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Best answer

We can write, the unit’s place in $26591749^{110016} = $ the unit’s place in $9^{110016}$

Now, we know that the unit place of $9^{1} = 9,9^{2} = 1,9^{3} = 9,\ldots$

$\therefore$ The cyclicity of $9$ is $2.$

And, $110016$ is divisible by $2,$ so the unit place of $9^{2} = 1.$

$\text{(Or)}$

The unit place of $9^{\text{odd}} = 9,9^{\text{even}} = 1.$

The number $110016$ is even, so the unit place of $9^{\text{even}} = 1.$

So, the correct answer is $(A).$

Now, we know that the unit place of $9^{1} = 9,9^{2} = 1,9^{3} = 9,\ldots$

$\therefore$ The cyclicity of $9$ is $2.$

And, $110016$ is divisible by $2,$ so the unit place of $9^{2} = 1.$

$\text{(Or)}$

The unit place of $9^{\text{odd}} = 9,9^{\text{even}} = 1.$

The number $110016$ is even, so the unit place of $9^{\text{even}} = 1.$

So, the correct answer is $(A).$