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

Math100 said:

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 said:

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.##

Nanitf said:

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?