What are the last three digits of
N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)?
please help..
N = 2^1 + 2^2! + 2^(1000!)!)
The last number has no ending digits as small as 2 or 4
006 would be the last three digits of the complete number.
Since any number n! where n>=5 ends in zero,
(2^1)! + (2^2)! + ...
= 2 + 24 + ...
ends in 6
To find the last three digits of N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!), we need to calculate the value of each term and then add them together.
First, let's break down the problem step by step:
1. Calculate the factorial of each number from 1 to 1000:
Factorial of 1: 1! = 1
Factorial of 2: 2! = 2 * 1 = 2
Factorial of 3: 3! = 3 * 2 * 1 = 6
... continuing up to ...
Factorial of 1000: 1000! = 1000 * 999 * ... * 3 * 2 * 1
2. Calculate 2 raised to the power of each factorial obtained in step 1:
2^(1!) = 2^1 = 2
2^(2!) = 2^2 = 4
2^(3!) = 2^6 = 64
... continuing up to ...
2^(1000!) = 2^(a very large number)
Now, to find the last three digits of N, we need to add up all these values. Since we only need the last three digits, we can ignore all the digits before the last three.
1. Start summing up the values:
N = (2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)
2. To simplify the calculation, take the remainder when dividing each value by 1000, as we are interested in the last three digits only:
(2^(1!)!) mod 1000
(2^(2!)!) mod 1000
...
(2^(1000!)!) mod 1000
3. Finally, add up all the values:
S = ((2^(1!)!) mod 1000) + ((2^(2!)!) mod 1000) + ... + ((2^(1000!)!) mod 1000)
By performing these calculations, we can find the last three digits of N. However, due to the complexity of the calculations involved and the large numbers in the factorials, it would require a significant amount of computational power and time.