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I have asked a similar question before on Chinese Remainder Theorem.

Now concepts are getting clear. Thinking of a possible case where there are no solutions. Suppose the question is

x ≡ 2 (mod 88)      
x ≡ 3 (mod 99)

Then there will not be any solutions, correct?

Since 88 and 99 can be written as co-prime products 8*11 and 9*11,

x ≡ 2 (mod 8)      
x ≡ 2 (mod 11) 
x ≡ 3 (mod 9)   
x ≡ 3 (mod 11)

x ≡ 2 (mod 11) and x ≡ 3 (mod 11) will not come together. hence there are no solutions.

Is this right? Your comments are really helpful for me

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Kiran
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    Yes, you are right. Your analysis shows that the two congruences are incompatible. – André Nicolas Aug 19 '15 at 21:05
  • thanks @André Nicolas, this is really helping in my studies. – Kiran Aug 19 '15 at 21:06
  • @André Nicolas, whether the rearrangements must be as product of co-primes? For example,can x ≡ 1(mod 24) be rearranged as x ≡ 1(mod 2) and x ≡ 1(mod 12) ? I believe this can not be done and only product of co-primes must be possible. Am I right here? – Kiran Aug 19 '15 at 21:14
  • When you split a mod, they must be co-prime. – 1-___- Aug 19 '15 at 21:17
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    You are right. The system $x\equiv 1\pmod{2}$, $x\equiv 1\pmod{12}$ is not equivalent to the congruence $x\equiv 1\pmod{24}$. But the system $x\equiv 1\pmod{3}$, $x\equiv 1\pmod{8}$ is. – André Nicolas Aug 19 '15 at 21:17
  • @André Nicolas thank you very much. May I know the reason for this? I think the reasoning for this is "if a number can is divisible by co-primes, it is divisible by their products also" whereas the this cannot be true with all factors. right? – Kiran Aug 19 '15 at 21:21
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    You describe the reasoning correctly. One can give a proof using Unique Factorization. Or else use the Bezout "Identity" which says that if $a$ and $b$ are coprime there exist integers $x$ and $y$ such that $ax+by=1$. Now suppose $a$ and $b$ both divide $c$. We have $axc+byc=c$. Now we can show easily that if $a$ and $b$ divide $c$ then $ab$ divides each of $axc$ and $byc$, and hence it divides $c$. – André Nicolas Aug 19 '15 at 21:26
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    thx again, I am getting better and better in Maths due to this site. – Kiran Aug 19 '15 at 21:28
  • I encourage you to write an answer to this question containing what you've learned. This will help you cement your understanding, and my help others who find your question in the future. – davidlowryduda Aug 23 '15 at 04:48
  • thx @mixedmath. will do so now. – Kiran Aug 25 '15 at 07:34

1 Answers1

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Consider the system of congruences given below

x ≡ 2 (mod 88)      
x ≡ 3 (mod 99)

88 and 99 as product of co-prime numbers can be written as

88 = 8*11
99 = 9*11

Therefore, the given system must meet the following conditions

x ≡ 2 (mod 8)      
x ≡ 2 (mod 11) 
x ≡ 3 (mod 9)   
x ≡ 3 (mod 11)

x ≡ 2 (mod 11) and x ≡ 3 (mod 11) cannot happen together and these congruences are incompatible

Therefore, there are no solutions to the given system.

Kiran
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