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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.

Re: Unit's digit of "a" [#permalink]
05 Dec 2009, 11:40

1

This post received KUDOS

"Obviously 12! has the last digit 0"

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

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

Re: Unit's digit of "a" [#permalink]
05 Dec 2009, 12:43

2

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

sharkk wrote:

"Obviously 12! has the last digit 0"

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. _________________

Re: Unit's digit of "a" [#permalink]
05 Dec 2009, 23:34

Bunuel wrote:

sharkk wrote:

"Obviously 12! has the last digit 0"

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.

Re: Unit's digit of "a" [#permalink]
06 Jun 2013, 02:03

1

This post received KUDOS

Expert's post

kp1811 wrote:

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). _________________

Re: Unit's digit of "a" [#permalink]
24 Jun 2013, 09:11

Bunuel wrote:

sharkk wrote:

"Obviously 12! has the last digit 0"

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? _________________

Re: Unit's digit of "a" [#permalink]
24 Jun 2013, 09:18

Expert's post

prateekbhatt wrote:

Bunuel wrote:

sharkk wrote:

"Obviously 12! has the last digit 0"

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. _________________

Re: Unit's digit of "a" [#permalink]
24 Jun 2013, 20:03

1

This post received KUDOS

Expert's post

prateekbhatt wrote:

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]
24 Jun 2013, 23:56

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]
22 Apr 2015, 09:49

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