BN1989 wrote:
In regards to question 13, how do you know that (8!)^2 will have two 0's at the end?
You figure it out by knowing that there is a 5 as a factor in the final number.
example 8! = 8*7*6*5...*1
Now if u know how many fives are there you will have that number of zeros in the end because at least that many number of 2 will there for sure ...
Now IF u want to figure out how may fives are there in (which is equivalent to finding how many zeros in the end of the number )
then simply keep dividing the factorial by 5 and adding it to the result until 5^x exceeds the factorial, i know i have used very confusing language but a example will simplify things for sure
how many zeros at the end of 312!
answer :- 312!/5 + 312!/5^2 + 312!/5^3 = 62 + 12 + 2 (now note i stopped at 5^3, because 5^4 =25*25=625 which is more that 312 ) hence :- 62 + 12 + 2 =76 number of 5s in the factorial and same number of zeros in the end because these many number of 2s will obviously be there in factorial and will form 10s .
IF you are still confused I would suggest you to download Bunuel's math's notes and even if you are not confused i would still suggest you to do the same ,