Because abab is the same as aabb. I was how to solve these problems with the blank slot method, i.e. _ _ _ _. If I do this manually, it's clear to me the answer is 6, aabb abab abba baba bbaa baab Which is …

I've found and proven the following extensions to palindromes of the usual divisibility rules for 3 and 9: A palindrome is divisible by 27 if and only if its digit sum is. A palindrome is divisible by 81 if and only if …

For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of: $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are

Jan 26, 2026 · That is, there seems to be fairly strong symbolic evidence that for $n=4$, if $ABBA-BAAB = A-B$ and $A$ is nilpotent, then $B^4 = \lambda I$ for some $\lambda$.

There must be something missing since taking $B$ to be the zero matrix will work for any $A$.

Hint: Here you can limit the size of the possible solutions reasonably easily - suppose $a\ge b \gt 3$ then $a^b\gt a^3 \gt 3ab$

The minimum number of states in min DFA of regular expression $(a+b)^*b(a+b)$ is _____. My attempt: I have drawn this DFA and found the answer to be 3: Somewhere it was explained that we need "trap"

Apr 19, 2022 · Although both belong to a much broad combination of N=2 and n=4 (AAAA, ABBA, BBBB...), where order matters and repetition is allowed, both can be rearranged in different ways: …

I get the trick. Use the fact that matrices "commute under determinants". +1

1.How many sequences of 4 songs are there where no song plays twice in a row? If we label the songs {A, B, C, D}, then examples are ABCD and ABAB but not ABBA. For this problem I just thought the …