1. For how many $n=2,3,4,\ldots,99,100$ is the base-$n$ number $235236_n$ a multiple of $7$?
2.What is the smallest base-10 integer that can be represented as $AA_5$ and $BB_7$, where $A$ and $B$ are valid digits in their respective bases?
+9519 1.
Obviously, \(n\geq 7\).
\(\quad 235236_n\\ =2n^5 + 3n^4 + 5n^3 + 2n^2 + 3n + 6\)
This holds true for any \(n\geq 7\).
Then, clearly, \(2n^5+3n^4+5n^3+2n^2+3n\equiv 1\pmod7\).
Define \(P(n) = 2n^5+3n^4+5n^3+2n^2+3n-1\).
Consider all possible values of P(n) mod 7 when \(0 \leq n \leq 6, n\in \mathbb Z\).
\(P(0) \equiv 6\pmod 7\\ P(1) \equiv 0 \pmod 7\\ P(2) \equiv 4 \pmod 7\\ P(3) \equiv 1 \pmod 7\\ P(4) \equiv 1 \pmod 7\\ P(5) \equiv 1\pmod 7\\ P(6) \equiv 1 \pmod 7\\ \)
Generally, \(P(7k + 1) \equiv 0 \pmod 7 \quad\forall \;k\in \mathbb Z^+\).
All possible values of n are: 8,15,22,29,36,43,50,57,64,71,78,85,92,99.
So there are 14 possible values of n such that \(235236_n\) is divisible by 7.
+9519 2.
\(AA_5 = 5A + A = 6A \quad \forall \; A \in [0,4] \land A\in \mathbb Z\\ BB_7 = 7B + B = 8B \quad \forall \; B \in [0,6] \land B\in \mathbb Z\)
From this, we know that the required base-10 integer must be divisible by both 6 and 8.
Least common multiple of 6 and 8 = 24.
So the answer is 24.
Check: \(24_{10} = 44_5 = 33_7\)
