find a solution of 9x = 24 (mod 21) I need help finding a solution of $9x\equiv {24}\pmod {21}$.
Here is what I tried, but it's wrong.
mod x is the positive value of x. mod $21 = 21.$
$9x\equiv {24}\pmod {21}$.
$9x = 24*21$
$x = 24*21/9 = 56$
 A: $$9x=24\pmod{21}\iff 9x=24+21k=3+21m\;,\;k\,,\,m\in\Bbb Z\implies$$
$$3x=1+7k\implies x=5\pmod 7$$
since $\,3\cdot 5=15=1\pmod 7\iff 3^{-1}=5\pmod 7\;$
A: Since $21=3\cdot 7$ and $(3,7)=1$, we have:
$$9x\equiv 24\pmod{21}\Leftrightarrow 9x\equiv 24\pmod{3}\wedge9x\equiv 24\pmod{7}$$
Now, $9x\equiv 24\pmod{3}\Leftrightarrow 0x\equiv 0\pmod{3}$, which is always true and 
$$9x\equiv 24\pmod{7}\Leftrightarrow 2x\equiv 3\pmod{7}\Leftrightarrow 4\cdot2x\equiv 4\cdot 3\pmod{7}\Leftrightarrow x\equiv 5\pmod{7}$$
A: This essentially becomes a Diophantine equation:
$$9x\equiv{24}\pmod{21}  \\
9x-24=21y \\
9x-21y=24 \\
3x-7y=8$$
Solve the Diophantine equation: $3u-7v=1$ using the Euclidean algorithm:
$$3(2)+1=7\\
3(-2)-7(-1)=1\\
u=-2;\,v=-1$$
Now multiply out by $8$ to get the original Diophantine equation:
$$3(8u)-7(8v)=1(8)$$
$$x=8u=-16$$
Therefore $x=-16$.
A: I'm not sure what you mean by "mod x is the positive value of x", but you should not multiply $24$ and $21$.  What you want is a number $x$ such that $21$ will divide $9x - 24$.  We write this as $21 \mid 9x - 24$.  Note that $3$ divides both sides, so it must be that $21/3 \mid (9x - 24)/3$ or $7 \mid 3x - 8$.  From there you can try and calculate an inverse to $3$ modulo $7$ or you can just do a little guess and check to find that $x = 5$ works.
A: To solve 9x≡24(mod21).First check solvability of equation. To consider the congruence is solvable, the GCD of (9, 21) = 3 must be able to divide residue (24). It does. Therefore, simplify the equation by dividing each member (9, 24, 21) by 3; we get 3, 8 and 7 in that order; put these values into the equation; that now becomes : 3x≡8(mod7); but residue 8 is bigger than modulus 7 so simplify it by dividing 8 by 7 and collect the residue and use that instead of 8: 8 % 7 = 1. This makes the equation: 3x≡ 1(mod 7). Now apply Extended Euclidean Algorithm to 3 and 7; to get quotients in order as : 2 & 3 ; the last 2 columns of Magic Table will yield coefficients of dividend and divisor as 1 and 2: 2 is coefficient of 3 (as temporary value of x) and 1 is coefficient of modulus (7 is dividend with respect to 3 while applying EEA). The difference of two is checked to find algebraic sign of their difference: whether 1 obtained is +1 or -1. This stage is critical. In this case: (2*3) - (1*7) = -1. {Since we are interested in solving for x; the term containing x sould be kept on LHS (in this case: 3 x is to be kept on left).} Now, multiply 2 by -1; we get -2; it must be converted to positive integer by adding modulus (7); -2 + 7 = +5. The first particular solution of x has been located as 5. Check its validity by applying to modified and original equations: 3*5 - 1= 14 which is divisible by 7; 9*5 -24 = 21 that is divisible by 21. Both are multiples of 7. Since GCD was 3; there are exactly three valid solutions. Three solutions or values of x are obtained by adding 5 to multiples of 7  within limits of modulus 21 (original equation) i.e. 5+7*0 = 5 (x=5); 5+7*1 = 12 (x= 12); 5+7*2 = 19 (x=19). Three solutions for x are 5, 12, 19,....., 5+7k. You may go to any limit by adding 7 to preceeding value, but that is pointless, keep results within modulus (i.e. 21 in the instant case).
Prof.(Dr.)Shabir Ahmad Mir
A: $21|9x-24\iff7|3x-8\iff7|3x-1\iff7|15x-5\iff7|x-5$
so solutions are $x=5, 12, 19,$ etc.
A: By trial and error method:
\begin{align*}
9x &\equiv24\mod21\\
\implies 3x&\equiv24 8\mod7  \text{ (dividing by} 3)\\
\implies 7 &| (3x-8)
\end{align*}
(Put the values: $x=0,1,2,3,4,5\cdots$ or $-1,-2,-3,-4,-5\cdots$ and check it.
So for the value $x=5$ or $-2$  it satisfies $7 | (3x-8).$)
Thus $x\equiv5\mod7$ or $x\equiv-2\mod7.$ This implies $x\equiv5\mod7.$
[Hint: For  entrance exams, put the given options directly and check it.]
©by Ramesh Bhat, Nakre
