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# Baker's Dozen

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Math Expert
Joined: 02 Sep 2009
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08 Mar 2012, 13:27
25
KUDOS
Expert's post
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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

Solution: baker-s-dozen-128782-20.html#p1057502

2. If $$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4

Solution: baker-s-dozen-128782-20.html#p1057503

3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k?
A. 6
B. 8
C. 9
D. 10
E. 12

Solution: baker-s-dozen-128782-20.html#p1057504

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

Solution: baker-s-dozen-128782-20.html#p1057505

5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?
A. 460
B. 490
C. 493
D. 455
E. 445

Solution: baker-s-dozen-128782-20.html#p1057507

6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?
A. $$\frac{yz}{x+y+z}$$

B. $$\frac{yz}{yz+xz-xy}$$

C. $$\frac{yz}{yz+xz+xy}$$

D. $$\frac{xyz}{yz+xz-xy}$$

E. $$\frac{yz+xz-xy}{yz}$$

Solution: baker-s-dozen-128782-20.html#p1057508

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: baker-s-dozen-128782-20.html#p1057509

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

Solution: baker-s-dozen-128782-20.html#p1057512

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

Solution: baker-s-dozen-128782-20.html#p1057514

10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

Solution: baker-s-dozen-128782-20.html#p1057515

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12

Solution: baker-s-dozen-128782-40.html#p1057517

12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33

Solution: baker-s-dozen-128782-40.html#p1057519

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: baker-s-dozen-128782-40.html#p1057520
_________________
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Re: Baker's Dozen [#permalink]

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13 Oct 2012, 02:58
2
KUDOS
Awesome post Bunuel !!
Looking forward for more such posts !!

cheers
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GMAT 2: 720 Q48 V40
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Re: Baker's Dozen [#permalink]

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09 Mar 2012, 12:28
1
KUDOS
In regards to question 13, how do you know that (8!)^2 will have two 0's at the end?
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Re: Baker's Dozen [#permalink]

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09 Mar 2012, 23:41
1
KUDOS
BN1989 wrote:
In regards to question 13, how do you know that (8!)^2 will have two 0's at the end?

You figure it out by knowing that there is a 5 as a factor in the final number.

example 8! = 8*7*6*5...*1

Now if u know how many fives are there you will have that number of zeros in the end because at least that many number of 2 will there for sure ...

Now IF u want to figure out how may fives are there in (which is equivalent to finding how many zeros in the end of the number )
then simply keep dividing the factorial by 5 and adding it to the result until 5^x exceeds the factorial, i know i have used very confusing language but a example will simplify things for sure

how many zeros at the end of 312!
answer :- 312!/5 + 312!/5^2 + 312!/5^3 = 62 + 12 + 2 (now note i stopped at 5^3, because 5^4 =25*25=625 which is more that 312 ) hence :- 62 + 12 + 2 =76 number of 5s in the factorial and same number of zeros in the end because these many number of 2s will obviously be there in factorial and will form 10s .

IF you are still confused I would suggest you to download Bunuel's math's notes and even if you are not confused i would still suggest you to do the same ,
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Re: Baker's Dozen [#permalink]

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10 Mar 2012, 00:01
1
KUDOS
2)e
3)b
4)d
6)e
7)d
8)b
9)c
10)c
11)b
12) b
13)d

Bunuel could you please provide us/me with the answer keys so that I can cross check
thanks once again for the questions and a amazing post .
Math Expert
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Posts: 33061
Followers: 5773

Kudos [?]: 70800 [1] , given: 9857

Re: Baker's Dozen [#permalink]

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10 Mar 2012, 00:22
1
KUDOS
Expert's post
utkarshlavania wrote:
2)e
3)b
4)d
6)e
7)d
8)b
9)c
10)c
11)b
12) b
13)d

Bunuel could you please provide us/me with the answer keys so that I can cross check
thanks once again for the questions and a amazing post .

I'll provide OA's with detailed solutions after some discussion in a couple of days. If you need OA's asap pm me and I'll send you them.
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Re: Baker's Dozen [#permalink]

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10 Mar 2012, 00:38
1
KUDOS
Bunuel wrote:
utkarshlavania wrote:
2)e
3)b
4)d
6)e
7)d
8)b
9)c
10)c
11)b
12) b
13)d

Bunuel could you please provide us/me with the answer keys so that I can cross check
thanks once again for the questions and a amazing post .

I'll provide OA's with detailed solutions after some discussion in a couple of days. If you need OA's asap pm me and I'll send you them.

