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|A| = 2

$\underset{2024times}{\underset{?}{adj\left(adj\left(adj....\left(a\right)\right)\right)}}={\left|A\right|}^{{\left(n-1\right)}^{2024}}$            

$={\left|A\right|}^{{\left(n-1\right)}^{2024}}$                                             

$=2^{2^{2024}}$                                          

$2^{2024}=\left(2^2\right)2^{2022}=4{\left(8\right)}^{674}=4{\left(9-1\right)}^{674}$             

-> $2^{2024}\equiv 4\left(mod9\right)$

->,  $2^{2024}\equiv 9m+4$  m ¬ even

$2^{9m+4}\equiv 16\cdot {\left(2^3\right)}^{3m}\equiv 16\left(mod9\right)$

7

  $?$  R is symmetric relation

⇒   (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

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After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
^15+3–1C₂ = ¹⁷C₂ ways

  $=\frac{17*16}{2}=136$          

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