What are the last three digits of the number N=2(1!)!+2(2!)!+...+2(1000!)!?
let's look at the first few terms
2(1!) = 2
2(2!) = 4
2(3!) = 12
2(4!) = 48
2(5!) = 240
2(6!) = 1440
2(7!) = 10080
2(8!) = 80640
2(9!) = 725760
2(10!) = 7257600
2(11!) = .....3600 , since we are only concerned about the last 3 digits
2(12!) = ..... 3200
2(13!) = ... 1600
2(14!) = ...2400
2(15!) = ...36000
after that no result will contribute to the last 3 digits.
so add them up , to the end of the first 3 digits.
Unless I made a silly arithmetic error I got 626
N=2^(1!)!+2^(2!)!+...+2^(1000!)!?
this is the actual series..
stop this.
To find the last three digits of the number N=2(1!)!+2(2!)!+...+2(1000!)!, we can consider the last three digits of each term individually and then sum them up.
To get the last three digits of a factorial n!, we only need to consider the last three digits of the numbers being multiplied. For instance, in 10! = 10 * 9 * 8 * ... * 2 * 1, only the final three digits of each number matter.
Let's compute the last three digits of each term:
1. 2(1!)! = 2(1) = 2. The last three digits are 002.
2. 2(2!)! = 2(2 * 1) = 2(2) = 4. The last three digits are 004.
3. 2(3!)! = 2(3 * 2 * 1) = 2(6) = 12. The last three digits are 012.
As we can observe, after the third term, every factorial term greater than or equal to 5! will have a last three-digit ending in 000 (since 5! = 120, and any number multiplied by 120 has the last three digits as 000).
Now, we only need to consider the first three terms since the rest will have a last three-digit ending in 000.
N = 2 + 4 + 12 + ...
To find the sum of these terms, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(a + l),
where Sn is the sum, n is the number of terms, a is the first term, and l is the last term.
In this case, a = 2, and the common difference is d = 4 - 2 = 2.
The number of terms, n, can be found by dividing the last number (1000) by the common difference and adding 1 (since we want to include the first term):
n = (1000 - 2)/2 + 1 = 1000.
Now we can find l, the last term:
l = a + (n - 1)d,
l = 2 + (1000 - 1)2,
l = 2 + 1998,
l = 2000.
Now we can substitute these values into the formula to find the sum:
Sn = (n/2)(a + l),
S1000 = (1000/2)(2 + 2000),
S1000 = (500)(2002),
S1000 = 1,001,000.
So, the sum N = 1,001,000.
The last three digits of this number are 000. Therefore, the last three digits of N=2(1!)!+2(2!)!+...+2(1000!)! are 000.