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I am trying to get an idea of how the Chinese Remainder Theorem (CRT) can be used to finish up this problem, in which the problem

$$7^{30}\equiv x\pmod{ 100}$$

is attempted by splitting the modulus into relatively prime factors $25$ and $4,$ arriving at

$$\begin{align} 7^{30}&\equiv1\pmod4\\ 7^{30}&\equiv-1\pmod{25} \end{align}$$

I understand that the CRT may be called upon because $m=\prod m_i,$ and we have the same $7^{30}$ value on the LHS, but I don't know how to carry it out.

The question was touched upon in this post as the second entry:

How do I efficiently compute $a^b \pmod c$ when $b$ is less than $c.$ For instance $5^{69}\,\bmod 101.$

However, I don't see this particular point clearly worked out, perhaps because it is a multi-pronged question.

Following this presentation online, this seems to be the verbatim application of the CRT without any added concepts or shortcuts:

From @gimusi's answer (upvoted):

$$\begin{cases} x \equiv 7^{30} \pmod4\\ x\equiv 7^{30} \pmod{25} \end{cases}$$

rearranged into

\begin{cases} x \equiv 1 \pmod4\\ x\equiv -1 \pmod{25} \end{cases}

Given the general form of the equations above as $x\equiv a_i \pmod {m_i},$ the CRT states $x\equiv a_1 b_1 \frac{M}{m_1}+a_2 b_2 \frac{M}{m_2}\pmod M$ with $M=\prod m_i,$ and with

$$b_i =\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}.$$

The inverse of $\frac{M}{m_i}$ is such that $\frac{M}{m_i}\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}\equiv 1.$

Calculating the components:

$$\begin{align} a_1&=1\\ a_2&=-1\\ M&=4\times 25 =100\\ \frac{M}{m_1} &= \frac{100}{4}=25\\ \frac{M}{m_2} &= \frac{100}{25}=4\\ b_1 &= \left(\frac{M}{m_1}\right)^{-1} \pmod 4 = (25)^{-1}\pmod 4 =1\\ b_2 &= \left(\frac{M}{m_2}\right)^{-1} \pmod {25}= (4)^{-1} \pmod{25}=19 \end{align}$$

Hence,

$$x=1\cdot 25 \cdot 1 + (-1)\cdot 4 \cdot 19 = -51 \pmod{100}\equiv 49.$$

Antoni Parellada
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4 Answers4

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Welcome to Math SX! You have to use Euler's theorem as $\varphi(4)=2$, $\;\varphi(25)=20$ we have $$ 7^{30}\equiv7^{30\bmod2}=1\mod 4,\qquad 7^{30}\equiv7^{30\bmod20}=7^{10}\mod 25$$ To find the latter power, you can use the modular fast exponentiation algorithm, but here, it will be simpler: modulo $25$, $$7^2\equiv -1\enspace\text{so}\enspace 7^4=1,\enspace\text{hence } \;7^{30}\equiv 7^{30\bmod 4}=7^2\equiv -1.$$ Finally, since $\;25-6\cdot 4=1$ (Bézout's identity), $$7\equiv \begin{cases}\phantom{-}1\mod4\\-1\mod 25\end{cases}\iff 7\equiv 1\cdot 25-(-1)\cdot 6\cdot 4=49\mod 100.$$

