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In every numeration system (base b) there is a three-digit integer that is (b+1) times the sum of its digits. In base ten the number is 198. In base three the number is 121. Find the number in base eight that has this property

 Dec 30, 2021
 #1
avatar+20 
+1

176.  (in base 8)

 

or 126 base 10

 Dec 30, 2021
 #3
avatar+115920 
+2

Thanks OrcSlop

 

In every numeration system (base b) there is a three-digit integer that is (b+1) times the sum of its digits. In base ten the number is 198. In base three the number is 121. Find the number in base eight that has this property

 

I decided to look for a general solution to this problem.

My whole answer is presented in base b (not in base 10)

 

Let the number be XYZ  where X,Y and Z are digits of the 3 digit, base b number.

by definition,   0<=y,z

 

We are asked to find solutions such that :   (b is a given constant)

 

\((b+1)(X+Y+Z)=Xb^2+Yb+Z\\~\\ b(X+Y+Z)+(X+Y+Z)=Xb^2+Yb+Z\\~\\ b(X+Z)+(X+Y)=Xb^2\\~\\ Xb^2=(X+Z)b+(X+Y)\)

 

I viewed this much like a basic addition:

 

b^2

b1
 X+ZX+Y
carry 1carry 1 
 

X+Z+1 must be 10 (base b)

there will be a carry 1

X+Y must be 10 (base b)

there will be a carry 1

X00


So from this additon table I can see that    X=1    

Also  X+Y = 1+Y = b   so       Y = b-1

Also  X+Z+1 = 2+Z = b   so   Z = b-2

 

X=1,    Y = b-1,   Z = b-2

 

So base2  the number would be  110

So base3  the number would be  121

So base4  the number would be  132

So base5  the number would be  143

So base6  the number would be  154

So base7  the number would be  165

So base8  the number would be  176

So base9  the number would be  187

So base10  the number would be  198

So base11  the number would be  1A9

etc

 

That was an interesting little challenge cool

 

 

 

 

 

 

LaTex:

(b+1)(X+Y+Z)=Xb^2+Yb+Z\\~\\
b(X+Y+Z)+(X+Y+Z)=Xb^2+Yb+Z\\~\\
b(X+Z)+(X+Y)=Xb^2\\~\\
Xb^2=(X+Z)b+(X+Y)

 Dec 31, 2021

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