# How to prove that if $m = 10t+k$ and $67|t - 20k$ then 67|m?

m, t, k are Natural numbers.

How can I prove that if $m = 10t+k$ and $67|t - 20k$ then 67|m ?

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from first equation $k=m-10*t$,now put into equation we get that 67 divides $t-20*(m-10*t)$ or $67$ divides $201*t-20*m$, because $201/67=3$ ,it means that $201*t$ is divisible by $67$,now we have to have $-20*m$ have to divisible by $67$ which is possible if $67/m$

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We have $$10t+k=10t -200k+201k=10(t-20k)+(3)(67)k.$$ The result now follows from the fact that $67\mid(t-20k)$.

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We have $k=m-10t$ and so $67 | t - 20(m - 10t) = 201t - m$. But $201 = 3 \times 67$, and so

$$67 | 67 \times 3t - m$$

And so $67 | m$.

This can also be seen using modular arithmetic: we have $m=10t+k$ and $t \equiv 20k \pmod {67}$, so $m \equiv 201k \pmod {67}$. But $201 = 3 \times 67$, so $m \equiv 0 \pmod {67}$, and so $67|m$.

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Hint $\rm\ mod\ 67\!:\ t\,\equiv\, 20\,k\:\Rightarrow\: \color{#C00}{10\,t}\,\equiv\, 200\,k\, \equiv\, (3\!\cdot\! 67\!-\!1)\,k\,\equiv\, \color{#C00}{-k}\:\Rightarrow\: \color{#C00}{10\,t\!+\!k}\,\equiv\, 0$

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