aslan wrote:@kvcpk:ahh..your method makes sense of taking the numbers ending in *5 and *2 their product and basically the numbers containing *0...and the ultimate product till tenths is sufficient..what if it asks ror 100th digit...I hope it doesn't do that in the 2 min question !
@Night _reader:I saw the foil method being used someplace..too complicated to do in the stipulated time frame and not error free
btw both got the answer right as '0'
Hi aslan -
The test would be FAR more likely to ask you for the number of trailing 0s (zeros at the end of the number) in a high value factorial, or ask you for the 1000s digit (knowing that it will likely be 0 as well).
As a fun little side exercise - can you find how many trailing zeros are in 100! - you should be able to do so quite quickly
[spoiler]To find the number of trailing zeros you just need to find the total number of "10" factors in the number. That means that every 5 and every 2 would result in another 10, as would every actual 10 itself. Because there are SO many 2s in our factorial, we can just count the 5s and 10s. So let's count:
5 = (5*1) = 1 five, 15 = (5*3) = 1 five, 25 = (5*5) = 2 fives, 35 = 1 five, 45 = 1 five, 55 = 1 five, 65 = 1 five, 75 = (5*5*3) = 2 fives, 85 = 1 five, 95 = 1 five
Then the 10s
10=1 ten, 20=1 ten, 30=1 ten, 40=1 ten, 50=1 ten and 1 five, 60=1 ten, 70=1 ten, 80=1 ten, 90=1 ten, 100= 2 tens
So we just add them all up: 24 trailing 0s [/spoiler]
Whit
