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KarishmaB
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Baker's Dozen 22 Dec 2014, 21:01
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Quote:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000


I see that (9*9*1*1*1)5C3 is a popular answer method. I agree with this approach, but i was wondering, since permutation is choosing and arranging, 'why not permutation?'. I must be wrong here and i'd really like to know where/ how.
I did read all the other explanations. The repetition approach(5!/3!) looked good too. But i did not understand why a few of em used the probability approach (10*1/10*1/10....).
What would your process for this be?
Show more

Responding to a pm:

Permutation is choosing and arranging - correct! But you need to know what you are arranging. Permutation (written as nPr) implies choose r things out of n and then arrange the r things among themselves. Say you select 3 chocolates out of 5 to give as prizes in a game and then arrange the 3 selected chocolates in three spots - first, second and third. The one who stands first, gets the first chocolate and so on. So if you use permutation here, you are selecting 3 spots where you will put 6 and then arranging them in 3! ways. What we actually need to do is select 3 spots out of 5 and put 6 in them. We have to then select 2 digits for the leftover 2 spots, wherever they are!

Hence we select the three spots in 5C3 ways and then choose a digit for the remaining two places in 9*9 ways.

Probability approach:
There are 5 spots ___ ___ ___ ___ ___

What is the probability that the number looks like 6N6N6? (1/10)*(9/10)*(1/10)*(9/10)*(1/10)
N is any number other than 6.

What is the probability that the number looks like N66N6? (9/10)*(1/10)*(1/10)*(9/10)*(1/10)

Similarly there are many other ways of making the password with three 6s. You need to add all some probabilities. The probability in each case will be \((1/10)^3 * (9/10)^2\). How many such distinct cases will be there? It depends on the number of ways in which you can arrange the 6s and Ns = 5!/3!*2! = 10
Total probability \(= 10*(1/10)^3 * (9/10)^2. = 810/10^5\)
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Quote:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

I see that (9*9*1*1*1)5C3 is a popular answer method. I agree with this approach, but i was wondering, since permutation is choosing and arranging, 'why not permutation?'. I must be wrong here and i'd really like to know where/ how.
I did read all the other explanations. The repetition approach(5!/3!) looked good too. But i did not understand why a few of em used the probability approach (10*1/10*1/10....).
What would your process for this be?

Responding to a pm:

Permutation is choosing and arranging - correct! But you need to know what you are arranging. Permutation (written as nPr) implies choose r things out of n and then arrange the r things among themselves. Say you select 3 chocolates out of 5 to give as prizes in a game and then arrange the 3 selected chocolates in three spots - first, second and third. The one who stands first, gets the first chocolate and so on. So if you use permutation here, you are selecting 3 spots where you will put 6 and then arranging them in 3! ways. What we actually need to do is select 3 spots out of 5 and put 6 in them. We have to then select 2 digits for the leftover 2 spots, wherever they are!

Hence we select the three spots in 5C3 ways and then choose a digit for the remaining two places in 9*9 ways.

Show more
Can I use permutation for the 2 non-6 digits? 5C3 is for choosing 3 spots to place 6s in 5 spots, for the remaining 2 spots one of 9 digits are used. Can this be expressed with '9C2' and something?
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Baker's Dozen 25 Dec 2014, 22:56
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VeritasPrepKarishma
Quote:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

I see that (9*9*1*1*1)5C3 is a popular answer method. I agree with this approach, but i was wondering, since permutation is choosing and arranging, 'why not permutation?'. I must be wrong here and i'd really like to know where/ how.
I did read all the other explanations. The repetition approach(5!/3!) looked good too. But i did not understand why a few of em used the probability approach (10*1/10*1/10....).
What would your process for this be?

Responding to a pm:

Permutation is choosing and arranging - correct! But you need to know what you are arranging. Permutation (written as nPr) implies choose r things out of n and then arrange the r things among themselves. Say you select 3 chocolates out of 5 to give as prizes in a game and then arrange the 3 selected chocolates in three spots - first, second and third. The one who stands first, gets the first chocolate and so on. So if you use permutation here, you are selecting 3 spots where you will put 6 and then arranging them in 3! ways. What we actually need to do is select 3 spots out of 5 and put 6 in them. We have to then select 2 digits for the leftover 2 spots, wherever they are!

Hence we select the three spots in 5C3 ways and then choose a digit for the remaining two places in 9*9 ways.

