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What should $n$ be equal to, so that the number: $$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$ is completely divisible by 19? I broke it into this: $$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i do next?

Bill Dubuque
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manabou11
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  • You could test a few small values of $n$ and look for a pattern. If you find a pattern you can try to confirm it by the method of induction. – sammy gerbil Jul 21 '20 at 20:16
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    Hint: $20\cdot50^{n}+18\cdot12^{n} = (19+1)50^{n}+(19-1)12^{n} = 19(50^{n}+12^{n}) + (50^{n}-12^{n})$ – mwt Jul 21 '20 at 20:17

4 Answers4

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Okay, so you have shown the expression to be equal to $20\cdot 2^n\cdot 25^n+18\cdot3^n\cdot4^n=20\cdot 50^n+18\cdot12^n$ and as @mwt as shown in comments , write $20=19+1$ and $18=19-1$ to get the expression equal to $19(50^n+12^n)+50^n-12^n$. Now we know that $a^n-b^n$ is divisible by $a-b$ for any natural number $n$. If you don't know that, you can prove it by factorizing $a^n-b^n$.

So $50^n-12^n$ is divisble by $38$ and so $19$ divides the whole expression for any natural number $n$.

V.G
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  • what do you mean by factorizing – manabou11 Jul 21 '20 at 20:52
  • Well, you can divide $a^n-b^n$ by $a-b$ by doing normal division. You will end up with the quotient $\displaystyle a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}$. So that means $a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}\right)$.

    You can also see factor theorem.) or see this.

     – V.G Jul 22 '20 at 05:42
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For any natural $n$ we obtain: $$\begin{aligned}5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}&=2^{n+1}\left(10\cdot5^{2n}+9\cdot6^n\right) \\ &= 2^{n+1}\left(10\cdot(6+19)^{n}+9\cdot6^n\right) \\ & \equiv2^{n+1}\left(10\cdot6^n+9\cdot6^n\right) \pmod{19} \\ &\equiv 0 \pmod{19}.\end{aligned}$$

V.G
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Michael Rozenberg
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Having gotten to $20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n$, note that $20\equiv1$ mod $19$ while $18\equiv-1$ and $25\equiv6$. It follows that

$$20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n\equiv2^n\cdot6^n-3^n\cdot4^n\equiv12^n-12^n\equiv0\mod 19$$

Remark: This approach assumes you have modular arithmetic in your toolbox. (Someone added a tag indicating you do, but it may not be the case.)

Barry Cipra
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You can prove by induction that $20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n$

$=20\cdot50^{n}+18\cdot12^{n}$ is divisible by $19$.

It's obviously true for $n=0$, since $20+18=38=2\cdot19$.

Now assume $19$ divides $20\cdot50^{n}+18\cdot12^{n}$.

Then $20\cdot50^{n+1}+18\cdot12^{n+1}=50\cdot20\cdot50^{n+1}+12\cdot18\cdot12^{n+1}$

$=12\cdot(20\cdot50^n+18\cdot12^{n})+38\cdot20\cdot50^n$ is divisible by $19$.

QED

J. W. Tanner
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