Baker's Dozen : Problem Solving (PS)

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11 May 2013, 21:27

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Bunuel

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 (n-1).
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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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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Bunuel

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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Bunuel

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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18 Sep 2013, 07:19

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Bunuel

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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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

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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01 Dec 2013, 06:54

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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

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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14 Feb 2014, 01:39

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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.

Shouldn't -1-|y| be -1-y? since |y| will be positive always?

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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.

Shouldn't -1-|y| be -1-y? since |y| will be positive always?

No.

When $x\leq{0}$ then $|x|=-x$, or more generally when $some \ expression\leq{0}$ then $|some \ expression|={-(some \ expression)}$.

When $x\geq{0}$ then $|x|=x$, or more generally when $some \ expression\geq{0}$ then $|some \ expression|={some \ expression}$.

Since given that y is a negative number (y<0), then $|y| = -y$, hence $-1-|y|=-1-(-y)=-1+y$.

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.

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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.

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,

Bunuel

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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.

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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Bunuel

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!

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Bunuel

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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04 Dec 2014, 07:33

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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.

Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number,"
Why does it have to be powers of 2*7

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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.

Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number,"
Why does it have to be powers of 2*7

Let me ask you: by what positive integer we should multiply $126=2*3^2*7$ for the product to be the square of an integer?

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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

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22 Dec 2014, 03:19

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Bunuel

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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