Show that $ 7\mid (a + 3) $

Question

Being $ a $ and $ b \in \mathbb {Z} ^ {+} $ and $ 2560a = 10 ^ b $, show that $ 7 \mid (a + 3) $

$a=5^8*10 \equiv 2560 \equiv 5$ Does it serve as a counterexample?

or

$256a =10^{b-1}$ $2^8a= 2^{b-1} \cdot b^{b-1}$

$a+3 \equiv 0 $ (mod 7)

$5^8+3 \equiv 0$ (mod 7)

$(-2^8)+3 \equiv 0$ mod 7

$256 + 3 \equiv 0$ mod 7

$ 259 \equiv 0$ mod 7

Answer 1

Yes $a=2 \times 5^9$ serves as a counterexample. BUT your reasoning why it is a counterexample is not a well-written proof. Here is how I would do it:

First note that $2560 a = 10 \times 2^8 \times 2 \times 5^9 = 10^{10}$. So $2560a$ is indeed of the form $10^b$ for $b=10$.

So now it suffices to show that 7 does not divide $a+3$. To this end, note that $a = 10 \times 5^8$, and that $5^8 \equiv_7 4$ whereas $10 \equiv_7 3$ so $a = 5^8 \times 10 \equiv_7 4 \times 3 \equiv_7 5$. So $a+3 \equiv_7 1 \not = 0$. So indeed, 7 does not divide $a+3$.

Answer 2

\begin{align} 2560a &= 10^b \\ 5a &\equiv 3^b \pmod 7 \\ a &\equiv 3^{b+1} \pmod 7 \\ a+3 &\equiv 3^{b+1}+3 \pmod 7 \\ \hline 3^{b+1}+3 &\equiv 0 \pmod 7 \\ 3 \cdot 3^b &\equiv 4 \pmod 7 \\ 3^b &\equiv 6 \pmod 7 \\ b &\equiv 3 \pmod 6 &\text{See table below}\\ a &\equiv 4 \pmod 7 &\text{See table below} \end{align}

\begin{array}{|r|rrrrrr|} \hline b \pmod 6 & 0 & 1 & 2 & 3 & 4 & 5 \\ 3^b \pmod 7 & 1 & 3 & 2 & 6 & 4 & 5 \\ a \pmod 7 & 3 & 2 & 6 & 4 & 5 & 1 \\ \hline \end{array}

Since $a \not \equiv 4 \mod 7$, then $7 \not \mid a+3$.

Answer 3

If 7|(a+3) then $a+3\equiv 0 \pmod 7$

Let $a = 5^9\cdot 2$

$5^9\cdot 2+3\equiv 5^3\cdot 2 + 3 \pmod 7$ by Fermat's Little Theorem.

$253\equiv 1 \pmod 7$