Let A = (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

=> 2A = (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3¹⁶ - 1)(3¹⁶ + 1)(3³² + 1)

           = (3³² - 1)(3³² + 1)

           = 3⁶⁴ - 1

\(\Rightarrow A=\dfrac{3^{64}-1}{2}\)

(a) 5^m.25^2m = 125^{2m - 1}

=> 5^m.5^4m = (5³)^{2m - 1}

=> 5^5m = 5^{6m - 3}

=> 5m = 6m - 3

=> m = 3

(b) 5^m.25^m.125^m.625^m = 5³⁰

=> (5.25.125.625)^m = 5³⁰

=> (5.5².5³.5⁴)^m = 5³⁰

=> 5^10m = 5³⁰

=> 10m = 30

=> m = 3

We have : (2x² + 1)(5x + 2) = 10x³ + 4x² + 5x + 2

So a = 4

\(\left\{{}\begin{matrix}3x+y=9\\5x-4y=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=9-3x\\5x-4\left(9-3x\right)=32\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=9-3x\\17x-36=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=4\end{matrix}\right.\)

Replace x = 5 and y = 4 into the given equations,we have :

\(\left\{{}\begin{matrix}5k-4m=7\\5m+4k=22\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}20k-16m=28\\20k+25m=110\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}41m=82\\k=\dfrac{4m+7}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=2\\k=3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x+4y=5\\4x-2y=14\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+4y=5\\4y-8x=-28\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}11x=33\\y=\dfrac{5-3x}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x+3y=5\\y-3x=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3\left(3x+9\right)=5\\y=3x+9\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}11x+27=5\\y=3x+9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

=> 5(2x + y) + 2(3y - 3x) + x + y = 5.(-1) + 2.15 - 2 + 3 = 26

\(\left\{{}\begin{matrix}3x+2y=12\\3x-y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3x+2y\right)-\left(3x-y\right)=12-3\\3x=y+3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3y=9\\x=\dfrac{y}{3}+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)

=> 4x - 8y = -16 => 2a = 16 => a = 8

\(a+b⋮7\Rightarrow2a+2b⋮7\)but\(3a-b⋮7\)

So we have :\(\left(2a+2b\right)+\left(3a-b\right)⋮7\)or\(5a+b⋮7\)

\(3a+5b⋮13\Rightarrow2\left(3a+5b\right)⋮13\)or\(6a+10b⋮13\)

We have :\(13a+13b⋮13\Rightarrow\left(13a+13b\right)-\left(6a+10b\right)⋮13\)

\(\Rightarrow7a+3b⋮13\)

We have : 882 = 2 x 3² x 7² ; 1134 = 2 x 3⁴ x 7.So :

GCD(882 ; 1134) = 2 x 3² x 7 = 126

LCM(882 ; 1134) = 2 x 3⁴ x 7² = 7938

Let n be the smallest value of that number. When n is divided by 2,3,4,5,6,the remainder is always 1,so n - 1 is divisible by 2,3,4,5,6

=> n - 1 = LCM(2,3,4,5,6) = 60 => n = 61

We have : 3 - 2 = 4 - 3 = 5 - 4 = 6 - 5 = 1

So n + 1 is divisible by 3,4,5,6 but we need to find the smallest value of n.

=> n + 1 = LCM(3,4,5,6) = 60 => n = 59

(m ; n) = (2 ; 7)