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Baker's Dozen [#permalink]
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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: bakersdozen12878220.html#p10575022. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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: bakersdozen12878220.html#p10575033. 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: bakersdozen12878220.html#p10575044. 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: bakersdozen12878220.html#p10575055. 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: bakersdozen12878220.html#p10575076. 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. \(\frac{yz}{yz+xzxy}\) C. \(\frac{yz}{yz+xz+xy}\) D. \(\frac{xyz}{yz+xzxy}\) E. \(\frac{yz+xzxy}{yz}\) Solution: bakersdozen12878220.html#p10575087. 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: bakersdozen12878220.html#p10575098. 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: bakersdozen12878220.html#p10575129. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)?A. 1+y B. 1y C. 1y D. y1 E. xy Solution: bakersdozen12878220.html#p105751410. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x2y?A. 7 B. 11 C. 13 D. 17 E. 19 Solution: bakersdozen12878220.html#p105751511. 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: bakersdozen12878240.html#p105751712. 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: bakersdozen12878240.html#p105751913. 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: bakersdozen12878240.html#p1057520
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Re: Baker's Dozen [#permalink]
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02 Sep 2012, 12:55
Can you explain how you got from (3^53^2)^2 to 3^4*(3^31)^2? When extracting 3^2 from (3^53^2), do we also square the 3^2? I know it may be simple algebra but its a bit confusing, I got 3^2(3^31)^2. I know thats wrong but can you explain why? Thanks. Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
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]
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03 Sep 2012, 05:33
ctiger100 wrote: Can you explain how you got from (3^53^2)^2 to 3^4*(3^31)^2? When extracting 3^2 from (3^53^2), do we also square the 3^2? I know it may be simple algebra but its a bit confusing, I got 3^2(3^31)^2. I know thats wrong but can you explain why? Thanks. Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
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. Consider this: \((abac)^2=(a(bc))^2=a^2*(bc)^2\), so \((3^53^2)^2=(3^2*(3^31))^2=(3^2)^2*(3^31)^2=3^4*(3^31)^2\). Hope it helps.
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If x and y are negative numbers, [#permalink]
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04 Sep 2012, 09:40
If x and y are negative numbers, what is the value of square root of x2/xsquare root of y∗ya)1+y b)1−y c)−1−y d)y−1 e)x−y 1lyl= 1+Y..this is ans as it is given Y is negative.. absolute value of y will be +ve.as we know whatever the sign inside the absolute value , after taking out of absolute value it wil becum positive.. (1y) i think should be the ans... because when we take out y from absolute value it will be +ve...and negative sign outside absolute value will make the sign negative and the solution should be 1y.. where i m wrong??
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Re: If x and y are negative numbers, [#permalink]
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04 Sep 2012, 10:38
sanjoo wrote: 1lyl= 1+Y..this is ans
as it is given Y is negative.. absolute value of y will be +ve.as we know whatever the sign inside the absolute value , after taking out of absolute value it wil becum positive..
(1y) i think should be the ans... because when we take out y from absolute value it will be +ve...and negative sign outside absolute value will make the sign negative and the solution should be 1y..
where i m wrong?? If you plug in a negative number for y in the equation 1+y you will see that the term will get smaller, because you'll subtracte something from 1.



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Re: Baker's Dozen [#permalink]
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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]
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01 Oct 2012, 09:57
voodoochild 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. 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 Since we are talking about a 5digit password and not a 5digit number, then it can start with zero. For example a password can be 00123 or 00001. Hope it's clear.
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Re: Baker's Dozen [#permalink]
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13 Oct 2012, 00:23
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* 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, How it would be 810 ways. 9*9*1*1*1/10^5 Why the order iwll matter in above question ? thanks



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Re: Baker's Dozen [#permalink]
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13 Oct 2012, 02:58
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Awesome post Bunuel !! Looking forward for more such posts !! cheers



