Largest base $b$ such that sum of digits of base $b$ of $11^4$ do not add up to $16$ When $11^4$ is written out in base $10$, the sum of its digits is $16=2^4$. What is the largest base $b$ such that the base-$b$ digits of $11^4$ do not add up to $2^4$?
Is there an elegant solution? I tried plugging in bases, but this obviously does not work. I am thinking Binomial Theorem, but I am not sure how that will translate to other bases.
Assuming you are asking about $11_b^4$ in base $b$, the largest is base $6$, where
$11_6^4=7^4_{10}=15041_6$ with a digit sum of $11_{10}$ or $15_6$
If we don't worry about carries, we have $11^4=14641$ in any base.  This comes from $(b+1)^4=b^4+4b^3+6b^2+4b+1$.  It is only when the $6$ carries that the sum will not be $16_{10}$
Answer appears to be any base &gt;14626.
$$
11^4_{10} = 14641_{10}
$$
$$
11^4_{14626} = 1F_{14626}
$$
Sum of digits in base 14626 appears to be 16. All bases higher than that have a different sum for the digits.