300$ cases.
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If the 4th digit is deleted then the remaining number would have to be a multiple of $3$. That means that there are $10 \cdot 10 \cdot 3 = 300$ cases.
So there are $1200$ cases but this is wrong.
Best Answer
A number is divisible by 3 iff its sum of digits is divisible by 3. Using this fact, it is easy to see that our 4-digit number satisfies the given condition iff the four digits all have the same remainder modulo 3. Therefore, we have $3*(9/3)^4=243$ cases when each digit is nonzero.