Question

The last digit of (1! + 2! + …. + 2005!)^500 is ____

Answer 1

Answer:

1

Step-by-step explanation:

1!=1

2!=2

3!=6

4!=24

5!=120, 6!=720, ....

the higher factorials have 0 in their one's digit place

Therefore the sum of factorials above 4! have 0 in their one's place.

1!+2!+3!+4! = 33

Therefore the sum of factorials till 2005! have 3 in their one's place.

That number can be represented as 10a + 3.

Now consider 3¹=3

3²=9

3³=27

3⁴=81

3⁴*3 = 243, 3⁴*3²=729

You can observe that the ones digit in the powers of 3 is repeating in the pattern of 3,9,7,1, 3,9,7,1...

since 500 is divisible be 4, the last digit in (10a +3)^500 the one's digit is 1