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Manager
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10 Apr 2007, 15:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi everyone,
I have three questions that I have trouble to solve them. If you know how do them, please let me know. By the way, these questions are from GMATPrep which should be very similar to the real GMAT.

Thanks,

1)For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a)between 2 and 10
b)between 10 and 20
c)between 20 and 30
d)between 30 and 40
e)greater than 40

2)An equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

a)5
b)8
c)11
d)15
e)19

3)For which of the following functions f is f(x) = f(1-x) for all x?

a)f(x) = 1 â€“x
b)f(x) = 1 â€“ x^2
c)f(x) = x^2 â€“ (1-x)^2
d)f(x) = x^2(1 â€“ x)^2
e)f(x) = x / (1-x)
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10 Apr 2007, 20:23
Quote:
2)An equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

a)5
b)8
c)11
d)15
e)19

An equitlateral triangle will means the arcs are equal in length. Arc ABC is 2/3rds of the circle and equals 24. So the whole circle is 24 +12, or 36.

2pi(r)=36 - and to get the diameter, divide by pi

which will give you 36/3.14 which is about 11
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10 Apr 2007, 21:29
Quote:
ECA

Why don't you explain #1 and #3?
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11 Apr 2007, 00:17
2)An equilateral triangle ABC is inscribed in the circle. If the length of arc ABC is 24, what is the approximate diameter of the circle?

a)5
b)8
c)11
d)15
e)19

Now its given that a tirangle ABC is an equilateral trianlgle i.e the measure of all its angle is equal and equal to 60 degrees. lets say O is the center of the cirlce. Now measure angle ABC is 60. So measure angle AOC is twice the measure angle ABC that is equal to 120. So arc ABC is 240. given all that we can calculate the radius and thus diameter of the circle. Here's how we do it. Lets assume #= 3.14

(240/360)*2#r = 24 (given in the statement)
therefore r= 18/3.14 Lets say 18/3 = 12.
So the answere should be closer to 12 but it should be less than 12. The onle answere choice that satisties this condition is answer choice C. So the answer is C.

Javed

Cheers!
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11 Apr 2007, 00:19
3)For which of the following functions f is f(x) = f(1-x) for all x?

a)f(x) = 1 â€“x
b)f(x) = 1 â€“ x^2
c)f(x) = x^2 â€“ (1-x)^2
d)f(x) = x^2(1 â€“ x)^2
e)f(x) = x / (1-x)[/quote]

Simply put the value of x=1-x in all of the above equations. The only equation yeilds the same answere is equation D). so answer is D)
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11 Apr 2007, 05:42
Let's try out a few values.

h(4) = 2*4 --> 2 * 2 * 2 = 2^2(1*2)
h(6) = 2*4*6 --> 2 * 2 * 2 * 2 * 3 = 2^3(1*2*3)
h(8) = 2*4*6*8 --> 2 * 2 * 2 * 2 * 3 * 2^3 = 2^4 (1*2*3*4)
h(10) --> we can almost predictably write 2^5(1*2*3*4*5)

So if n = 10, then the general form would be 2^(n/2)[(n/2)!]

So h(100) = 2^50(50!) ----- [1]

Now, let's go back to the examples.
If we had h(4) + 1, we would get 9 --> largest prime is 3, smallest prime is 3 as well
If we had h(6) + 1, we would get 49 --> largest prime is 7, smallest prime is 7 as well
If we had h(8) + 1, we would get 385 --> largest prime is 11, smallest prime is 5

You get the picture now:
The biggest possible prime of h(n)+1 is always going to be bigger than the biggest prime of h(n).

So back to h(100) = 2^50(50!). The largest prime of this function is 47. We know h(100)+1 must have a larger prime than 47 so the answer is e.

Note:
I know you'll be asking, why are we testing the largest possible prime. This is because when we are asked for the smallest possbile value of anything, we want to have a buffer. So if you can pass the largest value, surly your smallest value will be in the range. I can get you an example if you're not clear on this note.
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11 Apr 2007, 07:11
javed wrote:
3)For which of the following functions f is f(x) = f(1-x) for all x?

a)f(x) = 1 â€“x
b)f(x) = 1 â€“ x^2
c)f(x) = x^2 â€“ (1-x)^2
d)f(x) = x^2(1 â€“ x)^2
e)f(x) = x / (1-x)

Simply put the value of x=1-x in all of the above equations. The only equation yeilds the same answere is equation D). so answer is D)[/quote]