Can I use permutation for the 2 non-6 digits? 5C3 is for choosing 3 spots to place 6s in 5 spots, for the remaining 2 spots one of 9 digits are used. Can this be expressed with '9C2' and something?
Show more

No. With 9C2, you are selecting 2 digits out of 9. There are two problems with it:
1. You are assuming that the two digits are distinct. Nothing stops the password from being 66622.
2. You are not considering that the two digits are for two distinct places. 66623 is different from 66632 but with 9C2, you are assuming that it is the same password.
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Bunuel
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.

Answer: D.
Show more

Hi Bunuel,

Thanks for your great workouts. I have a question on this one though. If K= B (36), the number would become a square of a positive integer as well. For that matter any perfect square number will do -- say 4?!

\(126*\sqrt{36}\) = 756. Why can't B be an answer. Your comment is highly appreciated.
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Baker's Dozen 20 Feb 2015, 03:27
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falewman
Bunuel
4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.

Answer: D.

Hi Bunuel,

Thanks for your great workouts. I have a question on this one though. If K= B (36), the number would become a square of a positive integer as well. For that matter any perfect square number will do -- say 4?!

\(126*\sqrt{36}\) = 756. Why can't B be an answer. Your comment is highly appreciated.
Show more

\(126*\sqrt{k}\) must be a prefect square. If k=36, \(126*\sqrt{k}=756\), which is not a perfect square.
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Baker's Dozen 11 Jan 2016, 06:33
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Bunuel
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.
Show more

Hi Bunuel,

I have a doubt in 5C3 part in this problem.
I solved something like this:

Code will be something like this

6,6,6,A,B

this is like arranging 5 letters where 3 are exactly similar. It would give:

5! / 3! which is not equal to 5C3 .
Therefore I am getting a wrong answer.

Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation).

Thanks in advance.

SR
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Baker's Dozen 11 Jan 2016, 08:47
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solitaryreaper
Bunuel
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.
Show more

Hi solitaryreaper,

there are various Queries on this Q that why don't we use permutations and use combinations..
that is why 5!/3!2! instead of 5!/3!..

firstly, yes the Question is of permutation, but we still require to use combination formula..
WHY?
we have choosen three 6 digits and 2 digits as any of the remaining three..
answer is 9*9*5C3.... and not 9*9*5P3
because the permutations will be ok till the time we have two separate digits...
and it will be combinations the moment the other two digits are same....
so how can we use permutations even when digits are same, and that is what we are doing when we take other two digits as 9*9..

then how do we arrive at 9*9*5!/3!2!..
there are two ways we can take the remaining two digits..

1) three 6s and both other digits are different..
aaabc b and c can be chosen out of 9 digits
ways = 9C2*5!/3!=(9*8)/2! * 5!/3!...

2)three 6s and both other digits are same..
aaabb.. b can be selected in 9 ways
ways = 9*5!/3!2!..

total ways = add the two=9*8*5/3!2! + 9*5!/3!2!= 9*5!/3!2!* (8+1)=9*9*5!/3!2!=9*9*5C3..

I hope this clears the air around the use of permutation or combination in this Q..
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solitaryreaper
Bunuel
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

Hi Bunuel,

I have a doubt in 5C3 part in this problem.
I solved something like this:

Code will be something like this

6,6,6,A,B

this is like arranging 5 letters where 3 are exactly similar. It would give:

5! / 3! which is not equal to 5C3 .
Therefore I am getting a wrong answer.

Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation).

Thanks in advance.

Now I am having a hard time to grasp how combination is giving the same answer. Can you please help me with the combination approach(how the problem translates to combination which is equivalent to the one the we got via permutation) ?

Moreover , how to decide when to use Permutation or Combination (and one that would lead to the solution in lesser time)


SR
Show more

Responding to a pm:

This is a question that haunts many - how do I know when to use permutation and when to use combination?
There is a simple solution - never use the permutation formula. The permutation formula finds very little direct use but leads to too many complications.

Always think in terms of selecting and arranging. For selecting r distinct elements out of n distinct elements, use nCr formula. Then arrange depending on whether the r elements selected need to be all distinct (r!) or some need to be same (a!/b!*c!) etc

Here, out of 9 digits you can select 2 in 9C2 ways and get a case such as 666AB. This will be arranged in 5!/3! ways.
Or you can select 1 digit out of 9 in 9C1 ways and get a case such as 666AA. This will be arranged in 5!/2!*3! ways.

9C2*5!/3! + 9C1*5!/2!*3! = 9*5!/3! * 9/2 = 810

Probability = 810/10^5

Note that you are using the Combinations formula only in this method too.

