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Problem Solving for 780+ Aspirants

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Re: Problem Solving for 780+ Aspirants [#permalink]

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16 Apr 2009, 04:39
Here is the next one.....

If $$x,y$$are positive integers such that $$5 \times x^3 = 6 \times y^4,$$ find the minimum possible value of $$x+y.$$

A. 1240

B. 1260

C. 1060

D. 1080

E. None of these

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Re: Problem Solving for 780+ Aspirants [#permalink]

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16 Apr 2009, 21:32
Can someone explaine the way how to solve it?

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Re: Problem Solving for 780+ Aspirants [#permalink]

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16 Apr 2009, 22:32
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cicerone wrote:
Here is the next one.....

If $$x,y$$are positive integers such that $$5 \times x^3 = 6 \times y^4,$$ find the minimum possible value of $$x+y.$$

A. 1240
B. 1260
C. 1060
D. 1080
E. None of these

Given that: x and y are +ve integers and
5x^3 = 6y^4
5x^3 = 2x3 y^4
y = (5x^3/6)^(1/4)

x = 5x6x6x6 = 6x180= 1080

y = [5 (5x6x6x6)^3 /6)]^(1/4)
y = [5^4 (6)^8)]^(1/4)
y = 5x6x6=180

x+y = 1260

Guess I encountered such question earlier too.
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Re: Problem Solving for 780+ Aspirants [#permalink]

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17 Apr 2009, 03:36
GMAT TIGER wrote:
cicerone wrote:
Here is the next one.....

If $$x,y$$are positive integers such that $$5 \times x^3 = 6 \times y^4,$$ find the minimum possible value of $$x+y.$$

A. 1240
B. 1260
C. 1060
D. 1080
E. None of these

Given that: x and y are +ve integers and
5x^3 = 6y^4
5x^3 = 2x3 y^4
y = (5x^3/6)^(1/4)

x = 5x6x6x6 = 6x180= 1080

y = [5 (5x6x6x6)^3 /6)]^(1/4)
y = [5^4 (6)^8)]^(1/4)
y = 5x6x6=180

x+y = 1260

Guess I encountered such question earlier too.

so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm
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Re: Problem Solving for 780+ Aspirants [#permalink]

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17 Apr 2009, 05:13
nightwing79 wrote:
so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get $$y$$'s value in integer. To do that, $$x$$ has to be $$5*6*6*6$$. Since $$y = 4sqrt(5x^3 / 6)$$ has one $$5$$, one $$6$$ in denominator, and three $$x$$, make the value of $$(5x^3 / 6)$$ a square of a square. To do that, $$x$$ requires $$5^3$$ to make $$5^4$$ and $$6^9$$ to make $$6^8$$ after cancealing a 6. Hence 4th sqrt power of $$(5x^3 / 6)$$ is a minimum integer vale.
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Re: Problem Solving for 780+ Aspirants [#permalink]

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17 Apr 2009, 17:37
GMAT TIGER wrote:
nightwing79 wrote:
so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get $$y$$'s value in integer. To do that, $$x$$ has to be $$5*6*6*6$$. Since $$y = 4sqrt(5x^3 / 6)$$ has one $$5$$, one $$6$$ in denominator, and three $$x$$, make the value of $$(5x^3 / 6)$$ a squre of a square. To do that, $$x$$ requires $$5^3$$ to make $$5^4$$ and $$6^9$$ to make $$6^8$$ after cancealing a 6. Hence 4th sqrt power of $$(5x^3 / 6)$$ is a minimum integer vale.

Good stuff!

However - why does it have to be 6^9 why not 6^5 to make 6^4 hence square root of would still be an integer value.

The key is root to the power of 4 - have to remember that and then looking to understand what a min integer value would look like - good learning.

che
dg
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Re: Problem Solving for 780+ Aspirants [#permalink]

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18 Apr 2009, 16:32
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nightwing79 wrote:
GMAT TIGER wrote:
nightwing79 wrote:
so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get $$y$$'s value in integer. To do that, $$x$$ has to be $$5*6*6*6$$. Since $$y = 4sqrt(5x^3 / 6)$$ has one $$5$$, one $$6$$ in denominator, and three $$x$$, make the value of $$(5x^3 / 6)$$ a squre of a square. To do that, $$x$$ requires $$5^3$$ to make $$5^4$$ and $$6^9$$ to make $$6^8$$ after cancealing a 6. Hence 4th sqrt power of $$(5x^3 / 6)$$ is a minimum integer vale.