Agree. Some how messed it up when I worked the problem first time. thanks!
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11 Apr 2007, 07:21
ywilfred wrote:
Let's try out a few values.

h(4) = 2*4 --> 2 * 2 * 2 = 2^2(1*2)
h(6) = 2*4*6 --> 2 * 2 * 2 * 2 * 3 = 2^3(1*2*3)
h(8) = 2*4*6*8 --> 2 * 2 * 2 * 2 * 3 * 2^3 = 2^4 (1*2*3*4)
h(10) --> we can almost predictably write 2^5(1*2*3*4*5)

So if n = 10, then the general form would be 2^(n/2)[(n/2)!]

So h(100) = 2^50(50!) ----- [1]

Now, let's go back to the examples.
If we had h(4) + 1, we would get 9 --> largest prime is 3, smallest prime is 3 as well
If we had h(6) + 1, we would get 49 --> largest prime is 7, smallest prime is 7 as well
If we had h(8) + 1, we would get 385 --> largest prime is 11, smallest prime is 5

You get the picture now:
The biggest possible prime of h(n)+1 is always going to be bigger than the biggest prime of h(n).

So back to h(100) = 2^50(50!). The largest prime of this function is 47. We know h(100)+1 must have a larger prime than 47 so the answer is e.

Note:
I know you'll be asking, why are we testing the largest possible prime. This is because when we are asked for the smallest possbile value of anything, we want to have a buffer. So if you can pass the largest value, surly your smallest value will be in the range. I can get you an example if you're not clear on this note.

Very well explained !!
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11 Apr 2007, 07:23
thanks!
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14 Apr 2007, 01:04
ywilfred wrote:
Let's try out a few values.

h(4) = 2*4 --> 2 * 2 * 2 = 2^2(1*2)
h(6) = 2*4*6 --> 2 * 2 * 2 * 2 * 3 = 2^3(1*2*3)
h(8) = 2*4*6*8 --> 2 * 2 * 2 * 2 * 3 * 2^3 = 2^4 (1*2*3*4)
h(10) --> we can almost predictably write 2^5(1*2*3*4*5)

So if n = 10, then the general form would be 2^(n/2)[(n/2)!]

So h(100) = 2^50(50!) ----- [1]

Now, let's go back to the examples.
If we had h(4) + 1, we would get 9 --> largest prime is 3, smallest prime is 3 as well
If we had h(6) + 1, we would get 49 --> largest prime is 7, smallest prime is 7 as well
If we had h(8) + 1, we would get 385 --> largest prime is 11, smallest prime is 5

You get the picture now:
The biggest possible prime of h(n)+1 is always going to be bigger than the biggest prime of h(n).

So back to h(100) = 2^50(50!). The largest prime of this function is 47. We know h(100)+1 must have a larger prime than 47 so the answer is e.

Note:
I know you'll be asking, why are we testing the largest possible prime. This is because when we are asked for the smallest possbile value of anything, we want to have a buffer. So if you can pass the largest value, surly your smallest value will be in the range. I can get you an example if you're not clear on this note.

Excellent

Javed.

Cheers!
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23 Apr 2007, 13:45
ywilfred wrote:
Let's try out a few values.

h(4) = 2*4 --> 2 * 2 * 2 = 2^2(1*2)
h(6) = 2*4*6 --> 2 * 2 * 2 * 2 * 3 = 2^3(1*2*3)
h(8) = 2*4*6*8 --> 2 * 2 * 2 * 2 * 3 * 2^3 = 2^4 (1*2*3*4)
h(10) --> we can almost predictably write 2^5(1*2*3*4*5)

So if n = 10, then the general form would be 2^(n/2)[(n/2)!]

So h(100) = 2^50(50!) ----- [1]

Now, let's go back to the examples.
If we had h(4) + 1, we would get 9 --> largest prime is 3, smallest prime is 3 as well
If we had h(6) + 1, we would get 49 --> largest prime is 7, smallest prime is 7 as well
If we had h(8) + 1, we would get 385 --> largest prime is 11, smallest prime is 5

You get the picture now:
The biggest possible prime of h(n)+1 is always going to be bigger than the biggest prime of h(n).

So back to h(100) = 2^50(50!). The largest prime of this function is 47. We know h(100)+1 must have a larger prime than 47 so the answer is e.