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Re: Baker's Dozen [#permalink]
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13 Oct 2012, 04:06
154238 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, How it would be 810 ways. 9*9*1*1*1/10^5 Why the order iwll matter in above question ? thanks Order matters because 12666 password differs from 66612. Check this post: bakersdozen12878260.html#p1079496 it might help.
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Re: Baker's Dozen [#permalink]
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03 Dec 2012, 21:12
Bunuel wrote: 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
Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders > Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares; Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders > Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares; Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders > Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;
Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).
Answer: D. can you explain this one... I didnot understand the way you are writing fractions of each person down..... thanks



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Re: Baker's Dozen [#permalink]
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04 Dec 2012, 03:25
Amateur wrote: Bunuel wrote: 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
Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders > Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares; Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders > Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares; Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders > Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;
Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).
Answer: D. can you explain this one... I didnot understand the way you are writing fractions of each person down..... thanks Explained here: bakersdozen12878240.html#p1059585
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Re: Baker's Dozen [#permalink]
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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: bakersdozen12878220.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]
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21 Apr 2013, 08:19
sujit2k7 wrote: 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: bakersdozen12878220.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... Responding to a pm: From the options, it is obvious that the marbles are considered unique i.e. the red marbles are all different, say they are numbered, and the blue marbles are all different. The problem with your approach is that when you leave 1 red and 1 blue, you don't provide for this. Also, once you account for the uniqueness, there will be double counting  when you leave Red1 and Blue1 and then choose 8 out of the remaining 10 and say you are left with Red2 and Red3 too, this will be counted as different from when you leave Red2 and Blue1 and then choose 8 out of the remaining 10 and you are left with Red1 and Red3. These two cases are actually the same. Rule of thumb is that 'at least one' cases are best done using the complement approach which is 'Total  None' (as shown by Bunuel in the solution).
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Re: Baker's Dozen [#permalink]
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11 May 2013, 17:15
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 .. can you please elaborate more ? thanks in advance
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Re: Baker's Dozen [#permalink]
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11 May 2013, 17:25
Bunuel wrote: 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
Total ways to select 8 marbles out of 7+5=12 is \(C^8_{12}\); Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\); Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);
Hence ways to select 8 marbles so that at least one red marble and at least one blue marble to remain the jar is \(C^8_{12}(C^7_7*C^1_5+C^5_5*C^3_7)=495(5+35)=455\).
Answer: D. I appreciate if you provide further classification pleaaaaaaaaaaaaase
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Re: Baker's Dozen [#permalink]
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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]
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11 May 2013, 21:27
TheNona wrote: 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? When m is divided by n, the remainder can be anything from 0 to (n1). Take examples: When m is divided by 8, the remainder can 0 or 1 or 2 till 7. Can the remainder be 8? No because that means the number is divisible by 8. Say you divide 20 by 8, you get remainder 4. You divide 21 by 8, you get remainder 5. You divide 22 by 8, you get remainder 6. You divide 23 by 8, you get remainder 7. You divide 24 by 8, you get remainder 0, isn't it? The remainder will not be 8  it will be 0. Hence if you know that the remainder when you divide m by n is 7, you can say that n must be at least 8. Similarly, when a number is divided by z and remainder is 8, z must be at least 9.
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Re: Baker's Dozen [#permalink]
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04 Jul 2013, 03:03
Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=[color=#ff0000]3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).[/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]
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04 Jul 2013, 03:24
BankerRUS wrote: Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), 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
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=[color=#ff0000]3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).[/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: \((abac)^2=(a(bc))^2=a^2*(bc)^2\), so: \((3^53^2)^2=(3^2*(3^31))^2=(3^2)^2*(3^31)^2=3^4*(3^31)^2=3^4*26=3^4*2^2*13^2\). \((5^75^4)^2=(5^4*(5^31))^2=(5^4)^2*(5^31)^2=5^8*(5^31)^2=5^8*124^2=5^8*2^4*31^2\) \((3^4*2^2*13^2)(5^8*2^4*31^2)=2^6*3^4*5^8*13^2*31^2\). Hope it's clear.
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Re: Baker's Dozen [#permalink]
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