Question

what is the smallest possible value of N which satisfy: ("==" means congruent)
N==3(mod 4)
N==2(mod 5)
N==6(mod 7)

Answer 1

do u know chineese remainder theorem

Answer 2

\(\large \color{black}{\begin{align}
\huge \text{CRT}\hspace{.33em}\\~\\
N & \equiv 3\pmod 4\hspace{.33em}\\~\\
N & \equiv 2\pmod 5\hspace{.33em}\\~\\
N & \equiv 6\pmod 7\hspace{.33em}\\~\\
N& =\left[3\cdot (5\cdot7) \left((5\cdot7)^{-1}\pmod 4\right)\\~\\
+2\cdot (4\cdot7) \left((4\cdot7)^{-1}\pmod 5\right)\\~\\
+6\cdot (4\cdot5) \left((4\cdot5)^{-1}\pmod 7\right)\right]\pmod {4\cdot 5\cdot7}\\~\\
\end{align}}\)

Answer 3

yes I do know CRT. I got 52 as my answer which is wrong. I actually didn't used the CRT, I took the two equations, N==2(mod 5) and N==6(mod7)
writing them algebraically: N=5a+2=7b+6 which simplifies 5a=7b+6
taking modulo 5: 0==2b+1
now, subtracting 1 from both sides: 2b==0(mod5)
b=5! 2*5=10==0(mod 5)
plugging in the value of b in 7b+1: 7*5=35, +6 = 41,
N==41(mod 35) AND the remaining one was:
N==3(mod 4)
applying the whole same method

Answer 4

\(\large \color{black}{\begin{align}
\huge \text{CRT}\hspace{.33em}\\~\\
N \equiv 3\pmod 4\hspace{.33em}\\~\\
N \equiv 2\pmod 5\hspace{.33em}\\~\\
N \equiv 6\pmod 7\hspace{.33em}\\~\\
N =\left[3\cdot (5\cdot7) \left((5\cdot7)^{-1}\pmod 4\right)\\~\\
+2\cdot (4\cdot7) \left((4\cdot7)^{-1}\pmod 5\right)\\~\\
+6\cdot (4\cdot5) \left((4\cdot5)^{-1}\pmod 7\right)\right]\pmod {4\cdot 5\cdot7}\\~\\
(5\cdot7)^{-1}\pmod 4\\~\\
35x\equiv 1\pmod 4\\~\\
3x\equiv 1\pmod 4\\~\\
\fbox{x=3}\\~\\
(4\cdot7)^{-1}\pmod 5\\~\\
28y\equiv 1\pmod 5\\~\\
3y\equiv 1\pmod 5\\~\\
\fbox{y=2}\\~\\
(4\cdot5)^{-1}\pmod 7\\~\\
20z\equiv 1\pmod 7\\~\\
6z\equiv 1\pmod 7\\~\\
\fbox{z=6}\\~\\
N =\left[3\cdot (5\cdot7) \left(3\right)\\~\\
+2\cdot (4\cdot7) \left(2\right)\\~\\
+6\cdot (4\cdot5) \left(6\right)\right]\pmod {4\cdot 5\cdot7}\\~\\
N =\dfrac{3\cdot (5\cdot7) (3)}{4\cdot 5\cdot7}\\~\\
+\dfrac{2\cdot (4\cdot7) (2)}{4\cdot 5\cdot7}\\~\\
+\dfrac{6\cdot (4\cdot5) (6)} {4\cdot 5\cdot7}\\~\\
=\dfrac{3\cdot (3)}{4}\\~\\
+\dfrac{2\cdot (2)}{5}\\~\\
+\dfrac{6\cdot) (6)} {7}\\~\\
=\dfrac{1}{4}\\~\\
+\dfrac{-1}{5}\\~\\
+\dfrac{1} {7}\\~\\
=\dfrac{35-28+20}{4\cdot5\cdot7}\\~\\
=\dfrac{27}{4\cdot5\cdot7}\\~\\
\Large =27



\end{align}}\)