Bunuel i have messaged you, looking forward
thanks once again ...
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Re: Baker's Dozen [#permalink]

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10 Mar 2012, 01:24
1
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[quote2]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

Answer to 9th question If x and y are negative numbers, what is the value of $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$?[/b]
Sqrtx^2/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y
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Re: Baker's Dozen [#permalink]

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10 Mar 2012, 02:38
1
KUDOS
-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but
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Re: Baker's Dozen [#permalink]

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10 Mar 2012, 07:42
1
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Quote:
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

x ;x+2;x+2*2;x+2*3....x+2*6

since the set has consecutive odd integers, then mean =median

mean of 5 largest integers=-185/5= -37
median of 5 largest integers is the 3d of the largest or the 5th of the whole set A= x+2*4
x+8=-37 ;x=-45

median of the first 5 small integers =-45+2*2=-41

the sum of the first small odd integers=number of integers*mean (or median)=5*median=5*(-41)=-205
answ is
[Reveal] Spoiler:
D

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Joined: 23 Oct 2010
Posts: 386
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Concentration: Finance
Schools: HEC '15 (A)
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Kudos [?]: 264 [1] , given: 73

Re: Baker's Dozen [#permalink]

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10 Mar 2012, 07:54
1
KUDOS
Quote:
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

x=((8!)^6*((8!)^4-1))/((8!)^3*((8!)^2-1))=(8!)^3*((8!)^2+1)

((8!)^3*((8!)^2+1) /(8!)^3)) -39=(8!)^2+1)-39
note, that 8! has 5 and 2. so 8! ends with 0. then (8!)^2 will end by two 0s.
***00-38=62
6*2=12
answ is
[Reveal] Spoiler:
D

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*\sqrt{k}=3*7*3*2*sqrt{k}=3^2*7*2
so we need one more 7*2 (14)
sqrootk=14
k=196
answ is
[Reveal] Spoiler:
D

what an irony. all the 3 questions are
[Reveal] Spoiler:
D
heh

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
-x/x-|y|=-1-(-y)=y-1 since y<0 ; x<0
[Reveal] Spoiler:
D
is the answ
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Last edited by LalaB on 10 Mar 2012, 09:08, edited 1 time in total.
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10 Mar 2012, 08:03
1
KUDOS
12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33

x/y the remainder is 3 . then the smallest x is 3

y/z the remainder is 8. then the smallest y is 8 and z is 9

3+8+9=20

answ is B
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Re: Baker's Dozen [#permalink]

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10 Mar 2012, 08:17
1
KUDOS
10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

-9<x<9
-5<y<5

-9<x<9
-10<2y<10
x-2y<9-(-10)
x-2y<19

the answer is 17. D again the mystical letter for today heh

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12

1, 3, 9, 27 ok please pay attention to the fact that the term of sequence has 3 one less than the number of the term,i.e.
the 3d term is 9, but the number has two 3s (one less than the term). the 4th term is 3*3*3 ,which has three 3s.

ok, then we have the following-
the q. asks us to find out -
(13th term+15th term) -(12th term +14th term)

(3^12+3^14)-(3^11-3^13)=
3^11(3-1)+3^13(3-1)=
2*3^11(1+3^2)=
20*3^11
the answer is B

p.s. also note, that answer choices B and D are leaders of the day heh
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Re: Baker's Dozen [#permalink]

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11 Mar 2012, 00:34
1
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I am not so certain whether I am right but I would go with D in #9.

IF x<0 and y<0 then sqrt(x^2)/x=|x|/x, then -x/x=-1; because there is a minus in front of x, than -(-x) should be positive.

sqrt two negative y gives square root of y^2, than because y<1, y is negative as well. -y.

From the expression we've got -1+y, thus D should be a correct answer.
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Re: Baker's Dozen [#permalink]

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11 Mar 2012, 01:04
1
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Expert's post
LalaB wrote:
Bunuel, could u please tell us which question is still unanswered or answered wrong by all of us? I want to think on such questions more, before u post solutions.

and btw, do u read our solutions or just only answers? I wonder whether my way of thinking was ok. in some cases I tried to use another method not to repeat others. I just wonder whether it worked, or it was just coincidence

I do read all the solutions and award +1 Kudos if there is at least one correct explanation in the post.

