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So basically we should calculate the last digit of \((12!)^4-1\). Obviously 12! has the last digit 0, so has (12!)^4, hence \((12!)^4-1\) has the last digit 9.

How do we determine the last digit of 12! is 0? Please explain. _________________

--------------------------------------------------------------- Check-out the following: --------------------------------------------------------------- 1. Math Divisibility Test

How do we determine the last digit of 12! is 0? Please explain.

\(12!=1*2*3*...*10*11*12\), as there is \(10\) among the multiples the last digit will be \(0\). Basically if we have \(n!\) and \(n>=5\) the last digit will be \(0\), as there are \(2\) and \(5\) among the multples and they make zero when multiplied.

Actually \(12!\) has \(2\) zeros in the end, as \(\frac{12}{5}=2\).

If you want to know more about trailing zeros, see link about the factorials below. _________________

How do we determine the last digit of 12! is 0? Please explain.

\(12!=1*2*3*...*10*11*12\), as there is \(10\) among the multiples the last digit will be \(0\). Basically if we have \(n!\) and \(n>=5\) the last digit will be \(0\), as there are \(2\) and \(5\) among the multples and they make zero when multiplied.

Actually \(12!\) has \(2\) zeros in the end, as \(\frac{12}{5}=2\).

If you want to know more about trailing zeros, see link about the factorials below.

If \(\frac{(12!)^{16} - (12!)^8}{(12!)^8 + (12!)^4} = a\) , what is the unit’s digit of \(\frac{a}{(12!)^4}\)?

(A) 0 (B) 1 (C) 3 (D) 5 (E) 9

Responding to a pm:

Once you understand that you need to find the last digit of (12!)^4- 1, there isn't much left to do.

What will be the last digit of 12! Look: 1! =1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 and so on... The last digit will always be 0 (starting 5!) because there will always be a 2 and 5 to make a 10. So 12! will end with a 0 too. When you take it to fourth power, it will end with 0 again (\(10^4 = 10000, 20^4 = 160000\) etc).

Hence, \((12!)^4\) ends with a 0. When you subtract 1 from it, it must end with a 9 (it is 1 less than a multiple of 10). _________________

How do we determine the last digit of 12! is 0? Please explain.

\(12!=1*2*3*...*10*11*12\), as there is \(10\) among the multiples the last digit will be \(0\). Basically if we have \(n!\) and \(n>=5\) the last digit will be \(0\), as there are \(2\) and \(5\) among the multples and they make zero when multiplied.

Actually \(12!\) has \(2\) zeros in the end, as \(\frac{12}{5}=2\).

If you want to know more about trailing zeros, see link about the factorials below.

There is also a rule which says that till 4! you can calculate the last digit and when it goes beyond 5! it always be 0.Am i right? _________________

How do we determine the last digit of 12! is 0? Please explain.

\(12!=1*2*3*...*10*11*12\), as there is \(10\) among the multiples the last digit will be \(0\). Basically if we have \(n!\) and \(n>=5\) the last digit will be \(0\), as there are \(2\) and \(5\) among the multples and they make zero when multiplied.

Actually \(12!\) has \(2\) zeros in the end, as \(\frac{12}{5}=2\).

If you want to know more about trailing zeros, see link about the factorials below.

There is also a rule which says that till 4! you can calculate the last digit and when it goes beyond 5! it always be 0.Am i right?

Yes, for integer n>4, then the units digit of n! is 0. _________________

There is also a rule which says that till 4! you can calculate the last digit and when it goes beyond 5! it always be 0.Am i right?

Also, don't think of it as a "rule". When you have 5!, you have both a 2 and a 5 so they multiply to give 10. Hence 5! will end with a 0. This is true for every n after 5 since there will always be at least one 2 and one 5 for all the factorials: 6! = 1*2*3*4*5*6; 7! = 1*2*3*4*5*6*7 etc _________________

Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]

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25 Jun 2013, 00:56

E is correct, here is my 2 cents

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Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]

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22 Apr 2015, 10:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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