The other method using the combination formula just shows a different way of thinking.

You have 5 spots: _____ _____ _____ _____ _____

You choose any two spots out of these 5 in 5C2 ways.
For the first spot you choose, select a digit in 9 ways. For the second spot, select a digit in 9 ways.
In all remaining spots, just put 6.
This 5C2*9*9 directly gives you the total number of ways.
Note that 5C2 is the same as 5C3 (either you choose 2 spots for non 6 digits or you choose 3 spots for 6).
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7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000

Solution: https://gmatclub.com/forum/baker-s-dozen ... l#p1057509




Let F represents Fritz, L for Luis, A for Alfred and W for Werner

Fritz owns 2/3rd of the shares of the other three shareholders --> F = 2/3 (3,600,000 - F) ---> 3F = 7,200,000 - 2F ---> F = 1,440,000

Luis owns 3/7th of the shares of the other three shareholders --> L = 3/7 (3,600,000 - L) ---> 7L = 10,800,000 - 3L ---> L = 1,080,000

Alfred owns 4/11th of the shares of the other three shareholders --> A = 4/11 (3,600,000 - A) ---> 11A = 14,400,000 - 4A ---> A = 960,000

W= 3,600,000 -(F + L + A) ---> 3,600,000 - (1,440,000 + 1,080,000 + 960,000) = 120,000

Answer: D
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Baker's Dozen 27 Mar 2017, 18:07
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Here's my solution for #2 (see visual below). It helps to use brackets!
Attachments

Screen Shot 2017-03-27 at 6.07.16 PM.png
Screen Shot 2017-03-27 at 6.07.16 PM.png [ 182.8 KiB | Viewed 5340 times ]

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Bunuel niks18

Quote:
A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215
Show more


Quote:

Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Show more

I solved the same problem taking hint from number theory that a consecutive odd number is represented by:

2x+1 , 2x+3

My approach had slightly complex calculations but it ensured that my first no in series is always odd.

Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)
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adkikani
Bunuel niks18

Quote:
A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215


Quote:

Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).

I solved the same problem taking hint from number theory that a consecutive odd number is represented by:

2x+1 , 2x+3

My approach had slightly complex calculations but it ensured that my first no in series is always odd.

Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)
Show more

In this question, as shown, you can take 7 consecutive odd integers to be \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), and \(x+12\). For some number properties questions, you should represent odd integers, as 2k + 1, 2k + 3, ... Here you can also take the integers, to be 2k + 1, 2k + 3, ... but it's not necessary.
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Baker's Dozen 26 Mar 2018, 04:59
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nahid78
Bunuel
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\)

Answer: B.

I know this is how we solve this question. But can anyone please clarify why i don't get the same result if i work differently ....

Suppose other two numbers are also same 6,6,6, and 1,1 or 6,6,6 and 2,2 so on... or other two are different. 6,6,6,1, and 2, or 6,6,6,2,and 4
Show more

You do get the same result in this case too.

If the other 2 numbers are same, number of ways = 9C1 * 5!/3!*2! = 90
If the other 2 numbers are different, number of ways = 9C2 * 5!/3! = 720

Total = 90+720 = 810
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Bunuel
9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y

Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).

So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)

Answer: D.
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Just Excellent Question Bunuel...
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Baker's Dozen 22 Dec 2019, 04:35
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Bunuel
8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).

Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)

Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185\) --> \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185\) --> \((5x+20)+20=-185\) --> \(5x+20=-205\)

Answer: D.
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Hello Bunuel, I tried to solve it like this.. Can u please tell me where did I go wrong?

Let the smallest no. be 2n+1, so the numbers of the set will be :
2n+1, 2n+3, 2n+5, 2n+7, 2n+9, 2n+11, 2n+13

given that sum of largest 5 no.s in the set is -185.
so, 2n+5+2n+7+2n+9+2n+11+2n+13 = -185
solving, n=-23

so the no.s are -45,-47,-49,-51,-53,-55,-57

sum of smallest 5 are -49-51-53-55-57 = --265

Also, if we take into consideration that these are negative no.s, from the start we can write that 2n+1+2n+3+2n+5+2n+7+2n+9 = -185
solving n= -21
so the no.s are -41, -43, -45, -47, -49, -51, -53
sum of 5 smallest = -245

Really confused.. Please help.
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Baker's Dozen 22 Dec 2019, 08:57
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Bunuel
8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).

Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)

Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185\) --> \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185\) --> \((5x+20)+20=-185\) --> \(5x+20=-205\)

Answer: D.


