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 \(\large{3}\) ?

Thanks~

Any help appreciated peeps!

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 \(\large{3}\) ?

Thanks~

Any help appreciated peeps!

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

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– Ansh Bhatt · 2 years, 4 months ago

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

– Rajdeep Dhingra · 2 years, 4 months ago

Nice thinking! I think your answer is right sir.Log in to reply

– Siddhartha Srivastava · 2 years, 4 months ago

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

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– Rajdeep Dhingra · 2 years, 4 months ago

Even I got the sameLog in to reply

– Anik Mandal · 2 years, 4 months ago

Yep I know.Log in to reply

– Anik Mandal · 2 years, 4 months ago

Think so...Well what is your method?Log in to reply