Question

Find the remainder when a is divided by m

a = 207^321 + 689! ; m=7

Answer 1

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,

`(a)/(m)=(207^(321)+689!)/(7)\\\\=(207^(321))/(7)+(689!)/(7)\\\\=((210-3)^(321))/(7)+0\\\\=\frac{_(0)^(321)\textrm{C}* (210)^(321)* (3)^(0) -_(1)^(321)\textrm{C}* (210)^(320)* (3)^(1)+_(2)^(321)\textrm{C}* (210)^(319)* (3)^(2)-----(-1)* _(321)^(321)\textrm{C}* (321)^(0)* (3)^(321)}{7}\\\\ \text{As ,210 is divisible by 7}\\\\=( (3)^(321))/(7)`

⇒
`3^7` ,when divided by 7, gives remainder 3.

`( (3)^(321))/(7)=( (3)^(45* 7+6))/(7)\\\\=3+(3^6)/(7)\\\\=3+1\\\\=4`

So,Remainder when

`(207^(321)+689!)/(7)` is 4.