Respuesta :
Answer with explanation:
⇒689!=689×688×687×686×........×7×6×5×4×3×2×1
So, 689! is divisible by 7.
We have to find the remainder when,
[tex]\frac{a}{m}=\frac{207^{321}+689!}{7}\\\\=\frac{207^{321}}{7}+\frac{689!}{7}\\\\=\frac{(210-3)^{321}}{7}+0\\\\=\frac{_{0}^{321}\textrm{C}\times (210)^{321}\times (3)^{0} -_{1}^{321}\textrm{C}\times (210)^{320}\times (3)^{1}+_{2}^{321}\textrm{C}\times (210)^{319}\times (3)^{2}-----(-1)\times _{321}^{321}\textrm{C}\times (321)^{0}\times (3)^{321}}{7}\\\\ \text{As ,210 is divisible by 7}\\\\=\frac{ (3)^{321}}{7}[/tex]
⇒[tex]3^7[/tex] ,when divided by 7, gives remainder 3.
[tex]\frac{ (3)^{321}}{7}=\frac{ (3)^{45\times 7+6}}{7}\\\\=3+\frac{3^6}{7}\\\\=3+1\\\\=4[/tex]
So,Remainder when
[tex]\frac{207^{321}+689!}{7}[/tex] is 4.
