Let X=[ 2 1 0 3 ] be a matrix. If X6=[ a b c d ] , then the number of divisors of (a + b + 2020 c + d) is equal to

Question

Let X=[ 2 1 0 3 ] be a matrix. If X6=[ a b c d ] , then the number of divisors of (a + b + 2020 c + d) is equal to (A) 7 (B) 14 (C) 21 (D) 28

Tardigrade Admin · Accepted Answer

X2=[ 2 1 0 3 ][ 2 1 0 3 ]=[ 4 5 0 9 ] X3=X2X=[ 4 5 0 9 ][ 2 1 0 3 ]=[ 8 19 0 27 ] X6 = X3 ⋅ X3 = [ 8 19 0 27 ] [ 8 19 0 27 ] = [ 64 665 0 729 ] ⇒ a=64,b=665,c=0,d=729 ⇒ a+b+2020c+d=1458=2× 36 Number of divisors =(1 + 1)(6 + 1)=14

Q. Let $X = [2013]$ be a matrix. If $X^{6} = [a c b d]$ , then the number of divisors of $(a + b + 2020 c + d)$ is equal to

$7$

$14$

$21$

$28$

$X^{2} = [2013] [2013] = [4059]$
$X^{3} = X^{2} X = [4059] [2013] = [801927]$
$X^{6} = X^{3} \cdot X^{3} = [801927] [801927] = [640665729]$
$\Rightarrow a = 64, b = 665, c = 0, d = 729$
$\Rightarrow a + b + 2020 c + d = 1458 = 2 \times 3^{6}$
Number of divisors $= (1 + 1) (6 + 1) = 14$

Q. Let X=[20​13​] be a matrix. If X6=[ac​bd​] , then the number of divisors of (a+b+2020c+d) is equal to

7

14

21

28

X2=[20​13​][20​13​]=[40​59​] X3=X2X=[40​59​][20​13​]=[80​1927​] X6=X3⋅X3=[80​1927​][80​1927​]=[640​665729​] ⇒a=64,b=665,c=0,d=729 ⇒a+b+2020c+d=1458=2×36 Number of divisors =(1+1)(6+1)=14

Q. Let $X = [2013]$ be a matrix. If $X^{6} = [a c b d]$ , then the number of divisors of $(a + b + 2020 c + d)$ is equal to

$7$

$14$

$21$

$28$

$X^{2} = [2013] [2013] = [4059]$
$X^{3} = X^{2} X = [4059] [2013] = [801927]$
$X^{6} = X^{3} \cdot X^{3} = [801927] [801927] = [640665729]$
$\Rightarrow a = 64, b = 665, c = 0, d = 729$
$\Rightarrow a + b + 2020 c + d = 1458 = 2 \times 3^{6}$
Number of divisors $= (1 + 1) (6 + 1) = 14$