How many four digit numbers containing \(\text{NO}\) zeros have the property that whenever any one of its four digits is removed, the resulting three-digit number is divisible by \(\large{3}\) ?

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## Comments

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We can easily prove that all the digits must have the same remainder modulo 3, and that any such contruction will satisfy the property.

There are three groups with the same remainder modulo 3, \( (1,4,7) ; (2,5,8) ; (3,6,9) \). After selecting the group, there are three possible digits to select for four places. Hence the answer is \( 3* (3^4) = 243 \)

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did it the same way. this should be correct, i don't see any alternative.

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Nice thinking! I think your answer is right sir.

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Why does everyone call me Sir? I'm still in school -.-

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Comment deleted May 22, 2015

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Even I got the same

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Yep I know.

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Think so...Well what is your method?

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