J. W. Tanner
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Bernard
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  • Thank you. I'm not sure how to interpret expressions such as $7^{30}\equiv7^{30\bmod2}=1.$ I looked up Euler's theorem, and it led me to things like $7^{\phi(4)}\equiv 1 \pmod 4.$ – Antoni Parellada Jan 21 '18 at 21:59
  • So you can reduce the exponent modulo φ(4), which is equal to $2$. This means $7^n\equiv 7$ if $n$ is odd, ${}\equiv 1$ if $n$ is even. – Bernard Jan 21 '18 at 22:02
  • So we left it at $7^{\phi(4)}=7^2\equiv 1 \pmod 4$ and $7^{\phi(25)}=7^{20}\equiv 1 \pmod 25.$ How do you go from this to $7\equiv \begin{cases}\phantom{-}1\mod4\-1\mod 25\end{cases}$ and the subsequent final identity? – Antoni Parellada Jan 21 '18 at 22:34
  • For the last case, I had a shortcut: $7$ has order $4$ mod. $25$, not $20$, so I simply had to reduce the exponent mod. $4$, which makes the computation, considerably simpler. – Bernard Jan 21 '18 at 22:39
  • Well, if $7^4\equiv 1$, $7^5=7^4\cdot 7\equiv 1\cdot 7=7 $, $7^6=7^4\cdot 7^2\equiv 1\cdot (-1)=-1 $, $7^7=7^4\cdot 7^3\equiv 1\cdot (-7)=-7 $, $7^8=(7^4)^2\equiv 1^2=7 $, and so on. More generally $7^{4k+r}=7^{4k} \cdot7^r\equiv 1\cdot 7^r=7^r $. – Bernard Jan 21 '18 at 23:23
  • I've come across other posts by you on the same topic, and it seems as though you always go through Bézout's identity to set up the final linear equation. Is that your choice, or something implicit, yet not universally included in explanations of the CRT? – Antoni Parellada Jan 22 '18 at 14:45
  • Well, Bézout's identity is fundamental tool in arithmetic questions, especially through the extended Euclidean algorithm, which gives effective tool to compute modular inverses, gcds, lcms. I often use it because many questions are ’How do I compute, or solve, this or that’, and there are not so many general tools. But I also gave some purely conceptual answers, which do not use this identity (e.g. proving that $\sqrt n$ can be rational only if $n$ is a perfect square – which requires only knowing that any non-empty subset of $\mathbf N$ has a smallest element). – Bernard Jan 22 '18 at 17:35
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$$7^{30}\equiv x\pmod{ 100}$$

Could be solved easily without Chinese Remainder Theorem. Note that $7^4=2401 \equiv 1\pmod {100} $ Thus $$ 7^{30} = 7^{28}\times 49 \equiv 49 \pmod {100}$$

Solving the system with Chinese Remainder Theorem requires finding a linear combination of $25$ and $4$ to equal 1.

Such a combination is $$ 1= 1(25) -6(4) $$

Therefore the answer to the system is $$ x\equiv (1)(1)(25) +(-1)(-6)(4) \pmod {100}$$

That is $$ x\equiv 49 \pmod {100}$$

Mohammad Riazi-Kermani
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From here

$$\begin{cases} x \equiv 7^{30} \pmod4\\ x\equiv 7^{30} \pmod{25} \end{cases}$$

by CRT we know that solutions exist $\pmod{100}$.

Then note that since $7^2=49\equiv 1 \pmod4$

$$x\equiv7^{30} \implies x\equiv 49^{15} \equiv 1\pmod4$$

and since $7^2=49\equiv -1 \pmod{25}$

$$x\equiv7^{30} \implies x\equiv 49^{15} \equiv -1\pmod{25}$$

Thus the system becomes

$$\begin{cases} x \equiv 1 \pmod4\\ x\equiv -1 \pmod{25} \end{cases}$$

Note that CRT guarantees that the solutions exist $\pmod{100}$ but doesn't give special shortcut to find the solution.

When you can't by inspection (in this case you can easily find $x=49$), in general to find the solution you can follow the procedure indicated here CRT -Case of two moduli

user
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  • There is a formula based on Bézout's identity. – Bernard Jan 21 '18 at 21:44
  • @Bernard Of course you can use Bezout, what I mean is that CRT is not aimed to give the solution but to prove that solution exist, am I wrong? Of course we can use the method applied for the proof to find the solutions. – user Jan 21 '18 at 21:49
  • Yes, but there exists an effective CRT, fairly simple in the case of two congruences. – Bernard Jan 21 '18 at 21:53
  • @Bernard Yes indeed it is precisely the link I've given! :) – user Jan 21 '18 at 21:55
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Chinese Remainder Theorem says:

$$\mathbb{Z}/100 \simeq \mathbb{Z}/25 \times \mathbb{Z}/4$$

where the isomorphism is given by mapping $x \pmod {100}$ to $(x \pmod {25}, x \pmod {4})$. Thus, the class of $x$ is a number from $0$ to $99$ that congruence to $1 \bmod 4$ and $24 \bmod 25$. The numbers $0 \le x \le 99$ and $x \equiv 24 \pmod {25}$ are: $24, 49, 74, 99$. Now which one is congruence to $1 \bmod 4$?

Huy Dang
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