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.)

mathleteig Feb 29, 2020

#2**0 **

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.)

\(11_{base \;b}=b+1\\ (b+1)^4=b^4+4b^3+6b^2+4b+1\\ \text{This will be true, as written, for any base where the base is greater than 6}\\ \text{so for all bases (b) greater than 6 the sum of the digits will be 16 base 10 }=2^4_b \)

IF b=6 then

\(b^4+4b^3+6b^2+4b+1\\ =b^4+4b^3+b*b^2+4b+1\\ =b^4+5b^3+0b^2+4b+1\\\)

so the sum of the digits will be 1+5+0+4+1=11 which is definitely not 2^4

**So the largest base b that does NOT work is 6**

Melody Mar 4, 2020