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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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Any help appreciated peeps!

Note by Anik Mandal
2 years, 2 months ago

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243?

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 \( 3* (3^4) = 243 \) Siddhartha Srivastava · 2 years, 2 months ago

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@Siddhartha Srivastava did it the same way. this should be correct, i don't see any alternative. Ansh Bhatt · 2 years, 2 months ago

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@Siddhartha Srivastava Nice thinking! I think your answer is right sir. Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra Why does everyone call me Sir? I'm still in school -.- Siddhartha Srivastava · 2 years, 2 months ago

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Comment deleted May 22, 2015

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@Soutrik Bandyopadhyay Even I got the same Rajdeep Dhingra · 2 years, 2 months ago

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@Rajdeep Dhingra Yep I know. Anik Mandal · 2 years, 2 months ago

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@Soutrik Bandyopadhyay Think so...Well what is your method? Anik Mandal · 2 years, 2 months ago

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