We offer a selection of arithmetic exercises with detailed solutions for high school students. The exercises on arithmetic are at different levels. In fact, some of the exercises are in the pre-algebra category, while others are in the algebra category.

**Exercise:** The greatest common divisor (gcd) of two integers $a$ and $b$ will be denoted by ${\rm gcd}(a,b)$, which is the largest positive integer that divides each of the integers. On the other hand, we say that $a$ is a divisor of $b$, or that $b$ is a multiple of $a$, and one writes ${\displaystyle a\mid b}$ if there exists an integer $k$ such that $b=ka$.

- Let $a,b,a',b'$ and $d$ be non nulls integers such that \begin{align*} d={\rm gcd}(a,b),\quad a=da'\quad\text{and}\; b=db'. \end{align*} Prove that ${\rm gcd}(a',b')=1$
- Let $a$ and $b$ be non nulls integers. Prove that \begin{align*} {\rm gcd}(a,b)=|a|\;\Longleftrightarrow\; {\displaystyle a\mid b}. \end{align*}
- Let $a,b$ and $c$ be non nulls integers. Prove that \begin{align*} {\rm gcd}(ab,ac)=|a|{\rm gcd}(b,c). \end{align*}
- Let $a,b$ and $n$ be non nulls integers such that ${\displaystyle a\mid n}$ and ${\displaystyle a\mid n}$ . Prove that \begin{align*} {\rm gcd}(a,b)=1\;\Longleftrightarrow\; {\displaystyle ab\mid n}. \end{align*}

**Solution:** 1) As $d={\rm gcd}(a,b),$ there exists $(p,q)\in\mathbb{Z}^2$ such that \begin{align*} d=pq+qb. \end{align*} But $a=da'$ and $b=db'$. Then $d=(pa'+qb')d$. As $d\neq 0,$ then one can simplify by $d$ and get $1=pa'+qb'$. Now Bezout Theorem implies that ${\rm gcd}(a',b')=1$

2) The direct implication: Let ${\rm gcd}(a,b)=|a|$. Then by definition ${\displaystyle |a|\mid b}$, which means that there exists $k\in\mathbb{Z}$ such that $b=k|a|$.Then $b=ka$ or $b=(-k)a$. In both cases, we have ${\displaystyle a\mid b}$.

Then converse implication: If ${\displaystyle a\mid b}$ then there exists $q\in\mathbb{Z}$ such that $b=qa$. Then $b\mathbb{Z}\subset a\mathbb{Z}$. This implies that \begin{align*} a\mathbb{Z}+b\mathbb{Z}\subset a\mathbb{Z}=|a|\mathbb{Z}. \end{align*} Evidently, we have $a\mathbb{Z}\subset a\mathbb{Z}+b\mathbb{Z}$. This yields \begin{align*} a\mathbb{Z}+b\mathbb{Z}=|a|\mathbb{Z}, \end{align*} which means that ${\rm gcd}(a,b)=|a|$.

3) We first observe that for any $(m,n)\in \mathbb{N}^\ast\times \mathbb{N}^\ast,$ $n\mathbb{Z}=m\mathbb{Z}$ is equivalent to $m=n$. In fact, ${\displaystyle n\mid m}$ and ${\displaystyle m\mid n}$ implies that $n=m$.

We have \begin{align*} \left({\rm gcd}(ab,ac)\right)\mathbb{Z}&=ab\mathbb{Z}+ac\mathbb{Z}\cr &= |a|\left(b\mathbb{Z}+c\mathbb{Z}\right)\cr &=|a|{\rm gcd}(b,c)\mathbb{Z}. \end{align*} With our fist observation, we conclude that \begin{align*} {\rm gcd}(ab,ac)=|a|{\rm gcd}(b,c). \end{align*}

4) As ${\displaystyle a\mid n}$ and ${\displaystyle b\mid n}$, then there exist $q,q'\in \mathbb{Z}$ such that $n=qa$ and $n=q'b$. On then other hand, since ${\rm gcd}(a,b)=1$, by Bezout theorem, there exists $(u,v)\in\mathbb{Z}^2$ such that $au+bv=1$. This implies that \begin{align*} n&=nau+nbv\cr &= q'ba u+qab v\cr &= (q'u+qv)ab. \end{align*} This means that ${\displaystyle ab\mid n}$.

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