Hello Bunuel, I tried to solve it like this.. Can u please tell me where did I go wrong?

Let the smallest no. be 2n+1, so the numbers of the set will be :
2n+1, 2n+3, 2n+5, 2n+7, 2n+9, 2n+11, 2n+13

given that sum of largest 5 no.s in the set is -185.
so, 2n+5+2n+7+2n+9+2n+11+2n+13 = -185
solving, n=-23

so the no.s are -45,-47,-49,-51,-53,-55,-57

sum of smallest 5 are -49-51-53-55-57 = --265

Also, if we take into consideration that these are negative no.s, from the start we can write that 2n+1+2n+3+2n+5+2n+7+2n+9 = -185
solving n= -21
so the no.s are -41, -43, -45, -47, -49, -51, -53
sum of 5 smallest = -245

Really confused.. Please help.
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Since n = -21, then the seven consecutive numbers are: -45, -43, -41, -39, -37, -35, -33.

The sum of the five largest is -41 + (-39) + (-37) + (-35) + (-33) = -185.
The sum of the five smallest is -45 +(-43) +(-41) + (-39) + (-37) = -205.
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Baker's Dozen Updated on: 02 Feb 2021, 09:16
Originally posted by sujoykrdatta on 02 Feb 2021, 08:45.
Last edited by sujoykrdatta on 02 Feb 2021, 09:16, edited 1 time in total.
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I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
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Total digits = 10 (from 0 to 9)
Of the 5 digits, we have 6 thrice.

Case 1: Let the other 2 digits be same:
Choose that digit 'n' in 9 ways and arrange the digits of the number 666nn in 5!/3!2! = 10 ways => Total ways = 90

Case 2: Let the other 2 digits be different:
Choose the digits 'n' in 9c2 = 36 ways and arrange the digits of the number 666mn in 5!/3! = 20 ways => Total ways = 720

Thus, total such numbers = 810
Total 5-digit numbers = 10 x 10 x 10 x 10 x 10 = 10000
=> Probability = 810/10000 - Answer B
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Baker's Dozen 02 Feb 2021, 09:05
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I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!
13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14

Solution: https://gmatclub.com/forum/baker-s-dozen ... l#p1057520
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x = [(8!)^10−(8!)^6] / [(8!)^5−(8!)^3] = {(8!)^6 x [(8!)^4 − 1]} / {(8!)^3 x [(8!)^2 − 1]} = (8!)^3 x [(8!)^2 + 1]

Thus, x/(8!)^3 - 39 = (8!)^2 − 38

8! ends in 0; thus, (8!)^2 ends in 00 i.e. it is a multiple of 100

Thus, the last 2 digits of (8!)^2 − 38 are 62 => Product = 12

Answer D
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Baker's Dozen 12 Aug 2023, 08:39
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Bunuel
SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

The total number of 5-digit codes is \(10^5\). It's important to note that it's not \(9*10^4\), as the first digit can be zero in a password.

The number of passwords with three digits as 6 can be calculated as \(9*9*C^3_5 = 810\). We have 9 choices for each of the two remaining digits (not 6), resulting in \(9*9\). The term \(C^3_5\) represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).

The probability is therefore, \(P=\frac{favorable}{total}=\frac{810}{10^5}\).


Answer: B
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why we didn't multiply it with 2! ? two non-six numbers can be arranged in 2 ways...it is a password so we have to go for arrangement...right? please someone help me to understand

regards

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Baker's Dozen 05 Mar 2024, 04:22
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Bunuel
13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14

Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).

Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).

Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.

Answer: D.
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First of all thanks fort he series
I approached the problem in a diff way
Replace 8! with A for simpler maths -> and write faster :)

\(x=\frac{(A)^{10}-(A)^6}{(A)^{5}-(A)^3}\frac\).
We extract a^3
\(x=\frac{(A)^{3}*(A)^{3}*((A)^{4}-1)}{(A)^{3}*(A)^{2}-1}\frac\).
Now on the top, we have:
\(a^2-b^2=(a-b)(a+b)\)
\(x=\frac{(A)^{3}*(A)^{3}*((A)^{2}-1)*((A)^{2}+1)}{(A)^{3}*((A)^{2}-1)}\frac\).
So we can simplify 
\(x={(A)^{3}*((A)^{2}+1)}\).
as we have x/8!^3
\({((A)^{2}+1)}-39\).
then we reach
\((8!)^2-38\) 
and we can proceed on the same way (2*5)^2 is 00 so 62 -> 6*2 = 12


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