I've been trying to see how to calculate this math problem: 10+9a{ mod } 23 =14.I know that a=3, but I don't know the steps to apply to these digits to get 'a' if I were to calculate it. 10+9*3{ mod }23=14. I'm just looking for the steps on how to find 'a'. If I were asked to calculate it. Thanks so much in advance. And Please I would understand more if you could implement the answer mathematically like the way it's in the question above
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It implies $9a \pmod{23}=4$ and as $gcd(9,23)=1$ it guarantees the existence of one $a$ satisfying this (To see why, suppose not, and $9a=9b \pmod{23}$ for some distinct $a,b$ such that $|a-b|<23$. As $gcd(9,23)=1$ you can cancel the $9$ both sides to obtain $|a-b| = 0 \pmod {23}$ which is false as $|a-b|<23$, thus one should have $a=b$). Now use brute force. (I am not sure if other methods exist, which is unlikely as they rely on computation, it is fine if you show the existence of such $a$, then you obtain infinitely many $a$)
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@Andrew bro I don't understand your calculations..... What happened to the mod? – Dave Kent Apr 15 '20 at 02:31
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$a=b\pmod n$ is actually another way to write $n | a-b$, so... – Apr 15 '20 at 02:32
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Please could you implement the answer mathematically like the way it's in the question above – Dave Kent Apr 15 '20 at 02:50
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Ok, i will edit it. Is it fine now? – Apr 15 '20 at 02:56
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$10+9a\equiv14\bmod23\implies9a\equiv4\bmod23\implies9a\equiv27\bmod23\implies a\equiv3\bmod23.$
J. W. Tanner
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J. W. Tanner thanks bro but your answer is incorrect..... 27 mod 23 is 4...... Because 94=36 not 27.... The only way this solution could be applied is when 4 is three which means 93 would be equal to 27. Thanks so much bro – Dave Kent Apr 20 '20 at 14:50
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My answer is correct. I agree that $27$ mod $23$ is $4$; that's why I wrote $9a\equiv4\bmod23\implies 9a\equiv27\bmod23$. And I agree that $9\times4=36$ not $27$; that's why I wrote $9a\equiv27\bmod23\implies a\equiv3$ (not $4$) $\bmod 23$ – J. W. Tanner Apr 20 '20 at 14:59
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Oh.... Smart one bro.... You divided 27 by 9 to get 3 mod 23. Thanks a lot bro – Dave Kent Apr 21 '20 at 02:47
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Makes sense.... But does your solution apply to this question: 8=17*a mod 23 – Dave Kent Apr 22 '20 at 22:25
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$17a\equiv8\equiv-15\equiv-38\equiv-61\equiv-84\equiv-107\equiv-130\equiv-153\bmod23$ so $a\equiv-9\equiv14\bmod23$ – J. W. Tanner Apr 22 '20 at 22:58
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Your answer is confusing.... Please could you break it down for me.... I'm confused especially on the part where the number adds/multiplies up.... Thanks in advance – Dave Kent Apr 24 '20 at 17:40
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I just keep adding or subtracting the modulus, which doesn't change the residue class, until I find a representative where I can divide both sides to solve for $a$ – J. W. Tanner Apr 24 '20 at 18:07
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Oh I now see.... It has a common additive value... Could you break down the part when you divided both sides to get 14. – Dave Kent Apr 24 '20 at 20:00
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We have $17a\equiv-153\bmod23$. Since $17$ is relatively prime to $23$, we can divide both sides by $17$ to get $a\equiv-9\bmod23$. Then I just added $0\equiv23$ to get $a\equiv-9+23=14\bmod23$ – J. W. Tanner Apr 24 '20 at 20:09
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Tanner Nice one..... i'm surprised on how you knew that 153 was the right number to tally with 17..... would that be a problem if the number was somewhat larger? Like Ten digits large – Dave Kent Apr 24 '20 at 21:32
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If the numbers were much larger, I would use the extended Euclidean algorithm – J. W. Tanner Apr 24 '20 at 21:39
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How does that work? Is it like an extended version of this answer.... Please I would love an example if you don't mind – Dave Kent Apr 24 '20 at 23:19
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I didn't see the extended Euclidean algorithm..... I did see your answer though and i couldn't understand yours either....... i'm new to this algorithm – Dave Kent Apr 26 '20 at 02:42