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If (12!^1612!^8)/(12!^16+12!^4)=a, what is the unit's digit
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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"
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14 Nov 2009, 01:08
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: \(a=\frac{x^{16}x^8}{x^8+x^4}=\frac{(x^8+x^4)*(x^8x^4)}{(x^8+x^4)}=(x^8x^4)\) \(a=(12!)^8(12!)^4\) \(\frac{a}{(12!)^4}=\frac{(12!)^8(12!)^4}{(12!)^4}=\frac{(12!)^8}{(12!)^4}1=(12!)^41\) So basically we should calculate the last digit of \((12!)^41\). Obviously 12! has the last digit 0, so has (12!)^4, hence \((12!)^41\) has the last digit 9. Answer: E.
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Re: Unit's digit of "a"
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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"
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05 Dec 2009, 13:43
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"
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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"
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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 lastdigitofapower70624.html#p520632
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Re: Unit's digit of "a"
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04 Aug 2011, 06:21
Bunuel wrote: \(\frac{a}{(12!)^4}=\frac{(12!)^8(12!)^4}{(12!)^4}=\frac{(12!)^8}{(12!)^4}1=(12!)^41\)
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"
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05 Aug 2011, 12:34
\(((12!)^8[(12!)^81])/((12!)^4[(12!)^4+1])\)
lets say 12!= x
x^4(x^81)/(x4+1) = x^4(x^4+1)(x^1)/(x^4+1)
= x^4(x^41)
a = (12!)^4[(12!)^41]
=> a/(12!)^4 = [(12!)^41]
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"
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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



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Re: Unit's digit of "a"
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06 Jun 2013, 03:03
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"
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24 Jun 2013, 10:11
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"
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24 Jun 2013, 10:18
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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Re: Unit's digit of "a"
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24 Jun 2013, 21:03
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!^1612!^8)/(12!^16+12!^4)=a, what is the unit's digit
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25 Jun 2013, 00:56
E is correct, here is my 2 cents
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Re: If (12!^1612!^8)/(12!^16+12!^4)=a, what is the unit's digit
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19 Apr 2014, 23:57
I did a mistake.However, below is the solution.
{(12!)^16 (12!)^8}/{(12!)^8 + (12!)^4}
{(12!)^4 [(12!)^41) *( (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
01 = 9
Unit Digit is 9



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Re: If (12!^1612!^8)/(12!^16+12!^4)=a, what is the unit's digit
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28 Jul 2016, 11:16
Aloha Everyone ..! Nice question. Here is my approach => Here we can use the property of a^2b^2=(a+b)*( ab) to simplify a to (12!)^8(12!)^4 a/(12!)^4=> (12!)^41 now 12! has a zero as its UD . Hence (12!)^4 also has a zero as its UD so the UD of a => UD0UD1=> UD 9 Smash that E
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Re: If (12!^1612!^8)/(12!^16+12!^4)=a, what is the unit's digit
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04 Feb 2019, 23:40
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 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
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Re: If (12!^1612!^8)/(12!^16+12!^4)=a, what is the unit's digit
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