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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.)

 Sep 29, 2020
 #1
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pls respond!!!

 Sep 29, 2020
 #2
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Try base 6.

Guest Sep 29, 2020
 #3
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Why base 6???

Guest Sep 29, 2020
 #4
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base 7 doesnt work either...

Guest Sep 29, 2020
 #5
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I think the answer is way bigger than 6

Guest Sep 29, 2020
 #6
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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. :)

 Sep 30, 2020

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