Find the smallest odd n,n>3 such that 3|n,5|n+2 and 7|n+4

Question

I usually solve problem by by inertia byut I found the solution that not met requirement of problem Then,I try to find the solution by CRT instead but I'm not sure the solution exist since

set

$x_{1}=0$ $M_{1}=35$ $y_{1}=?$ I discarded to solve this because by CRT it will be 0

Because I the problem asked for n>3 I must tried to manipulate sub-congruence

$x_{2}=-2$ $M_{2}=21$ $y_{2}=?$

reduce congruence to

$y_{2}\equiv 1 mod5$

Solve this and found $y_{2}=-4$

$x_{2}=-2$ $M_{2}=21$ $y_{2}=-4$

third congruence

$x_{3}=-4$ $M_{3}=15$ $y_{3}=?$

reduce the congruence to

$y_{3}\equiv 1 mod7$

Solve this and found $y_{3}=-6$

By CRT $(-2\times 21\times -4)+(-4\times 15\times-6 )$ and it is sum of even is even

Is it solution exits?

Answer 1

Hint: $n-3$ should be divisible by $2$, $3$, $5$ and $7$.

Answer 2

Well just write the system with congruences language.

$\begin{cases} n \text{ is odd} & n\equiv 1\pmod 2&\iff n \equiv 3\pmod {2}\\ 3\mid n & n\equiv 0\pmod 3&\iff n\equiv 3\pmod{3}\\ 5\mid n+2 & n+2\equiv 0\pmod 5&\iff n\equiv 3\pmod 5\\ 7\mid n+4 & n+4\equiv 0\pmod 7&\iff n\equiv 3\pmod 7\\ \end{cases}$

By application of Chinese remainder theorem we get $n\equiv 3\pmod {210}$, so the smallest integer greater than $3$ will be $213$.

Answer 3

Well, $n=3$ is obviously a solution. The CRT says that is unique in a residue system.

So we find another residue system....

The next solution will be $3+3*5*7$.

But, oops, that's not odd.

So the next is $3+2*3*5*7$.

Note: to formally solve note that it is a solution to

$n\equiv 1 \mod 2$

$n\equiv 0\mod 3$

$n\equiv -2\equiv 3 \mod 5$

$n\equiv -4\equiv 3 \mod 5$

Clearly answer is $3$. I'm not sure if your version of CRT states all solutions are congruent mod least common multiple, but that's a basic consequence.

So solution is $3+$ least common multiple of $2,3,5,7$.