Base Equation

Number Theory Level 4

Find the value of n n such that the following equality holds:

( n − 1 ) ( n − 2 ) ( 0 ) ‾ n = ( n − 2 ) ( n − 3 ) ( 1 ) ‾ n + 1 , ∣ n ∣ > 3 \large\overline{(n-1) (n-2) (0)}_n = \overline{(n-2) (n-3) (1)}_{n+1}, \qquad |n| > 3

Clarification : The subscript indicates number base . ( n − 1 ) , ( n − 2 ) , ( 0 ) , … (n-1),(n-2), (0),\ldots are digits of the number.

Try more questions on Bases .


The answer is 4.

Alex G by Alex G , 22, USA

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1 solution

Alex G
May 16, 2016

Relevant wiki: Number Base - Problem Solving

The numbers in the problem are equivalent to:

n 2 ( n − 1 ) + n ( n − 2 ) n = ( n − 2 ) ( n + 1 ) 2 + ( n − 3 ) ( n + 1 ) + 1 n^2(n-1)+n(n-2)n=(n-2){(n+1)}^2+(n-3)(n+1)+1

Expanding:

n 3 − 2 n = ( n − 2 ) ( n 2 + 2 n + 1 ) + n 2 − 2 n − 3 + 1 n^3-2n=(n-2)(n^2+2n+1)+n^2-2n-3+1

n 3 − 2 n = n 3 − 3 n − 2 + n 2 − 2 n − 2 n^3-2n=n^3-3n-2+n^2-2n-2

0 = n 2 − 3 n − 4 0=n^2-3n-4

0 = ( n − 4 ) ( n + 1 ) 0=(n-4)(n+1)

The possible answers are n = 4 n=4 and n = − 1 n=-1 . n = − 1 n=-1 is prohibited by the problem as the result numbers are nonsense (-3 as a digit in base -1?) The answer is therefore n = 4 n=4 .

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