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Baker's Dozen

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Bunuel
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Re: Baker's Dozen [#permalink] New post   04 Jul 2013, 03:24
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BankerRUS wrote:
Bunuel wrote:
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


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=[color=#ff0000]3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).[/color]

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.


Bunuel, could you please kindly explain this part? How did you come up with this factorization? Many thanks!


Consider this: \((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\), so:

\((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2=3^4*26=3^4*2^2*13^2\).
\((5^7-5^4)^2=(5^4*(5^3-1))^2=(5^4)^2*(5^3-1)^2=5^8*(5^3-1)^2=5^8*124^2=5^8*2^4*31^2\)

\((3^4*2^2*13^2)(5^8*2^4*31^2)=2^6*3^4*5^8*13^2*31^2\).

Hope it's clear.
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Re: Baker's Dozen [#permalink] New post   12 Sep 2014, 08:13
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DangerPenguin wrote:
Bunuel wrote:
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 was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from.

Thanks,


XX666

Each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.
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Re: Baker's Dozen [#permalink] New post   16 Nov 2014, 13:35
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Madrigal wrote:
Bunuel wrote:
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


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.


Hi Brunel,
Great set of questions! Could you please explain the logic behind how you factorised this:
\(y=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2\).
I am not too clear on how you pulled out factors of 3^4 and 5^8, and still have a 3^3 and a 5^3 left in the brackets.

Thanks!


\((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\). So, \((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2\).

Similarly \((5^7-5^4)^2=(5^4(5^3-1))^2=5^8*(5^3-1)^2\).

Hope it's clear.
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Re: Baker's Dozen [#permalink] New post   22 Dec 2014, 03:19
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deeuk wrote:
Bunuel wrote:
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. \(\frac{yz+xz-xy}{yz}\)

With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-xy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours (time is reciprocal of rate).

In \(\frac{xyz}{yz+xz-xy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}\) amount of the water into the pool.

Answer: B.

I did not get why 1/x is multiplied. Would you mind terribly, to explain?
Thanks


The question asks to find the amount of water in terms of the fraction of the pool which pump A pumped into the pool. We know that the pool will be filled in \(\frac{xyz}{yz+xz-xy}\) hours and the rate of A is 1/x pool/hour. The job done by A in that time is (job) = (time)*(rate) = \(\frac{1}{x}*\frac{xyz}{yz+xz-xy}\).

Hope it's clear.
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Re: Baker's Dozen [#permalink] New post   27 Mar 2017, 18:07
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Here's my solution for #2 (see visual below). It helps to use brackets!
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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 4369 times ]

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Re: Baker's Dozen [#permalink] New post   26 Mar 2018, 04:59
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nahid78 wrote:
Bunuel wrote:
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


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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Re: Baker's Dozen [#permalink] New post   21 Dec 2019, 08:27
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Bunuel wrote:
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.


Just Excellent Question Bunuel...
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Re: Baker's Dozen [#permalink] New post   22 Dec 2019, 08:57
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dips1122 wrote:
Bunuel wrote:
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.


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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Re: Baker's Dozen [#permalink] New post   11 Mar 2012, 05:36
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well
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Re: Baker's Dozen [#permalink] New post   11 Mar 2012, 20:52
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utkarshlavania wrote:
@karishma

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


You are right that |x|=-x since x is negative but tell me, what is -x, negative or positive? Negative of negative gives you positive, right? So -x must be positive. Now, if x is negative,
-x/x must be positive/negative giving you -1.

You are confusing yourself too much with negatives and positives. Just think of it this way:

|x|/x = -x/x (By definition, since |x| = -x when x < 0)
-x/x = -1 ( x and x get canceled here leaving you with -1)
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Re: Baker's Dozen [#permalink] New post   Updated on: 28 Apr 2012, 04:00
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)?
Why not : 9*9*1*1*1* 5!/2! ??

Originally posted by mofasser08 on 28 Apr 2012, 03:41.
Last edited by mofasser08 on 28 Apr 2012, 04:00, edited 1 time in total.
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Re: Baker's Dozen [#permalink] New post   02 Sep 2012, 12:55
Can you explain how you got from (3^5-3^2)^2 to 3^4*(3^3-1)^2? When extracting 3^2 from (3^5-3^2), do we also square the 3^2? I know it may be simple algebra but its a bit confusing, I got 3^2(3^3-1)^2. I know thats wrong but can you explain why? Thanks.


Bunuel wrote:
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


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.
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Re: Baker's Dozen [#permalink] New post   01 Oct 2012, 09:52
Bunuel wrote:
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.



Bunuel,
I did a mistake while calculating 10^5 - I calculated 9*10^4. I have a doubt : when the question says "5 digit" code, doesn't it mean that numbers start from 10000? I have seen some GMATClub questions were three digit codes start from 100. Can you please help?
Thanks
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Re: Baker's Dozen [#permalink] New post   19 Apr 2013, 00:20
Bunuel wrote:
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!

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



Actually need help from the moderators to know where my approach is going wrong:

it is it is told at least 1 red , at least one blue will be there.
So i am leaving 1 red , 1 blue=> Now i have 6 Red , 4 Blue ..I have to select 8 marbels...
So it means i have to select 8 marbels out of 10 marbels (colors don't matter)
= \(C^10_8\) = \(C^10_2\) = 45

I know i am not getting the ans...but what's wrong with this I m not able to spot...
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Re: Baker's Dozen [#permalink] New post   11 May 2013, 19:25
Bunuel wrote:
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.

Answer: B.


why z is minimum= 9?
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Re: Baker's Dozen [#permalink] New post   04 Jul 2013, 03:03
Bunuel wrote:
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


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=[color=#ff0000]3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).[/color]

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.


Bunuel, could you please kindly explain this part? How did you come up with this factorization? Many thanks!
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Re: Baker's Dozen [#permalink] New post   18 Sep 2013, 06:00
Bunuel wrote:
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


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.



Can someone please explain how the move from the second part of the equation to the third part was?
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Re: Baker's Dozen [#permalink] New post   18 Sep 2013, 07:19
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ronr34 wrote:
Bunuel wrote:
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


\(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).

Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).

Answer: E.



Can someone please explain how the move from the second part of the equation to the third part was?


Consider this: \((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\), so \((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2\).

Hope it helps.
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Re: Baker's Dozen [#permalink] New post   01 Dec 2013, 03:32
Hello Bunuel,

Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest?

\(x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = -185\)

\(x = -41\)

Then we take 5 smallest integers, which are the latter in our set

\((x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40\)

\(x= - 41\)

\(-41*5 + 40=-165\)

answer A?

Please correct me if I'm wrong.

Bunuel wrote:
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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Re: Baker's Dozen [#permalink] New post   01 Dec 2013, 06:54
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trailrunner wrote:
Hello Bunuel,

Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest?

\(x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = -185\)

\(x = -41\)

Then we take 5 smallest integers, which are the latter in our set

\((x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40\)

\(x= - 41\)

\(-41*5 + 40=-165\)

answer A?

Please correct me if I'm wrong.

Bunuel wrote:
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.


Five largest integers from {\(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\)} are {\(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\)} no matter whether x is negative or positive.
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Re: Baker's Dozen [#permalink]
01 Dec 2013, 06:54
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