N -1 -2 -3 -4 -47 -48 -49 -50 -51 - Then what is the unit digit of N N -A- 7B- 1C- 9D- 3

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Unit Digit of a Number

N =1 !-2 !+3 ...

Question

N=1! - 2! + 3! - 4! +...+ 47! - 48! + 49! - 50! + 51! Then what is the unit digit of $N^{N}$ ?

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Solution

The correct option is A 1
From 5! onwards, each of the numbers would have a units digit of 0. Hence, the units digit of the given number would depend on the units digit of 1! -2! +3!-4!, which would be -1(1-2+6-4) i.e. 9. But the rest of the numbers end up with with a higher number, which ends in Zero, than (1! -2! +3!-4!). Hence, N is a number that has a unit's digit of 0 - 9 = -9 i.e. 1 and when it is raised to any power, the digit would not change. Hence, the correct answer would be 1.

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N=1! - 2! + 3! - 4! +...+ 47! - 48! + 49! - 50! + 51! Then what is the unit digit of NN?

N=1! - 2! + 3! - 4! +...+ 47! - 48! + 49! - 50! + 51! Then what is the unit digit of $N^{N}$ ?