nidhis.1408 wrote:This is the solution in Veritas prep-
Solution: A.
g(f(100)) = g(100!) = the product of all the digits of whatever 100! equals. Since 100! is 100 times some integer, it will certainly end in (at least) two zeros. Even one of those zeros would be sufficient to render the product of all the digits 0, since 0 times anything is 0.
I couldn't understand the solution though!
Hi nidhis.1408!
So we want g(100!). Which means that after we multiply out 100 x 99 x 98 x 97 x 96 x 95 x 94 x ... x 5 x 4 x 3 x 2 x 1, we will multiply the digits of that number together.
Well here is a fun fact about factorials:
1! = 1
2! = 2x1 = 2
3! = 3x2x1 = 6
4! = 4x3x2x1 = 24
5! = 5x4x3x2x1 = 120
6! = 6x5x4x3x2x1 = 720
...
And every factorial over 5! will end in AT LEAST one zero! Why?
Well notice what happens every time we multiply a 5x2...we get a 10, and once we multiply by 10, we get what is called a trailing zero (a zero in the furthest right spaces before the decimal). And every 10 in the product simply adds another 0 to the list. So we know that any factorial over 5! will have these trailing zeros.
What will that do the product of the digits? Well, what happens when I multiply anything by 0? Thats right, that black hole of zero just makes the entire product 0!! So we don't really care what factorial they ask us to solve, if it is bigger than 5! it will have zeros in its digits.
And for an even faster shortcut...think about 100! as I wrote it above. It has 100 in the product - so what happens when I multiply ANY number by 100? try it.... I get 2 trailing zeros added to the end. So we can see just from the 100 that we will have 0s for some of the digits of the number, which would make the product of these digits go to zero as well!
Hope this helps!
Whit
, this is a trick problem
