# Find the unit digit in expression 7 7! (1! +2! +3! +……+2020!) In abov

| Find the unit digit in expression 7

^{7!}(1! +2! +3! +……+2020!)A. 0

B. 3

C. 1

D. 5

### Right Answer is: B

#### SOLUTION

In above expression unit digit for any factorial greater than 4 will be zero, so we will calculate unit digit till 4!

so 1! +2! +3! +4! = 1+2+6+24 = 33

and unit digit of 7^{7!}= 7^{5040}

We know,

Cyclicity of 2,3,7 and 8 is 4

So in case of any number with unit digit 2,3,7 and 8 raised to some power, power is divided by 4 and remainder is obtained. We can find unit digit in this case using following table

Here, 5040 is completely divisible by 4. So, unit digit of 7^{7!}= 1

So, unit digit of 7^{7!}(1! +2! +3! +……+2020!) = 1×3 = 3

That means unit digit of above expression will be 3.

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