x! is defined for all whole numbers x as the product of all the positive integers from 1 to x inclusive (where 0!, considered an "empty product," is defined as 1). For example, 5! = (1)(2)(3)(4)(5). Let two functions be defined over the set of whole numbers as follows:
f(x) = x! g(x) = the product of all the digits of x.
What is g(f(100))?
a. 0
b. 1
c. 100
d. 10,000
e. 36,288,000
factorial
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- anuprajan5
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This is 1 question where I would have to guess the answer is A
Logic:
f(x)= x! Therefore f(100) = 100!
g(100!) means you have to find the multiplication of all 100 numbers. But the trick here is that you might lose out on the fact that there could be a zero in this number.
for example 7! is 5040.
Regards
Anup
Logic:
f(x)= x! Therefore f(100) = 100!
g(100!) means you have to find the multiplication of all 100 numbers. But the trick here is that you might lose out on the fact that there could be a zero in this number.
for example 7! is 5040.
Regards
Anup
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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!
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!
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Nidhi,
7! is a subset of 100! - as in my example. And 7! has a zero in the number 5040. Since g(x) is the multiplication of all the numbers involved, if we took g(7) that would be zero (5*0*4*0).
So if we took g(100), then we would have a 0 that means the product would equate to 0.
REgards
Anup
7! is a subset of 100! - as in my example. And 7! has a zero in the number 5040. Since g(x) is the multiplication of all the numbers involved, if we took g(7) that would be zero (5*0*4*0).
So if we took g(100), then we would have a 0 that means the product would equate to 0.
REgards
Anup
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Hi nidhis.1408!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!
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
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veryyy vell xplained anup.... i must say even i got lost in the veritas xplaination the nidhi provided
anuprajan5 wrote:Nidhi,
7! is a subset of 100! - as in my example. And 7! has a zero in the number 5040. Since g(x) is the multiplication of all the numbers involved, if we took g(7) that would be zero (5*0*4*0).
So if we took g(100), then we would have a 0 that means the product would equate to 0.
REgards
Anup
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anuprajan5 wrote:Nidhi,
7! is a subset of 100! - as in my example. And 7! has a zero in the number 5040. Since g(x) is the multiplication of all the numbers involved, if we took g(7) that would be zero (5*0*4*0).
So if we took g(100), then we would have a 0 that means the product would equate to 0.
REgards
Anup
So anup is it that the question asks us to multiply all the factorials of 100?... As in u multilied 7! i.e. 5*0*4*0... similarly i multiplied 8!=2*0*3*2... is that what the question asks????
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Hi,
This is a 2 step solution.
First part is f(x) which we get as f(100) = 100!
Now the second part is g(100!). Now per the definition of the question, first we would have to find 100*99*98*97..... and then once we have that number we multiply the individual digits. There's an interesting part to this and that is that 100 is part of the mix. Therefore when we do get the final number it is bound to be in the form XXXXXXX...XXXXX00. So if we multiply the individual digits, there are 2 zeroes and hence the answer will be 0.
Now where I jumped the gun in my solution was that I saw a trend. If we know 7! which is a part of 100! (7!*8*9*10*....100) has already 1 zero in it because of 5040 (when we multiply this number with another the unit digit is going to amplify the other number by 10), then I concluded that this will become zero.
In your example 8! is 20320 (my point of 7! amplifying). If you multiply the individual digits, then you will get zero.
Regards
Anup
This is a 2 step solution.
First part is f(x) which we get as f(100) = 100!
Now the second part is g(100!). Now per the definition of the question, first we would have to find 100*99*98*97..... and then once we have that number we multiply the individual digits. There's an interesting part to this and that is that 100 is part of the mix. Therefore when we do get the final number it is bound to be in the form XXXXXXX...XXXXX00. So if we multiply the individual digits, there are 2 zeroes and hence the answer will be 0.
Now where I jumped the gun in my solution was that I saw a trend. If we know 7! which is a part of 100! (7!*8*9*10*....100) has already 1 zero in it because of 5040 (when we multiply this number with another the unit digit is going to amplify the other number by 10), then I concluded that this will become zero.
In your example 8! is 20320 (my point of 7! amplifying). If you multiply the individual digits, then you will get zero.
Regards
Anup
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Since we have to find 100!, which is not an easy calculation , this is a trick problemnidhis.1408 wrote:x! is defined for all whole numbers x as the product of all the positive integers from 1 to x inclusive (where 0!, considered an "empty product," is defined as 1). For example, 5! = (1)(2)(3)(4)(5). Let two functions be defined over the set of whole numbers as follows:
f(x) = x! g(x) = the product of all the digits of x.
What is g(f(100))?
a. 0
b. 1
c. 100
d. 10,000
e. 36,288,000
look for zeros
100!= 100x99x98......1
Since there are 2 zeros we can say definitely there will be zero in the result of the 100!
Hence answer is A
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Singh
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