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

Guest Dec 30, 2021

#3**+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 | 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)

Melody Dec 31, 2021