Is there any other divisibility test involving all the digits, like mod 9?
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For example, can mod 3 also test all the digits?
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@amusingminnow Yeah! You can pick any modulo you want for these types of tests. Although, he reason why Prof. Loh chose mod \(9\) is because it's really easy to find both \(8^8 \pmod{9}\) and the really huge numbers\(\pmod{9}\), since (digit sum of number) mod 9 = (number) mod 9. Other modulos might not be as convenient.
If you wanted to use\(\pmod{3}\), it would look something like this:
since \(8 \equiv 2 \equiv -1 \pmod{3}\),
\(8^8\pmod{3}\equiv(-1)^8\pmod{3}\equiv 1\pmod{3}\).So, out of the answer choices, we have to find the number that's also \(1\pmod{3}\).
The thing is, digit sums actually works for mod 3 too, it's the exact same process. Doing so will give you that \(16777216\equiv1\pmod{3}\). Same answer!\(3\) and \(9\) are nice because we can find big numbers mod 3 and mod 9 really easily. For numbers like \(7\) though, it might be tougher (if you want to, go ahead and try it out!)
For this problem though, I think mod 9 is best. Consider the fact that
\(1,4,7 \equiv 1\pmod{3}\), but they are \(\equiv1,4,7\pmod{9}\), respectively. Luckily, for this problem, we didn't run into any confusion, so mod 3 worked fine. However, other problems might be different, and the fact that mod 9 is larger can help in some situations. Hope this makes sense, and let me know if you have any other questions!
