# 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

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.

• So, calculate $m$ for your problem. And then calculate the number of positive integers $x<1000$. And then think about what's happening. – Gerry Myerson Oct 10 '14 at 11:46
• 27 and 37 are pairwise relatively prime so $m=999$ ,$x=a(mod\;999)$ and $x=b(mod\;999)$ then I think $x=999$ is only integer which is divisible by 999 and then we get integer $a=b=0$. Is it right ? @GerryMyerson – user114673 Oct 10 '14 at 18:31
• The Chinese Remainder Theorem says that for every $a$ and every $b$ there is a solution to $x\equiv a\pmod{27}$ and $x\equiv b\pmod{37}$, and what's more this solution is unique modulo 999. "Solution is unique modulo 999" is the same as saying there is exactly one solution among the numbers $1,2,\dots,999$. That should tell you what $N_{a,b}$ is. – Gerry Myerson Oct 11 '14 at 3:33
• So, how are we doing? – Gerry Myerson Oct 12 '14 at 6:37
• I am not well understand that if we take $a= 2$,$b= 3$ then we are getting the solution $x=299$ and if we take $a=3$, $b=4$ then $x=300$ then how can we say that the solution is unique. – user114673 Oct 12 '14 at 12:05

## 1 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 $a,b,\; N_{a,b}=1.$