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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 [#permalink] New post 13 Nov 2009, 23:49
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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
[Reveal] Spoiler: OA
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Re: Unit's digit of "a" [#permalink] New post 14 Nov 2009, 00:08
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kp1811 wrote:
If [(12!)^16 - (12!)^8]/[(12!)^8 + (12!)^4] = a , what is the unit’s digit of a/(12!)^4?

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


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.
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Re: Unit's digit of "a" [#permalink] New post 05 Dec 2009, 11:40
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"Obviously 12! has the last digit 0"

How do we determine the last digit of 12! is 0? Please explain.
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Re: Unit's digit of "a" [#permalink] New post 05 Dec 2009, 12:43
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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.
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Re: Unit's digit of "a" [#permalink] New post 05 Dec 2009, 13:57
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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.
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Re: Unit's digit of "a" [#permalink] New post 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.



This is the link which will give you details about finding last digit
last-digit-of-a-power-70624.html#p520632
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Re: Unit's digit of "a" [#permalink] New post 04 Aug 2011, 05:21
Bunuel wrote:
\frac{a}{(12!)^4}=\frac{(12!)^8-(12!)^4}{(12!)^4}=\frac{(12!)^8}{(12!)^4}-1=(12!)^4-1


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.
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Re: Unit's digit of "a" [#permalink] New post 05 Aug 2011, 11:34
((12!)^8[(12!)^8-1])/((12!)^4[(12!)^4+1])


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[(12!)^4-1]


=> a/(12!)^4 = [(12!)^4-1]

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.
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Re: Unit's digit of "a" [#permalink] New post 05 Aug 2011, 12:33
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
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Re: Unit's digit of "a" [#permalink] New post 06 Jun 2013, 02:03
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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).
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Re: Unit's digit of "a" [#permalink] New post 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?
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Re: Unit's digit of "a" [#permalink] New post 24 Jun 2013, 09:18
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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.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Unit's digit of "a" [#permalink] New post 24 Jun 2013, 20:03
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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
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Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink] New post 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] New post 19 Apr 2014, 22:57
I did a mistake.However, below is the solution.

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

{(12!)^4 [(12!)^4-1) *( (12!)^4 + 1)]}/{(12!)^4 *[(12!)^4 +1]

(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
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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
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