Good stuff!

However - why does it have to be 6^9 why not 6^5 to make 6^4 hence square root of would still be an integer value.

The key is root to the power of 4 - have to remember that and then looking to understand what a min integer value would look like - good learning.

che
dg

How do you get that? If you can, then thats the answer. But again, how?:roll:
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Re: Problem Solving for 780+ Aspirants [#permalink]

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18 Apr 2009, 18:54
nightwing79 wrote:
GMAT TIGER wrote:
nightwing79 wrote:
so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get $$y$$'s value in integer. To do that, $$x$$ has to be $$5*6*6*6$$. Since $$y = 4sqrt(5x^3 / 6)$$ has one $$5$$, one $$6$$ in denominator, and three $$x$$, make the value of $$(5x^3 / 6)$$ a squre of a square. To do that, $$x$$ requires $$5^3$$ to make $$5^4$$ and $$6^9$$ to make $$6^8$$ after cancealing a 6. Hence 4th sqrt power of $$(5x^3 / 6)$$ is a minimum integer vale.

How do you get that? If you can, then thats the answer. But again, how?:roll:

come to think of it .... i don't know either..... i think i went back to that special place....where the basic laws of math and physics don't apply....

the whole problem seems so much easier now...lol....

as is always....
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Re: Problem Solving for 780+ Aspirants [#permalink]

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26 Apr 2009, 23:08
No..You are absolutely right nightwing..you can use 6^5 BUT with x having an exponent of 3, you can never get that. The least figure you can get is 6^9. Thus GMAT tiger is correct.

I think this is quite a likely question of the highest level of the actual GMAT..ie, if you keep getting the questions right.

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Re: Problem Solving for 780+ Aspirants [#permalink]

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05 May 2009, 23:26
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cicerone wrote:
Here is the next one.....

If $$x,y$$are positive integers such that $$5 \times x^3 = 6 \times y^4,$$ find the minimum possible value of $$x+y.$$

A. 1240

B. 1260

C. 1060

D. 1080

E. None of these

Folks, nice to see a good discussion on this question and also our friends have come up with some quick and lucid solutions.

I just thought to put forward another lucid way ....

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Re: Problem Solving for 780+ Aspirants [#permalink]

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05 May 2009, 23:39
Folks, here is the next question..

From the set of positive factors of 840 one factor is chosen at random. What is the probability that it is divisible by 15?

A. 1/4

B. 7/32

C. 1/2

D. 1/56

E. None of these

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Re: Problem Solving for 780+ Aspirants [#permalink]

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06 May 2009, 16:45
Just a quick question: Why do you have "none of the above" in answer choice E in almost all question?

cicerone wrote:
Folks, here is the next question..

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A. 1/4
B. 7/32
C. 1/2
D. 1/56
E. None of these

840 = 2x2x2x3x5x7 = 2^3 x 3 x 5 x 7

No of factors = (3+1)(1+1)(1+1)(1+1) = 32
How many factors of 840 are divisible by 15? 16. How is 16?
No of factors divisible by 15 are 15 and multiples of 15 = 15, 15x2, 15x2x2, 15x2x2x2....................840.
No of factors divisible by 15 = 1 + (4c1+4c2+4c3+4c4) = 16

The prob = 16/32 = 1/2
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Re: Problem Solving for 780+ Aspirants [#permalink]

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06 May 2009, 20:27
cicerone wrote:
sureshbala wrote:
Folks, sorry for the delay (I am on a vacation).

Anyway, here is the next one.....

There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?

A. 4

B. 6

C. 8

D. 10

E. Cannot be determined

Folks, this can be answered very quickly if you can conclude the following two results.

Result 1: Since the number of chairs occupied by the girls in both the cases is same there is no need to consider the number of girls.

Result 2: In the second case we are able to seat 2 boys per chair and no boy is left unseated, hence the number of boys must be even.

Let the number of boys be 2x.

Seats occupied in the first case = 2x-1

Seats occupied in the second case = x

Given that 2x-1-x = 3

So x = 4

Hence the number of boys 2x = 8.

thanks. my approach:
Number of chair=B+G-1=B/2+G+3
solving it will give b=8

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Re: Problem Solving for 780+ Aspirants [#permalink]

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06 May 2009, 21:00
Specially the method to calculate the number of factors divisible by 15.
+1
GMAT TIGER wrote:
Just a quick question: Why do you have "none of the above" in answer choice E in almost all question?

cicerone wrote:
Folks, here is the next question..

