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 …

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

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 …

Nov 21, 2013 · Truly lost here, I know abba could look anything like 1221 or even 9999. However how do I prove 11 divides all of the possiblities?

Oct 4, 2016 · The algorithm is normally created by taking AB, then inverting each 2-state 'digit' and sticking it on the end (ABBA). You then take this entire sequence and repeat the process (ABBABAAB).

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: …

May 29, 2015 · @user1551 it feels like there should be a simpler justification, but I'll settle for the quick and dirty explanation.

I have tried using the change of base formula, but can't quite complete the equality: $$ a^{\\log{b}} \\\\ a^{\\frac{\\log_a{b}}{\\log_a{a}}} $$ How do I get the base of the exponent to be b?

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 …

My friend and I were discussing infinity and stuff about it and ran into some disagreements regarding countable and uncountable infinity. As far as I understand, the list of all natural numbers is