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Trying to prove this congruence: $$ \frac{34\times 10^{(6n-1)}-43}{99} \equiv-1~\text{ or }~5 \pmod 6,\quad n\in\mathbb{N}$$

Progress

Brought it to the form $$34\times 10^{6n-1}-43\equiv -99\pmod{6\cdot 99}$$ but how to proceed further? $6\cdot 99$ is not a small number.

R K
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  • I am stuck at what to do with that number $99$ in that denominator. – R K Feb 22 '15 at 20:46
  • @RK The problem is equivalent to $34\times 10^{6n-1}-43\equiv -99\pmod{6\cdot 99}, \forall n\in\mathbb N$. – user26486 Feb 22 '15 at 20:47
  • @Bernard $34\times 10^{6n-1}-43\equiv 1\times (-1)^{6n-1}-43\equiv -1-43\equiv -44\equiv 0\pmod {11}$. – user26486 Feb 22 '15 at 20:51
  • @user314: Sorry. I misused the criterion for divisibility by $11$. I'll withdraw my comment. – Bernard Feb 22 '15 at 20:53
  • you can use $10^6= 1 \mod 27$ and it's clear that the number is odd – Elaqqad Feb 22 '15 at 20:54
  • @user314 $34\times 10^{6n-1}-43\equiv -99\pmod{6\cdot 99}$ ,how to proceed further, the divisor is a large number. – R K Feb 22 '15 at 20:57
  • let $N=34.10^{6n-1}-43+99$ N is divisibly by $2$ and by $11$ and the only remaining case is $N$ divisibly by $3.9=27$ and to do this you have to use $10^6=1 mod 27$ – Elaqqad Feb 22 '15 at 21:00
  • @RK The $,99,$ cancels out of the term difference - see my answer. – Bill Dubuque Feb 22 '15 at 22:47

2 Answers2

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The problem is equivalent to $$34\times 10^{6n-1}-43\equiv -99\pmod{6\cdot 99},\ \ \ \forall n\in\mathbb N$$

Now,

$$34\times 10^{6n-1}-43\equiv -99\pmod{6\cdot 99}\iff 34\cdot 10^{6n-1}\equiv -56\pmod{6\cdot 99}$$

Then

$$34\cdot 10^{6(n+1)-1}-43\equiv 10^6\cdot 34\cdot 10^{6n-1}-43\equiv 10^6\cdot (-56)-43\equiv -99 \pmod{6\cdot 99}$$

Thus if the property holds for $n$, then it holds for $n+1$.

But it holds for $n=1$, so by the method of mathematical induction the property is true $\forall n\in\mathbb N$.

user26486
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${\ 6\mid f_{n+1}-f_n =\, \dfrac{34\cdot(\color{#c00}{10^6\!-1}) 10^{6n\!-1}}{99}\ }$ by ${\ 3\cdot 99\mid \color{#c00}{10^6\!-1} = 999999 = 99\cdot\!\!\!\!\!\!\!\!\!\!\!\underbrace{10101}_{\equiv\, 1+1+1\pmod 3}}$

So $\,{\rm mod}\ 6\!:\ f_{n+1}\!\equiv f_n\,\overset{\rm induct}\Rightarrow\,f_n\equiv f_1\equiv -1$

Bill Dubuque
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