Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

3

This post received KUDOS

Expert's post

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

Attachments

Math1.png [ 12.07 KiB | Viewed 769 times ]

_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Chris Bangle - Former BMV Chief of Design.

gmatclubot

Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit
[#permalink]
24 Jun 2013, 23:56