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I found this on a test, and it's under "Numerical sequences". I can't figure it out. I tried using the numbers of the alphabet: 1, 2, 9, 12, 1, 2. I don't see a clear pattern.

A,B,I,L,A,B,?

This has been solved by the great Rick Rosner.

It doesn't go by the pattern ABIL ABIL ABIL...

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The answer is

$Y$

Reason:

Assume that each element has an index (position) $i$. Now, the element at $i$ is $i^{i-1}\text{ mod }26$. For example,$\text{}\\\\$ $1^0 \equiv 1 \text{ (mod 26}) = A,\\ 2^1 \equiv 2 \text{ (mod 26}) = B,\\ 3^2 \equiv 9 \text{ (mod 26}) = I,\\ 4^3 = 64 \equiv 12 \text{ (mod 26}) = L,\\ 5^4 = 625 \equiv 1 \text{ (mod 26}) = A,\\ 6^5 = 7776 \equiv 2 \text{ (mod 26})= B,\\ 7^6 = 117649 \equiv 25 \text{ (mod 26}) = Y$

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    $\begingroup$ That's clever. How did you come up with that? $\endgroup$ – PutnamLegend Jun 27 at 23:34
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    $\begingroup$ PutnamLegend, he saw 1,2,9, he recognized n^(n-1) series. The alphabet is just like setting a modulus. This, he also recognized. And then it was just done. $\endgroup$ – FIreCase Jun 28 at 14:24
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    $\begingroup$ Please try using "mod" instead of "%". The "%" symbol may not be friendly to non-programmers. $\endgroup$ – Scratch---Cat Jun 29 at 10:21
  • $\begingroup$ Ok, @Scratch---Cat $\endgroup$ – John Brookfields Jun 29 at 10:28

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