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
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 | b | 1 |
X+Z | X+Y | |
carry 1 | carry 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 | |
X | 0 | 0 |
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
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)