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I am still struggling with the modulo operation and have the following two to prove:

  1. Prove that for all $a_1$, $a_2$, $a_3$ $\in \mathbb N \cup${$0$} this applies: $$100\cdot a_3 + 10\cdot a_2 + a_1 = a_3 + a_2 + a_1 \bmod 3.$$

I used proof by induction here, however got stuck substituting for $n+1$. This is as far as I got:

  1. Let $a_1$ = 1. Then, $a_{1+1}=a_2 = 2$ and $a_{1+1+1}=a_3 = 3$. Therefore, $$100\cdot (3) + 10\cdot (2) + (1) = 321 = {3+2+1} \bmod 3 = 0$$
  2. The condition is the following: Let $a_1$ = n. Then, $a_{n+1}=a_2$ and $a_{n+2}=a_3$. Therefore, $$ 100\cdot (a_{n+2}) + 10\cdot (a_{n+1}) + (a_n) = a_{n+2} + a_{n+1} + a_n \bmod 3$$

I hope someone can help me solve for $n+1$ and complete this proof.

  1. Prove that there exists no $x \in \mathbb Z_{111}$, s.t. $3x\equiv 1$ $\bmod 111.$

Unfortunately, I was unable to progress with this one and hope someone can show me how to approach and solve this problem!

Thank you!!

J. W. Tanner
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Rikk
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    There is no need for induction in the first one. Just note that $100\equiv 1 \pmod 3$ and $10\equiv 1 \pmod 3$. – lulu Feb 18 '19 at 11:58
  • Further, the point of proving $(1)$ first is that it helps to easily solve $(2)$ - see my answer (this was overlooked in the prior answers). The exercises are designed to highlight the utility of casting out threes in solving diophantine equations (similat to the better-known casting out nines). – Bill Dubuque Feb 18 '19 at 15:37

3 Answers3

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I can give you some HINTS:

For $1)$ you do not need induction, you can do like this:

$$ 100 = 99+1 \equiv_{3} 1 $$

Can you figure out the rest? :)

For $2)$ you can use the Chinese remainder theorem

$$ 111= 3\cdot 37 $$

Now look at the resulting congruences

$$ 3x\equiv_{3} 1 $$ for example.

Hope this helped :)

Vinyl_cape_jawa
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For part $1$ $$100a_3+10a_2+a_1\equiv(3\cdot33+1)a_3+(3\cdot3+1)a_2+a_1\equiv a_3+a_2+a_1 \mod{3}$$ For part $2$ $$3x\equiv1\mod{111}$$ $$\therefore 3x=1+111k,\,\,k\in\mathbb{Z}$$ $$x=\frac{1}{3}+37k,\,\,k\in\mathbb{Z}$$ But this contradicts the fact that $x\in\mathbb{Z}$, so there are no solutions for $x$.

J. W. Tanner
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Peter Foreman
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The point is that $(1)$ helps easily solve $(2)$. For $(1)$ the congruence Sum and Product Rules yield

$\!\bmod 3\!:\ 10\equiv 1\,\Rightarrow\, 10^2\equiv 1^2\equiv 1\,\Rightarrow\, 10^2a+10b+c\equiv 1a+1b+c$

So $\,3x\equiv 1\pmod{\!111}\,\Rightarrow\,3x = 1+111n\,\Rightarrow\,\bmod 3\!:\ 0 \equiv 1+ (1\!+\!1\!+\!1)n\equiv 1,\,$ contradiction

Bill Dubuque
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