As for your other question: I don't see a solution for #7, so you can try this one.
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Re: Baker's Dozen [#permalink]

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11 Mar 2012, 02:44
1
KUDOS
Expert's post
utkarshlavania wrote:
[quote2]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

Answer to 9th question If x and y are negative numbers, what is the value of $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$?[/b]
Sqrtx^2/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y

Responding to a pm:

[Reveal] Spoiler:
First of all, I would suggest you to put x = -2 and y = -3 and see what you get.

$$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$

$$\frac{\sqrt{(-2)^2}}{-2}-\sqrt{-(-3)*|-3|}$$

$$\frac{2}{-2}-\sqrt{9}$$ (because square root will always be positive)

-1 - 3 = -1 + y (since y = -3)

Coming to your solution:

I see you have everything correct till the last step: |x|/x-|y|

|x|/x will be -1 which is fine.
Do you remember how we define mods?
We say:
|x| = x if x >= 0
|x| = -x if x < 0

Why? If x >= 0 e.g. say x = 5,
|x| = x = 5
Instead, if x < 0 e.g. say x = -6,
is |x| = x? No! mods are never negative. So |x| = -x = -(-6) = 6 (Since x itself is negative, mod of x is negative of x which becomes positive.)

Therefore, if we know that y is negative, what is the value of |y|?
|y| = -y

Therefore, |x|/x-|y| = -1 -(-y) = -1 + y

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11 Mar 2012, 05:59
1
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Expert's post
utkarshlavania wrote:
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well

If $$x<0$$ then $$|x|=-x$$ and $$\frac{|x|}{x}=\frac{-x}{x}=-1$$, ($$-\frac{x}{x}=-1$$ no matter whether $$x$$ is negative or positive).
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13 Mar 2012, 03:56
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Expert's post
4
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BOOKMARKED
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: $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$$ --> $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$$ --> $$(5x+20)+20=-185$$ --> $$5x+20=-205$$

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13 Mar 2012, 03:59
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Expert's post
10
This post was
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10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

Notice that we are not told that $$x$$ and $$y$$ are integers.

$$x^2<81$$ means that $$-9<x<9$$ and $$y^2<25$$ means that $$-5<y<5$$. Now, since the largest value of $$x$$ is almost 9 and the largest value of $$-2y$$ is almost 10 (for example if $$y=-4.9$$), then the largest value of $$x-2y$$ is almost 9+10=19, so the actual value is less than 19, which means that the largest prime that can be equal to $$x-2y$$ is 17. For example: $$x=8$$ and $$y=-4.5$$.

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13 Mar 2012, 04:02
1
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Expert's post
20
This post was
BOOKMARKED
12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33

Given $$x=qy+3$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=3$$. This basically means that $$x$$ is less than $$y$$. For example 3 divided by 4 yields remainder of 3.

Thus we have that:
$$x$$ is divided by $$y$$ the remainder is 3 --> minimum value of $$x$$ is 3;
$$y$$ is divided by $$z$$ the remainder is 8 --> minimum value of $$y$$ is 8 and minimum value of $$z$$ is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);

So, the smallest possible value of $$x+y+z$$ is 3+8+9=20.

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Re: Baker's Dozen [#permalink]

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15 Mar 2012, 03:36
1
KUDOS
Bunuel wrote:
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.

Although there is hardly anything left after your's explanations but still I'll post my approach, which might help few people out here (esp. on the # trailing 0's as you people have kept in the Math book)--

first I simplified the whole x value as it looks so clumsy/heavy: given
$$x = (8!)^6 [(8!)^4 - 1]/ (8!)^3[(8!)^2 - 1] = (8!)^3[(8!)^2 + 1][(8!)^2 - 1]/[(8!)^2 - 1] so, (x/ (8!)^3) - 39 = [(8!)^2 + 1] -39 = (8!)^2 -38$$
Now as per the Q we need to know unit & tenth digit of the above number- as we know factorials of numbers > 4 have trailing zero's so here we can determine how many trailing 0's in 8!. Since $$8 < 5^2$$ so that means we have only 1 trailing 0 in here (alternately you can have manual calc in this case as well because it's small number). So $$(8!)^2$$ has 2 trailing 0's. Now it does not matter what hundredth position is becasue we need to subtract only 38- this would result in 6 in tenth pos and 2 in unit pos. so product is = 12...

P.S.- just for refrence # trailing 0's in factorial N is determined by- $$N/5^1+N/5^2+....+N/5^k (where 5^k < N)$$for each division we take integer part (quotient) and then add them up. this should give # trailing 0's
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Re: Baker's Dozen   [#permalink] 15 Mar 2012, 03:36

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