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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

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"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

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Expert's post

1

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]
19 Apr 2014, 22:57