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Ever since I started doing work on Bézout Identity and Extended Euclidean Algorithm, I have tried to understand how to do modular arithmetic of big numbers etc. Basically, I have come up with the following problem of my own that I wish to solve:

$t = 24^3 \mod 35$

We know that $35 = 7\cdot 5$ and both $7$ and $5$ are prime numbers. I believe that there is a way of doing modular arithmetic of this nature if the modulus number is coprime like in this case.

I got as far as this: $$t = 24^3 \mod 7$$ $$t = 24^3 \mod 5$$

What do I do after? It would be great if someone could help me compute this answer as I am very keen to better understand all of this using my own examples.

Please I am new to CRT and any help of steps of calculating this would mean a lot!

user23473433819233d
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  • Hint $\ t \equiv -1$ mod $5$ & $7$ so also mod $35$ by CCRT. – Bill Dubuque Aug 13 '19 at 21:11
  • Look up the Chinese Remainder Theorem to get more details about the technique. –  Aug 13 '19 at 21:12
  • @MatthewDaly Sorry Sir to be a pain, but I don't suppose you would be ever so kind to apply the Chinese Remainder Theorem for me please? So like a computation breakdown please. I really want to learn new things and believe the only way to best understand it is by using my own example – user23473433819233d Aug 13 '19 at 21:14
  • One might say that $35$ is composite (since it isn't prime) or even semi-prime (because it has two prime factors), but it is not "coprime". That refers to the relation between two numbers with no common factor greater than $1$. – hardmath Aug 13 '19 at 21:16

2 Answers2

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Because $24=21+3\equiv3\pmod7,$

$t\equiv24^3 \pmod 7\implies t\equiv3^3 \pmod 7\implies t\equiv27\equiv\color{purple}{-1}\pmod 7.$

Because $24=25-1\equiv-1\pmod5,$

$ t\equiv24^3 \pmod 5\implies t\equiv(-1)^3 \pmod 7\implies t\equiv\color{purple}{-1}\pmod 5.$

Therefore, since $7$ and $5$ are coprime, by the constant case of the Chinese remainder theorem,

$t\equiv\color{purple}-1\equiv34\pmod {7\times5=35}.$

J. W. Tanner
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  • There is a huge gap after "therefore" (esp. at the OP's level). – Bill Dubuque Aug 13 '19 at 21:15
  • Sorry Sir, is this the Chinese Remainder Theorem? I am still confused as to where you got 3^3 (mod 7) from. I don't suppose you can do a more concise breakdown please. It is my fault that I am not good at higher Math – user23473433819233d Aug 13 '19 at 21:15
  • The long way would be $\color{blue}5\times\color{green}3+\color{blue}7\times\color{green}{-2}=1$ so $t=4\times\color{green}{-2}\times\color{blue}7+6\times\color{green}3\times\color{blue}5$ – J. W. Tanner Aug 13 '19 at 21:16
  • @BillDubuque: I edited, adding a bit more detail – J. W. Tanner Aug 13 '19 at 21:23
  • @user23473433819233d: $\color{brown}{24}=21+3=7\times3+3\equiv\color{brown}3\pmod7,$ so $\color{brown}{24}^3\equiv\color{brown}3^3\pmod7$ – J. W. Tanner Aug 13 '19 at 21:24
  • I dont suppose you could edit the answer to include more detail please Sir. Would be amazing if I could see every step of the calculation – user23473433819233d Aug 13 '19 at 21:25
  • @user23473433819233d At this point, you might as well learn what CRT is again, and how to use it from an actual book. This ain't it. – Rushabh Mehta Aug 13 '19 at 22:09
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I can rewrite tour congruence system as: $$t\equiv 6 \mod 7$$ $$t\equiv 4 \mod 5$$

Because $24^3>7$ then I can calculate the true remainder by (for $7$): $\left (\frac{24^3}{7}-\left \lfloor \frac{24^3}{7} \right \rfloor \right )\cdot 7=6$; and (for $5$): $\left (\frac{24^3}{5}-\left \lfloor \frac{24^3}{5} \right \rfloor \right )\cdot 5=4$.

From the first congruences I obtain: $t\equiv 34 \mod 35$. The solutions are: $t=35k-1$ with $k \in Z$.

Matteo
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  • Sorry Sir but where did the 6 and 4 come from? I am now really lost – user23473433819233d Aug 13 '19 at 21:17
  • $24^3>7$, so the remainder is $\frac{24^3}{7}$ minus its integer part; all times $7$. The same with $5$. – Matteo Aug 13 '19 at 21:20
  • So sorry Sir I am still lost. I don't suppose you would be kind enough to edit your answer to accomodoate for all of this? I take it this is CRT and if so, it would be great if I could see all the calculation steps – user23473433819233d Aug 13 '19 at 21:22