Is this possible to solve through algebra?

Question

$$150 \equiv 17 \mod x, \qquad 100 \equiv 5 \mod x $$

Solve the simultaneous equation? Is this even a simultaneous equation? How do I find the value of $x$ too? I was doing a question and came up with these equations...I know the basics of modular arithmetic but don't really know too difficult ones

score 7 · Accepted Answer · answered Jun 19 '16 at 13:10

7

Since $150-17=133$, we have $$ 150\equiv17\pmod{x}\implies x\mid133 $$ Furthermore, since $100-5=95$, we have $$ 100\equiv5\pmod{x}\implies x\mid95 $$ The possibilities for $x$ can be derived from the fact that $133=7\cdot19$ and $95=5\cdot19$.

answered Jun 19 '16 at 13:10

robjohn

345,667

So we have use common factor to find the value of x? – Arc Neoepi Jun 19 '16 at 13:14
1

Well, we must have $x\mid19$. This leaves more than one possibility for $x$. – robjohn Jun 19 '16 at 13:15
I see...Thanks for your help! – Arc Neoepi Jun 19 '16 at 13:17

P Vanchinathan · Answer 2 · 2016-06-20T10:56:47.117

3

If $a,b,c,d$ are specific numbers and $x$ is unknown (all positive integers) then the equation $a\equiv b\pmod x$ and $c\equiv d\pmod x$ by definition means $x$ is a common divisor of the two numbers $a-b$ and $c-d$.

The solution need not be unique. So solving them is the same as finding the set of all common divisors. This is same as finding $g=\gcd(a-b,c-d)$ and then listing all divisors of $g$.

edited Jun 20 '16 at 10:56

answered Jun 19 '16 at 13:46

P Vanchinathan

19,504

So,can the solution be 1?since it is not unique? – Arc Neoepi Jun 19 '16 at 14:02
It should be called the trivial solution, it is always there. as the statement $1$ divides any any positive integer though correct and important, is not a profound statement. Unique solution means it has to be $1$. – P Vanchinathan Jun 19 '16 at 14:42
I see.Thanks for your help! – Arc Neoepi Jun 19 '16 at 14:43

score 2 · Answer 3 · answered Jun 19 '16 at 13:29

2

Hint:

The given equations imply $150-17=7\cdot19$ and $100-5=5\cdot19$ to be multiples of $x$. The $\gcd$ is $19$.

answered Jun 19 '16 at 13:29

score 2 · Answer 4 · edited Apr 13 '17 at 12:19

2

By definition your system is equivalent to $\ x\mid 133,\ x\mid 95,\,\ $ i.e. $\,\ x\mid 133,195$.

The universal property of the gcd tells us how to combine this pair of divisibility statements into an equivalent single divisibility statement, namely

$$ x\mid a,b \iff x\mid \gcd(a,b)$$

edited Apr 13 '17 at 12:19

Community

1

answered Jun 20 '16 at 13:13

Bill Dubuque

272,048

2

I don't understand why this answer is down voted. Strange. – P Vanchinathan Jun 22 '16 at 14:28

Is this possible to solve through algebra?

4 Answers4