[Math] For any integer $a,b$ let $N_{a,b}$ denote the number of positive integer $x<1000$ satisfying $x= a( mod\;27)$ and $x=b(mod\;37)$. Then

congruenceselementary-number-theory

For any integer $a,b$ let $N_{a,b}$ denote the number of positive integer $x<1000$ satisfying $x= a \mod 27$ and $x=b \mod 37$. Then which of them is correct

  1. there exist $a,b$ such that $N_{a,b} =0$
  2. for all $a,b$, $N_{a,b}=1$
  3. for all $a,b$, $N_{a,b}>1$
  4. there exist $a,b$ such that $N_{a,b}$ and there exist $a,b$ such that $N_{a,b}=2$.

Since we know that by Chinese Remainder Theorem if $m_1, m_2, .., m_k$ are pairwise relatively prime positive integers, and if $a_1, a_2, .., a_k$ are any integers, then the
simultaneous congruences $x ≡ a_1 (mod\;m_1), x ≡ a_2 (mod\;m_2), …, x ≡ a_k (mod\;m_k)$ have a solution, and the solution is unique modulo $m$, where
$m = m_1m_2⋅⋅⋅m_k$. Then how we will approach by acoording to this theorem. So please someone help me and explain it.

Best Answer

When $n=a (mod\;m_1)$ and $n=b (mod\;m_2)$ are only solvable when $a = b (mod\;(gcd\;(m_1,m_2))$. The solution is unique modulo $lcm\;(m_1,m_2)$. By this we get $a=b\;mod(1)$ which is true for all values of $a$ and $b$. So from here we can say that for all $a,b,\; N_{a,b}=1.$

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