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? (Note: here, 11^4 in base b means that the base-b number 11 is raised to the fourth power.)
+118667 I think it is 6
\((b+1)^4=b^4+4b^3+6b^2+4b+1 \)
The sum of the digits will always be 16 base 10 which is 2^4 for any base greater than 2
However this can only be true is 6 is allowed in the base.
The biggest base that does NOT allow the use of 6 is base 6
I have not tried it with examples, if someone wants to refute this with an example then I would like to see it. :)
