Find the smallest integer ## a>2 ## such that ## 2\mid a ##

[Math100]

Homework Statement

Find the smallest integer $a>2$ such that $2\mid a, 3\mid (a+1), 4\mid (a+2), 5\mid (a+3), 6\mid (a+4)$.

Relevant Equations

None.

Let $a>2$ be the smallest integer.
Then
\begin{align*}
&2\mid a\implies a\equiv 0\pmod {2}\implies a\equiv 2\pmod {2}\\
&3\mid (a+1)\implies a+1\equiv 0\pmod {3}\implies a\equiv -1\pmod {3}\implies a\equiv 2\pmod {3}\\
&4\mid (a+2)\implies a+2\equiv 0\pmod {4}\implies a\equiv -2\pmod {4}\implies a\equiv 2\pmod {4}\\
&5\mid (a+3)\implies a+3\equiv 0\pmod {5}\implies a\equiv -3\pmod {5}\implies a\equiv 2\pmod {5}\\
&6\mid (a+4)\implies a+4\equiv 0\pmod {6}\implies a\equiv -4\pmod {6}\implies a\equiv 2\pmod {6}.\\
\end{align*}
Observe that $lcm(2, 3, 4, 5, 6)=60$.
Thus $a\equiv 2\pmod {60}\implies a=62$.
Therefore, the smallest integer $a>2$ such that $2\mid a, 3\mid (a+1), 4\mid (a+2), 5\mid (a+3), 6\mid (a+4)$ is $62$.

 

[fresh_42]

Homework Statement:: Find the smallest integer $a>2$ such that $2\mid a, 3\mid (a+1), 4\mid (a+2), 5\mid (a+3), 6\mid (a+4)$.
Relevant Equations:: None.

Let $a>2$ be the smallest integer.
Then
\begin{align*}
&2\mid a\implies a\equiv 0\pmod {2}\implies a\equiv 2\pmod {2}\\
&3\mid (a+1)\implies a+1\equiv 0\pmod {3}\implies a\equiv -1\pmod {3}\implies a\equiv 2\pmod {3}\\
&4\mid (a+2)\implies a+2\equiv 0\pmod {4}\implies a\equiv -2\pmod {4}\implies a\equiv 2\pmod {4}\\
&5\mid (a+3)\implies a+3\equiv 0\pmod {5}\implies a\equiv -3\pmod {5}\implies a\equiv 2\pmod {5}\\
&6\mid (a+4)\implies a+4\equiv 0\pmod {6}\implies a\equiv -4\pmod {6}\implies a\equiv 2\pmod {6}.\\
\end{align*}
Observe that $lcm(2, 3, 4, 5, 6)=60$.
Thus $a\equiv 2\pmod {60}\implies a=62$.
Therefore, the smallest integer $a>2$ such that $2\mid a, 3\mid (a+1), 4\mid (a+2), 5\mid (a+3), 6\mid (a+4)$ is $62$.

Nice. And correct.

You could drop the first condition $a\equiv 0\pmod 2$ since it follows automatically from $4\,|\,(a+2).$ Same for $a\equiv 2\pmod 3.$ We only need $4,5,6.$

($6\,|\,(a+4)\Longrightarrow 3\,|\,(a+3+1)\Longrightarrow 3\,|\,(a+1)$)

 

[Nanitf]

You could drop the first condition $a\equiv 0\pmod 2$ since it follows automatically from $4\,|\,(a+2).$ Same for $a\equiv 2\pmod 3.$ We only need $4,5,6.$

I'm sorry but what do you mean by it follows automatically from 4 | a=(a+2) for the first condition? Could you show how you derived that?

 

[fresh_42]

I'm sorry but what do you mean by it follows automatically from 4 | a=(a+2) for the first condition? Could you show how you derived that?

Assume $a\not\equiv 0 \pmod{2}$ then $a=2k+1$ for some $k\in \mathbb{Z}$ and $a+2=2k+3$ is odd, and thus cannot be divided by $4.$ Or positively:
\begin{align*}
4\,|\,(a+2) \Longrightarrow 2\,|\,(a+2) \Longrightarrow 2\,|\,a \Longrightarrow a\equiv 0\pmod{2}
\end{align*}
where the only interesting step is $2\,|\,(a+2) \Rightarrow a+2=2\cdot b\Rightarrow a=2\cdot b-2=2\cdot(b-1) \Rightarrow 2\,|\,a.$