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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.)

 Feb 1, 2019
 #1
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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.)

 

Mmm I have to work out what is being asked first.

 

\((11_b)^4\\=((b+1)_{10})^4\\=(b^4+4b^3+6b^2+4b+1)_{10}\\ \qquad If \;\;b>6 \;\;then\\ =14641_b \)

 

1+4+6+4+1 = 16

If b =6 or less than 6  this would equal a different number and the digits would not add to 16

 

SO the largest b  where the digits DO NOT add to 16 is b=6

 Feb 1, 2019

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