The digit in the unit place of (2x41)2 +1 (where x is a positive integ

1.	The digit in the unit place of (2x41)2 +1 (where x is a positive integer(a) 1(b) 5(c) 3(d) none of these
Answer» (24ⁿ)² = 2²×4²ⁿ = 44²ⁿ = 4^(2n+1) = 4^(odd power) as 2n+1 will always be odd now 4^(odd power) will always have 4 as its last Digit.. => 4¹ = 4 , 4³ = 64 and so on.. so. now adding 1 to it.. , the last Digit will be 4+1 = 5 option B

The digit in the unit place of (2x41)2 +1 (where x is a positive integer(a) 1(b) 5(c) 3(d) none of these

Answer»

(2*4ⁿ)² = 2²×4²ⁿ = 4*4²ⁿ = 4^(2n+1) = 4^(odd power) as 2n+1 will always be odd

now 4^(odd power) will always have 4 as its last Digit.. => 4¹ = 4 , 4³ = 64 and so on..

so. now adding 1 to it.. , the last Digit will be 4+1 = 5

option B