AIME problem

[Maverick848]

Hello, I was trying to do an AIME problem as follows:

"LEt N be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when N is divided by 1000?

The solution for this said that:

"an integer is divisible by 8 if and only if the rightmost three digits are divisible by eight"

I do not understand how you know that is true?

 

[soroban]

Hello, Maverick8481

"An integer is divisible by 8 if and only if the rightmost three digits are divisible by eight"

I do not understand.
How you know that is true?

It is one of the "facts of life" in Mathmatics.

A number is divisible by 2 (even) if and only if the rightmost digit is divisible by 2.

A number is divisible by 4 if and only if the rightmost two digits form a number divisible by 4.

A number is divislble by 8 if and only if the rightmost three digits form a number divislble by 8.

. . . and so on.


Suppose we have a number \(\displaystyle N\) and its last three digits are \(\displaystyle xyz\).

Then \(\displaystyle N\) is of the form: \(\displaystyle \,N\:=\:1000a\,+\,xyz\)

Is \(\displaystyle N\) divisible by 8?

\(\displaystyle \;\;\)We have: \(\displaystyle \:\frac{N}{8}\;=\;\frac{1000a\,+\,xyz}{8}\;=\;125a\,+\,\frac{xyz}{8}\)

So \(\displaystyle N\) is divisible by 8 if and only if \(\displaystyle xyz\) is divisible by 8.


That's not a very rigorous proof, but I hope it gives you a general idea . . .