Find the sum of the last three digits of the number:
1 * 2^2 * 3^3 * 4^4 ...... * 2004^2004.
Please help; thanks in advance.
2007-12-07 02:53:11 · 5 answers · asked by PeterW 1 in Science & Mathematics ➔ Mathematics
Think for a second...
Inside the factors of that number (a really long product) there will be at least three 10s. In fact in 10^10 alone you will have 10 x 10 x 10 x 10... x 10, but you also have twos and fives as factors which also make 10. There will be *lots* of factors of 10 in that.
Let's just figure out 1^1 * 2^2 * 3^3 * 4^4 * 5^5 to confirm this:
1 x 2 x 2 x 3 x 3 x 3 x 4 x 4 x 4 x 4 x 5 x 5 x 5 x 5 x 5
Regroup them as follows:
(2 x 2 x 5) x (4 x 5 x 5) x (4 x 5 x 5) x (3 x 3 x 3 x 4 x 4 x 1)
= 100 x 100 x 100 x 432
= 432,000,000
So we already have 6 trailing zeroes. Continue to multiply and we will get more.
Each factor of 10 will add a trailing zero. You've only asked for the last 3 digits which will definitely be 0,0,0.
The sum of 0 + 0 + 0 = 0
Answer:
0
2007-12-07 02:58:03 · answer #1 · answered by Puzzling 7 · 4⤊ 0⤋
Even quicker than 100 x 100. Once you do 4 x 5, everything else will end in 0.
2007-12-07 03:04:54 · answer #2 · answered by tydvdtalk 3 · 1⤊ 0⤋
0
This is because 100^100 = 10000 times anything will result in the last 4 digits being 0.
2007-12-07 02:58:27 · answer #3 · answered by stevemorris1 5 · 1⤊ 1⤋
Call some friends over and urinate over your clothes and make it some sort of event/accomplishment in your life.
2016-05-22 00:19:14 · answer #4 · answered by reva 3 · 0⤊ 0⤋
000, because 1000^1000 = 1000000
2007-12-07 02:58:11 · answer #5 · answered by Anonymous · 1⤊ 0⤋