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Can someone prove that, $$7\space|\space 10^{(3+k)+6n}+3^k$$ $$i.e. 7 \space Divides \space 10^{(3+k)+6n}+3^k$$

$for$ $ all$ $k,n$ $\in$ $N_0$ $?$

ThanQ

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    Have you already studied some modular arithmetic? And what are those parentheses around that part of the exponent of $\;10\;$ for? – 2017-01-24
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    You can prove the case $n=1$ by induction on $k$ and then the general case by induction on $n$. But is it worthy the trouble when modular arithmetic gives you the proof in a pinch? – 2017-01-24
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    If you have a proof what was the reason for asking this question? – 2017-01-24

2 Answers 2

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Doing arithmetic modulo $\;7\;$ :

$$10^{3+k+6n}+3^k=3^{3+k+6n}+3^k=3^k\left(3^{3+6n}+1\right)=3^k\left(-3^{6n}+1\right)=3^k\left(-1^n+1\right)=0$$

You can also try induction on $\;n\;$ and for any $\;k\;$ ...

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    Which has the advantage of using no induction but only the elementary facts that $10=3\pmod{7}$ because $10=3+7$, that $3^3=-1\pmod{7}$ because $27=4\cdot7-1$, and its consequence $3^6=1\pmod{7}$. – 2017-01-24
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Note that:

$(10^{3+k+6n}+3^k)\bmod7=$

$((10\bmod7)^{3+k+6n}+3^k)\bmod7=$

$(3^{3+k+6n}+3^k)\bmod7=$

$(3^k\cdot(\color\green{3^{3+6n}+1}))\bmod7$

You can prove by induction that $7\mid\color\green{3^{3+6n}+1}$.

First, show that this is true for $n=0$:

$3^{3+6\cdot0}+1=28$

Second, assume that this is true for $n$:

$3^{3+6n}+1=7m$

Third, prove that this is true for $n+1$:

$3^{3+6(n+1)}+1=$

$3^{3+6n+6}+1=$

$3^{6+3+6n}+1=$

$3^{6}\cdot3^{3+6n}+1=$

$729\cdot3^{3+6n}+1=$

$729\cdot3^{3+6n}+729-728=$

$729\cdot(\color\red{3^{3+6n}+1})-728=$

$729\cdot\color\red{7m}-728=$

$7\cdot(729m-104)$

Please note that the assumption is used only in the part marked red.