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Task: find all natural numbers $n$ which give us $5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}$ divisible by $19$.

What I have done so far: $5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1} = 25^n\cdot 5\cdot2^n\cdot2^2+3^n\cdot3^2\cdot4^n\cdot2 = 20\cdot25^n\cdot2^n + 18\cdot3^n\cdot4^n$.

That's it. I've read about the rule of divisibility by $19$. First of all, it did not help much and secondly, I think it's impossible to memorize all the divisibility rules for such a variety of numbers up to $19$ and even greater. Isn't there a more general principle for these tasks?

Thanks in advance!

Kenta S
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George Zorikov
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1 Answers1

2

1. solution:

Let $$E:=5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}$$

So $$E= 20\cdot 25^n\cdot 2^n +18\cdot 3^n\cdot4^n$$ $$\equiv _{19}50^n-12^n$$ $$\equiv _{19}12^n-12^n \equiv 0$$

so it is true for all $n$.

2. solution

We can avoid modular arithmetic:

$$E = 20\cdot 50^n +18\cdot 12^n =$$ $$=(19+1)\cdot 50^n +(19-1)\cdot 12^n$$ $$ = 19\cdot (50^n + 12^n)+(50^n - 12^n)$$

Now since $$50^n - 12^n = (50-12)(50^{n-1}+...+12^{n-1}) = 38\cdot a$$ we get the same conclusion.

nonuser
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