Chinese Remainder Theorem Part 2 – Non Coprime Moduli

As promised on the last post, today we are going to discuss the “Strong Form” of Chinese Remainder Theorem, i.e, what do we do when the moduli in the congruence equations are not pairwise coprime. The solution is quite similar to the one we have already discussed in the previous post, so hopefully, it will be a lot easier to understand.


If you haven’t read my previous post, you can find it here: Chinese Remainder Theorem Part 1. I strongly suggest you read that one before proceeding onwards.


Given two sequences of numbers $A = [a_1, a_2, \dots, a_n]$ and $M = [m_1, m_2, \dots, m_n]$, find the smallest solution to the following linear congruence equations if it exists:

$$x \equiv a_1 \text{(mod } m_1) \\ x \equiv a_2 \text{(mod } m_2) \\ \dots \\ x \equiv a_n \text{(mod } m_n)$$

Note: The elements of $M$ are not pairwise coprime

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Euler Phi Extension and Divisor Sum Theorem

Previously we learned about Euler Phi Function. Today we are going to look at two theorems related to Euler Phi that frequently appears in CPPS. I am not sure whether these theorems have any official names, so I just made them up. These allow easy references so I will be using these names from now on.

Euler Phi Extension Theorem

Theorem: Given a number $N$, let $d$ be a divisor of $N$. Then the number of pairs $\\{a, N \\}$, where $1 \leq a \leq N$ and $gcd(a,N) = d$, is $\phi(\frac{N}{d})$.

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Euler Totient or Phi Function

I have been meaning to write a post on Euler Phi for a while now, but I have been struggling with its proof. I heard it required the Chinese Remainder Theorem, so I have been pushing this until I covered CRT. But recently, I found that CRT is not required and it can be proved much more easily. In fact, the proof is so simple and elegant that after reading it I went ahead and played Minecraft for 5 hours to celebrate.


Given an integer $N$, how many numbers less than or equal $N$ are there such that they are coprime to $N$? A number $X$ is coprime to $N$ if $gcd(X,N)=1$.

For example, if $N = 10$, then there are $4$ numbers, namely ${1,3,7,9}$, which are coprime to $10$.

This problem can be solved using Euler Phi Function, $phi()$. Here is the definition from Wiki:

In number theory, Euler’s totient function (or Euler’s phi function), denoted as $\phi(n)$, is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n. – Wiki

That’s exactly what we need to find in order to solve the problem above. So, how does Euler Phi work?

Euler Phi Function

Before we go into its proof, let us first see the end result. Here is the formula using which we can find the value of the $phi()$ function. If we are finding Euler Phi of $N = p_1^{a_1}p_2^{a_2}…p_k^{a_k}$, then:

$$\phi(n) = n \times \frac{p_1-1}{p_1} \times \frac{p_2-1}{p_2}… \times \frac{p_k-1}{p_k}$$

If you want you can skip the proof and just use the formula above to solve problems. That’s what I have been doing all these years. But I highly recommend that you read and try to understand the proof. It’s simple and I am sure someday the proof will help you out in an unexpected way.

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