3
$\begingroup$

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.

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

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

$\endgroup$
2
  • $\begingroup$ There has been some debate as to whether this question meets our attribution requirements. I've added a link to the exact test that this question is apparently from, which should put that debate to rest. $\endgroup$
    – F1Krazy
    Mar 19, 2021 at 20:23
  • $\begingroup$ @F1Krazy I don't really care about the feelings of the online IQ test creators. I've had terrible experiences with a lot of them. I just wanted to know the answer to the problem. $\endgroup$ Mar 20, 2021 at 21:21

1 Answer 1

5
$\begingroup$

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,$\text{}\\\\$ $1^0 \equiv 1 \text{ (mod 26}) = A,\\ 2^1 \equiv 2 \text{ (mod 26}) = B,\\ 3^2 \equiv 9 \text{ (mod 26}) = I,\\ 4^3 = 64 \equiv 12 \text{ (mod 26}) = L,\\ 5^4 = 625 \equiv 1 \text{ (mod 26}) = A,\\ 6^5 = 7776 \equiv 2 \text{ (mod 26})= B,\\ 7^6 = 117649 \equiv 25 \text{ (mod 26}) = Y$

To be exact, it is actually $(x-1) \text{ mod }26+1$ because there will be a conflict for $Z$

$\endgroup$
3
  • 3
    $\begingroup$ That's clever. How did you come up with that? $\endgroup$ Jun 27, 2020 at 23:34
  • 1
    $\begingroup$ 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, 2020 at 14:24
  • 2
    $\begingroup$ Please try using "mod" instead of "%". The "%" symbol may not be friendly to non-programmers. $\endgroup$ Jun 29, 2020 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.