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i need help

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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?

Feb 1, 2019
edited by Guest  Feb 1, 2019

#1
+8368
+4

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.

Feb 2, 2019
#2
+8368
+4

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.

Check: $$24_{10} = 44_5 = 33_7$$