Find an integer having the remainders ## 1, 2, 5, 5 ##

[Math100]

Homework Statement

Find an integer having the remainders $1, 2, 5, 5$ when divided by $2, 3, 6, 12$, respectively. (Yih-hing, died $717$).

Relevant Equations

None.

Let $x$ be an integer.
Then $x\equiv 1\pmod {2}, x\equiv 2\pmod {3}, x\equiv 5\pmod {6}$ and $x\equiv 5\pmod {12}$.
Note that $x\equiv 5\pmod {6}\implies x\equiv 5\pmod {2\cdot 3}$ and $x\equiv 5\pmod {12}\implies x\equiv 5\pmod {3\cdot 4}$.
Since $gcd(2, 3)=1$ and $gcd(3, 4)=1$, it follows that $x\equiv 1\pmod {4}$.
This means $x\equiv 2\pmod {3}$ and $x\equiv 1\pmod {4}$.
Now we have $x=2+3a$ where $a=1+4b$ for some $a, b\in\mathbb{N}$.
Thus $x=2+3a=2+3(1+4b)=5+12b=5+12(1)=17$.
Therefore, the integer is $17$.

 

[fresh_42]

Why is $a\equiv 1 \pmod{5}?$.

You get directly from the given conditions that $x=12k+5.$ From that, $x\equiv 1\pmod{2}\, , \,x\equiv 2\pmod{3}$ and $x\equiv 5\pmod{6}$ follows automatically. So $x=12k+5$ is actually all we have, and conversely, all numbers $12k+5$ fulfill the criteria, e.g. $x=5.$

 

[Math100]

Why is $a\equiv 1 \pmod{5}?$.

You get directly from the given conditions that $x=12k+5.$ From that, $x\equiv 1\pmod{2}\, , \,x\equiv 2\pmod{3}$ and $x\equiv 5\pmod{6}$ follows automatically. So $x=12k+5$ is actually all we have, and conversely, all numbers $12k+5$ fulfill the criteria, e.g. $x=5.$

I don't understand your question about why is $a\equiv 1\pmod {5}$. But I found out that all $2, 3, 6$ divide $12$, so the system of linear congruences is consistent. And that is why I got $x=5+12b$ at the end.

 

[fresh_42]

I don't understand your question about why is $a\equiv 1\pmod {5}$.

Typo, I meant $\pmod{4}.$

But I found out that all $2, 3, 6$ divide $12$, so the system of linear congruences is consistent. And that is why I got $x=5+12b$ at the end.

You can get this right from the start from $x\equiv 5\pmod{12}$ without any calculation.

 

[Math100]

Typo, I meant $\pmod{4}.$

You can get this right from the start from $x\equiv 5\pmod{12}$ without any calculation.

So how would you show the work for this problem?

 

[fresh_42]

So how would you show the work for this problem?

$x\equiv 5\pmod{12} \Longrightarrow x=12k+5$ for some $k\in \mathbb{Z}.$
Conversely, any number $x=12k+5$ fulfills all given conditions and are such valid solutions.

The smallest solution is for $k=0$ that means $x=5.$

 

[Math100]

$x\equiv 5\pmod{12} \Longrightarrow x=12k+5$ for some $k\in \mathbb{Z}.$
Conversely, any number $x=12k+5$ fulfills all given conditions and are such valid solutions.

The smallest solution is for $k=0$ that means $x=5.$

That was short.

 

[WWGD]

As a comment, you may also use the Chinese Remainder Theorem. Just need to make sure the remainder is indeed Chinese ;).