-1

Solve $6x \equiv 4 \pmod{10}$. I've tried the following:

Note that $\gcd(6,10) = 2$. Since $2 \mid 4$, a solution exists. We can rewrite the problem as $6x \equiv 24 \pmod{10}$, by adding $10$ to the right side twice. So, $x \equiv 4 \pmod{10}$.

Is this a correct approach? I'm going through the book for examples but it's minimal. The class lectures didn't cover it either.

Ѕᴀᴀᴅ
  • 34,263
Accribus
  • 67
  • 2
    It has solution 3x=2(mod 5) –  Nov 29 '22 at 02:19
  • can you elaborate on how you got that? – Accribus Nov 29 '22 at 02:22
  • 1
    @Accribus Note that $x \equiv 9 \pmod{10}$ also solves the original linear congruence of $6x \equiv 4\pmod{10}$. This, along with $x \equiv 4\pmod{10}$ that you stated, shows the valid values of $x$ can be written as $x \equiv 4\pmod{5}$, with this being the unique solution to $3x\equiv 2\pmod{5}$ that Safal Das Biswas's comment stated. – John Omielan Nov 29 '22 at 02:23
  • 1
    You can't say "Because 6x ≡ 24 (mod 10),so x ≡ 4 (mod 10)".In fact we have 6x ≡ 24 (mod 2), but 6 and 24 are "zero" module 2, so you can't just divide 24 by 6. – jdhejw Nov 29 '22 at 02:28
  • 1
    Well yes dear!!! Look what is the meaning of modulo? A=B(mod C) means C divides A-B good to go(By definition). In Question we have 6x=4(mod 10) so by definition 10 divides 6x-4 in other words the fraction $\frac{6x-4}{10}$ should be an integer okay that means it leaves no remainder while divisible by 10 so we need to search atleast one x that works now fraction can be reduced it it shares common factor 6x-4 and 10 Indeed does so it's 2 right? So we need to find that x with (3x-2)/5 okay guess which x works smallest positive such x is 4. –  Nov 29 '22 at 02:29
  • Now how to construct infinitely , you are rightly guessed I known now it's diophantine equation. NOW NOTE IF YOU HAVE DOUBT AGAIN ASK DONT SHY? OKAY –  Nov 29 '22 at 02:32
  • 1
    If u are in the beginning of number theory I suggest you to read Burton and if you have any doubt disturb mse but should clear it and don't shy!!! People will help don't worry. –  Nov 29 '22 at 02:36

1 Answers1

0

When you render $6x\equiv24\bmod10$, that is correct; but dividing by $6$ must be accompanied by dividing the modulus by the gcd of the modulus itself ($10$) and the divisor ($6$). Thus the modulus is divided by $10/(gcd(10,6))=2$ and so is reduced from $10$ to $5$.

$6x\equiv24\bmod10\implies x\equiv4\color{blue}{\bmod 5}.$

Oscar Lanzi
  • 39,403