If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit

Question

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

Bunuel · Accepted Answer

First let's simplify. We have:

a=\frac{x^{16}-x^8}{x^8+x^4}=\frac{(x^8+x^4)*(x^8-x^4)}{(x^8+x^4)}=(x^8-x^4)

a=(12!)^8-(12!)^4

\frac{a}{(12!)^4}=\frac{(12!)^8-(12!)^4}{(12!)^4}=\frac{(12!)^8}{(12!)^4}-1=(12!)^4-1

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.

Answer: E.

Bunuel · Answer

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.

sharkk · Answer

"Obviously 12! has the last digit 0"

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

sharkk · Answer

Bunnel:

Awesome!!! Excellent!!! Great!!!

The question I asked seemed so tough when I asked but after reading your explanation it seems not that tough.

Again, thanks you so much for clarifying this concept.
I will definitely spend some time reading the posts in your signature.

Thank you.

Bullet · Answer

This is the link which will give you details about finding last digit last-digit-of-a-power-70624.html#p520632

restore · Answer

Can you explain the this line? How you went from  \frac{(12!)^8-(12!)^4}{(12!)^4}=\frac{(12!^8}{(12!)^4}-1 ?

Im wondering how you simplified the powers in the numerator and where the  -1  came from.

Spidy001 · Answer

((12!)^8 )/((12!)^4 )

lets say 12!= x

x^4(x^8-1)/(x4+1) = x^4(x^4+1)(x^-1)/(x^4+1)

= x^4(x^4-1)

a = (12!)^4

=> a/(12!)^4 =

trailing zero's in 12! is 12/5 = 2
 =>units digit of 12!^4 = 0

=> units digit of a/(12!)^4 = 9

Answer is E.

gladvijay · Answer

very good problem .. first look I thought man .. this will take for ever.. but its quite simple..

lets assume (12!)^4 = x

a = (x^4 - x^2) / (x^2 + x) => x^2 - x

we are looking for a / (12!)^4 = (x^2 - x) / x = x - 1

12! units digit is 0.. as there is 10 in the factorial.. so - 1 .. units digit is 9

KarishmaB · Answer

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).

prateekbhatt · Answer

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?

Bunuel · Answer

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

KarishmaB · Answer

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

pqhai · Answer

E is correct, here is my 2 cents

TGC · Answer

I did a mistake.However, below is the solution.

{(12!)^16- (12!)^8}/{(12!)^8 + (12!)^4}

{(12!)^4  }/{(12!)^4 *

(12!)^4 - 1

Since 12! has one 0 contributed by 10

Unit digit (10)^4 = 0

Unit digit of (12!)^4 = 0

0-1 = 9

Unit Digit is 9

stonecold · Answer

Aloha Everyone ..!
Nice question.
Here is my approach =>
Here we can use the property of a^2-b^2=(a+b)*( a-b) to simplify a to (12!)^8-(12!)^4
a/(12!)^4=> (12!)^4-1
now 12! has a zero as its UD .
Hence (12!)^4 also has a zero as its UD
so the UD of a => UD0-UD1=> UD 9

Smash that E

KanishkM · Answer

Lets take 12! as x for sometime

x^8/x^4 * {x^8 - 1}/ {x^4 + 1} = a

a/x^4 = {(x^4 - 1) (x^4 + 1)} / {x^4 + 1} 
a/x^4 = x^4 - 1

Lets put back the value of 12! now

it will end on a zero 0 -1

9

E

rijain · Answer

Hi Burnel,

How we can do (x^8+x^4)(X^8-X^4) when the formula is a^2-b^2=a+b a-b we have first bring the number in the square format right.

Please explain....

Thanks in advance

rijain · Answer

Hi Burnel,

How we can do (x^8+x^4)(X^8-X^4) when the formula is a^2-b^2=a+b a-b we have first bring the number in the square format right.

Please explain....

Thanks in advance

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If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit

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If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit

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