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Consider the following congruences:

(1) $x \equiv 3\pmod 4$

(2) $x \equiv 11\pmod{8}$

(3) $x \equiv 13\pmod{16}$

Then

(A). (1) and (2) have a common solution but (1) and (3) do not have a common solution.

(B). (1) and (2) have a common solution but (2) and (3) do not have a common solution.

(C). (1) and (3) have a common solution but (1) and (2) do not have a common solution.

(D). (1) and (3) have a common solution but (2) and (3) do not have a common solution.

(E). (2) and (3) have a common solution but (1) and (2) do not have a common solution.

(F). (2) and (3) have a common solution but (1) and (3) do not have a common solution.

(G). All the three congruences have a common solution.

(H). Any two of the congruences have a common solution.

We know that if the integers $m_i$ where $i = 1,2,3,...,n$ are relatively prime in pairs then then congruences $x \equiv a_i\pmod m_i$ where $a_i$ are integers have one and only one common solution congruent $\pmod M$ where $M = m_1m_2...m_n$. But here $m_1 = 4, m_2 = 8, m_3 = 16$ are not relatively primes. So Chinese Remainder theorem is not applicable here. How we can solve the problem ?

Robert Z
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Sangam Academy
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1 Answers1

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If $x\equiv 13\pmod{16}$ then $x=13+16k$ for some $k\in\mathbb{Z}$ which implies that $$x=13+16k\equiv 5\pmod{8}\qquad, \qquad x=13+16k\equiv 1\pmod{4}.$$ Hence (3) does not have a common solutions with (1) (because and $x \equiv 3\pmod 4$) and (2) (because and $x \equiv 11\equiv 3\pmod 8$).

Does (1) and (2) have a common solutions? If $x\equiv 11\pmod{8}$ then what is $x$ congruent to modulo $4$?

Robert Z
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  • @Sangam Academy Any further doubt? Have you tried to compare (1) and (2)? – Robert Z Jul 17 '21 at 07:10
  • Yes, from first two, $x = 3+4p = 11+ 8q \implies 4(p-2q) = 8 \implies p-2q = 2$ which is possible for $p = 4$ and $q = 1$. So (1) and (2) have a common solution. – Sangam Academy Jul 17 '21 at 07:23
  • @Sangam Academy Yes, we may also say that $3+8k$ with $k\in\mathbb{Z}$ are all the common solutions of (1) and (2). – Robert Z Jul 17 '21 at 07:33
  • @Sangam Academy Was my answer of any help or did you find the solution in another way? – Robert Z Jul 17 '21 at 07:35
  • Yes, your answer is helpful. But your comment "$3+8k$ with $k \in \Bbb Z$ are all common solutions of (1) and (2)" is unclear for me. – Sangam Academy Jul 17 '21 at 07:55
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    Note that $x=3+8k$ are all the solutions of (2). Moreover $x=3+8k\equiv 3 \pmod{4}$ and therefore it solves also (1). – Robert Z Jul 17 '21 at 08:05
  • Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 17 '21 at 17:52