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Unit's digit of "a"

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Unit's digit of "a" [#permalink] New post 14 Nov 2009, 00:49
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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
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Re: Unit's digit of "a" [#permalink] New post 14 Nov 2009, 01:08
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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, 12:40
"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, 13:43
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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, 14:57
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 06 Dec 2009, 00:34
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, 06:21
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, 12:34
((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, 13:33
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
Re: Unit's digit of "a"   [#permalink] 05 Aug 2011, 13:33
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