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A. 1/4
B. 7/32
C. 1/2
D. 1/56
E. None of these

840 = 2x2x2x3x5x7 = 2^3 x 3 x 5 x 7

No of factors = (3+1)(1+1)(1+1)(1+1) = 32
How many factors of 840 are divisible by 15? 16. How is 16?
No of factors divisible by 15 are 15 and multiples of 15 = 15, 15x2, 15x2x2, 15x2x2x2....................840.
No of factors divisible by 15 = 1 + (4c1+4c2+4c3+4c4) = 16

The prob = 16/32 = 1/2

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Re: Problem Solving for 780+ Aspirants [#permalink]

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06 May 2009, 21:18
cicerone wrote:

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A factor could be not only a positive number but also negative one. That is why GMAC often writes "number of positive factors".

So, for 840 = 2^3*3*5*7:
N = 2 * (1+3)*(1+1)*(1+1)*(1+1) = 64
N15 = 2 * (1+3)*1*1*(1+1) = 16

p= 16/64 = 1/4
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Re: Problem Solving for 780+ Aspirants [#permalink]

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07 May 2009, 19:00
walker wrote:
cicerone wrote:

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A factor could be not only a positive number but also negative one. That is why GMAC often writes "number of positive factors".

So, for 840 = 2^3*3*5*7:
N = 2 * (1+3)*(1+1)*(1+1)*(1+1) = 64
N15 = 2 * (1+3)*1*1*(1+1) = 16

p= 16/64 = 1/4

Thats definitely a good point but in standerized tests such as GMAT "a factor is always a +ve one". In other word, the factors of a given number/integer is only natural numbers. I am further confirmed by the following statement that factors are only +ve numbers, at least in standarized tests.

HongHu wrote:
When we talk about multiples and factors, we are talking about natural numbers, I believe. Otherwise we will not be able to find the least common multiple for any two numbers. The LCM of two numbers is the smallest number (not zero) that is a multiple of both. If we could use negative number here, a negative number with a large absolute value is smaller than a negative number with a small absolute value. There won't be a limit to this. We won't be able to find the smallest number.

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Re: Problem Solving for 780+ Aspirants [#permalink]

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08 May 2009, 04:58
GMAT TIGER wrote:
Thats definitely a good point but in standerized tests such as GMAT "a factor is always a +ve one". In other word, the factors of a given number/integer is only natural numbers. I am further confirmed by the following statement that factors are only +ve numbers, at least in standarized tests.

Hi, Tiger!

Honestly, I have never seen any GMAT question that test factor / positive factor concept. So, to be GMAT-like the question should say "A positive factor of 840..." rather than "A factor of 840 ...".

By the way, it doesn't change answer.
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Re: Problem Solving for 780+ Aspirants [#permalink]

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08 May 2009, 19:58
GMAT TIGER wrote:
Just a quick question: Why do you have "none of the above" in answer choice E in almost all question?
cicerone wrote:
Folks, here is the next question..A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A. 1/4
B. 7/32
C. 1/2
D. 1/56
E. None of these

840 = 2x2x2x3x5x7 = 2^3 x 3 x 5 x 7
No of factors = (3+1)(1+1)(1+1)(1+1) = 32 How many factors of 840 are divisible by 15? 16. How is 16?
No of factors divisible by 15 are 15 and multiples of 15 = 15, 15x2, 15x2x2, 15x2x2x2....................840.
No of factors divisible by 15 = 1 + (4c1+4c2+4c3+4c4) = 16
The prob = 16/32 = 1/2

I noticed that no of factors divisible by 15 = 8 not 16. It would be 16 if the factors were not repeated.

So prob = 8/32 = 1/4.

But walker how did you get 64 and 16? Why you multiplied by 2 in both cases?
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Re: Problem Solving for 780+ Aspirants [#permalink]

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08 May 2009, 23:00
GMAT TIGER wrote:
But walker how did you get 64 and 16? Why you multiplied by 2 in both cases?

I remember that GMAT always says "number of positive factors" to avoid ambiguity. So, I used "2" to count both negative and positive set of factors. Anyway, it would be better to use word "positive" in the question.
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Re: Problem Solving for 780+ Aspirants [#permalink]

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10 May 2009, 22:36
Folks, the answer for this question remains the same even if you consider negative divisors. But as our friends pointed out, GMAT talks of only positive divisors and hence I have modified the question to reflect the same.

Regards
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Re: Problem Solving for 780+ Aspirants   [#permalink] 10 May 2009, 22:36

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