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

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Re: Unit's digit of "a" [#permalink]
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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]
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Bunnel:

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

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]
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 $$\frac{12}{5}=2$$.

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]
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]
$$((12!)^8[(12!)^8-1])/((12!)^4[(12!)^4+1])$$

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

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Re: Unit's digit of "a" [#permalink]
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]
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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]
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 $$\frac{12}{5}=2$$.

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]
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 $$\frac{12}{5}=2$$.

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" [#permalink]
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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]
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]
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]
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
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Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]
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!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]
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.

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Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]
Bunuel wrote:
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.

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.

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Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]
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Re: If (12!^16-12!^8)/(12!^16+12!^4)=a, what is the unit's digit [#permalink]
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