Help needed!

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!

Easy Math Editor

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Comments

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 - 6 years ago

Nice thinking! I think your answer is right sir.

Rajdeep Dhingra - 6 years ago

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

Siddhartha Srivastava - 6 years ago

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

Ansh Bhatt - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$