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Find the base of the numeration system in which 135/21 has a zero remainder.

 Feb 10, 2018
 #1
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+1

If your 135/21 are ALREADY converted to the desired base, then:

135_7  mod  21_7 =0 remainder.

 

This does make sense, since 135_7 = 75_10, and 21_7 = 15_10. So, 75/15 =5 with no remainder.

 Feb 10, 2018
edited by Guest  Feb 10, 2018
edited by Guest  Feb 10, 2018
 #2
avatar+129847 
+1

This was answered by Melody a few days ago...I believe that she found the answer to be base 7 as follows:

 

[ 1b^2 + 3b  + 5 ]  / [ 2b + 1]     and when b  = 7 we have that

 

[ 7^2 + 3*7 + 5 ] / [ 2*7 + 1  ]  =

 

[ 49 + 21  + 5 ] / [ 15 ]

 

[ 70 + 5 ]  / [ 15]    =

 

[ 75] / [ 15]   =

 

5

 

 

cool cool cool

 Feb 10, 2018
edited by CPhill  Feb 11, 2018
 #3
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0

Wow  !!  I didn't see Melody's answer, but that is very interesting way of doing it !!. So, you would arrive at the answer by trial and error? Or, is there a sure way of surmising the answer?

 Feb 10, 2018
 #4
avatar+129847 
0

She used trial and error....but..with the presence of the "5"  she knew that the base had to  be  6 or greater.......so...it didn't take her long to get to 7....LOL!!!!

 

 

cool cool cool

 Feb 11, 2018

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