We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
132
1
avatar+335 

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
avatar+101729 
+3

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

11 Online Users

avatar
avatar
avatar