Let $n$ be a positive integer. If the number $1998$ is written in base $n$, a three-digit number with the sum of digits equal to $24$ is obtained. Find all possible values of $n$.
Problem
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Tags: Bases, number theory
brmorgan82
30.04.2021 03:55
express $xn^2+yn+(24-x-y)=1998\to (n-1)(xn+x+y)=1974$ observe $1974=2\cdot3\cdot7\cdot47$ if $47|(n-1)$ then $n\geq48$ but $48^2>1998$ so it cannot be 3 digits, so $47|(xn+x+y)$ then we observe $n>12$ since $12^3<1998$ so we have $(n-1)=2\cdot7,3\cdot7,2\cdot3\cdot7$ and thus $\boxed{n=15,22,43}$
uptownmath
30.04.2021 03:57
We can write down the system of equations:
an^2+bn+c = 1998
a+b+c =24
Subtracting the two equations, we get a(n^2-1)+b(n-1) = (n-1)(an+n+b) = 1974. This means that n-1 is a factor of 1974.
Additionally, because a is not 0, an+n+b > n-1, so n-1 is less than sqrt(1974), or 45. In addition, we must have n^3 > 1988, so n > 12, or n-11 > 11.
1974 can be factored as 2*3*7*47. The factors between 11 and 45 exclusive, which correspond to n-1 values, are 14, 21, and 42. So the possible n values are 15, 22, and 43.
Edit: sniped but whatever