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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 $21 = 21.$

$9x\equiv {24}\pmod {21}$.

$9x = 24*21$

$x = 24*21/9 = 56$

Martin Sleziak
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Amber
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    Before trying to solve this problem, you may find it helpful to review again the fundamentals of modular arithmetic. $9x \equiv 24\mod 21$ does not mean $9x = 24\cdot 21.$ – user7530 Jul 31 '13 at 20:23
  • Fun fact: $\rm an\equiv bn~mod~cn\Leftrightarrow a\equiv b~mod~c$. Thus $$\rm 9x\equiv24\equiv3~mod~21\Leftrightarrow 3x\equiv1~mod~7\Leftrightarrow 2\equiv6x\equiv(-1)x\Leftrightarrow x\equiv-2\equiv5~mod~7.$$ – anon Jul 31 '13 at 22:23
  • See here. – Mhenni Benghorbal Aug 01 '13 at 09:30
  • If you want to find A solution of a specific congruence like yours, some try-and-error may lead soon to success. For instance, it takes only a few seconds to notice that $9\cdot5=45=21+24$. If you want to find ALL solutions or some way to decide if an arbitrary linear congruence has solutions, the development of some theory is necessary. Obviously, knowing the exact definition of what a congruence is helps a great deal. – Andrea Mori Aug 01 '13 at 11:29

7 Answers7

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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.

Jim
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    Jim, there are telltale signs that she is a programmer. For them "mod" is (unfortunate here) so called binary mod, i.e. a processor call (or an arithmetic operation) calculating the remainder. Usually they arrange the sign of the remainder to match that of the dividend. That's why she emphasized the +sign. – Jyrki Lahtonen Jul 31 '13 at 20:28
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$$9x=24\pmod{21}\iff 9x=24+21k=3+21m\;,\;k\,,\,m\in\Bbb Z\implies$$

$$3x=1+7k\implies x=5\pmod 7$$

since $\,3\cdot 5=15=1\pmod 7\iff 3^{-1}=5\pmod 7\;$

DonAntonio
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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 $$9x\equiv 24\pmod{7}\Leftrightarrow 2x\equiv 3\pmod{7}\Leftrightarrow 4\cdot2x\equiv 4\cdot 3\pmod{7}\Leftrightarrow x\equiv 5\pmod{7}$$

Tomas
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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 $x=-16$.

rurouniwallace
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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

Dr. Shabir Ahmad Mir
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$21|9x-24\iff7|3x-8\iff7|3x-1\iff7|15x-5\iff7|x-5$

so solutions are $x=5, 12, 19,$ etc.

J. W. Tanner
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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

Sahiba Arora
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Ramesh Bhat
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