Note:
I know you'll be asking, why are we testing the largest possible prime. This is because when we are asked for the smallest possbile value of anything, we want to have a buffer. So if you can pass the largest value, surly your smallest value will be in the range. I can get you an example if you're not clear on this note.

Wilfred - thanks for the solution. I was a little confused by your answer, but it got me going on the right track. Here is how I would think about it (same answer, just a slightly different explanation if you were confused by Wilfred's...)

Instead of thinking of h(n) as a product of a series of even numbers, think of it as a product of a series of PRIME numbers. In other words:

h(12) = 12*10*8*6*4*2 = (2*2*3)*(2*5)*(2*2*2)*(2*3)*(2*2)*(2)

Now, when you think of h(100), you could expand it as I have done above (this would be a very long number!!). Instead of fully expanding it, think about the prime numbers it would include. You know it would include 2, 3, and 5 from the expansion of h(12) above, but it would also include larger primes. So, h(100) would look something like this:

h(100) = 2*4*6*...*10...*14...*33..........*94....*100

h(100) = (2)*(2*2)*(2*3)*...*(2*5)...*(2*7)...*(3*11)............*(2*47)....*(2*2*5*5)

Note that the largest prime number in h(100) above is 47.

Now that we know the string of prime numbers in h(100), we can think about what divides h(100) + 1. Since all of the prime factors of h(100) > 1 (as are ALL primes), we know that when we add 1 to h(100), it is NO LONGER DIVISIBLE by ANY of its prime factors (dividing by these numbers would always give a remainder of 1). This means it is no longer divisible by 2, 3, 5, 7, .... 47.

We do not know if h(100) + 1 is divisible by any other number (it may be prime for all we know), but we DO know that all the primes 2, 3, ... 47 do NOT divide this number, so the least possible prime factor has to be greater than 47, and the answer is E.
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23 Apr 2007, 17:17
weird..I get A..can you explain how it is D??

goalsnr wrote:
javed wrote:
3)For which of the following functions f is f(x) = f(1-x) for all x?

a)f(x) = 1 тАУx
b)f(x) = 1 тАУ x^2
c)f(x) = x^2 тАУ (1-x)^2
d)f(x) = x^2(1 тАУ x)^2
e)f(x) = x / (1-x)

Simply put the value of x=1-x in all of the above equations. The only equation yeilds the same answere is equation D). so answer is D)

Agree. Some how messed it up when I worked the problem first time. thanks![/quote]
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23 Apr 2007, 19:53
fresinha12 wrote:
weird..I get A..can you explain how it is D??

goalsnr wrote:
javed wrote:
3)For which of the following functions f is f(x) = f(1-x) for all x?

a)f(x) = 1 тАУx
b)f(x) = 1 тАУ x^2
c)f(x) = x^2 тАУ (1-x)^2
d)f(x) = x^2(1 тАУ x)^2
e)f(x) = x / (1-x)

Simply put the value of x=1-x in all of the above equations. The only equation yeilds the same answere is equation D). so answer is D)

Agree. Some how messed it up when I worked the problem first time. thanks!
[/quote]

Substiture x=1-x in f(x)
1) f(1-x)=1-(1-x)=x not= f(x)
2) f(1-x)=1-(1-x)^2=2x-x^2 not=f(x)
3) f(1-x)=(1-x)^2-[1-(1-x)]^2 = 1-2x not=f(x)4) f(1-x)=(1-x)^2[1-(1-x)]^2=[(1-x)^2]x^2 = f(x)
5) f(1-x)= (1-x)/[1-(1-x)]=(1-x)/x not= f(x).I hope this helps.

Javed.

Cheers!
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24 Apr 2007, 18:31
ywilfred,

I admit that I do not understand what you mean about needing a buffer with the largest value of something. Can you provide me with the example you mention in your explanation post?

Thanks!

Bluebird
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24 Apr 2007, 18:54
ywilfred wrote:
Let's try out a few values.

h(4) = 2*4 --> 2 * 2 * 2 = 2^2(1*2)
h(6) = 2*4*6 --> 2 * 2 * 2 * 2 * 3 = 2^3(1*2*3)
h(8) = 2*4*6*8 --> 2 * 2 * 2 * 2 * 3 * 2^3 = 2^4 (1*2*3*4)
h(10) --> we can almost predictably write 2^5(1*2*3*4*5)

So if n = 10, then the general form would be 2^(n/2)[(n/2)!]

So h(100) = 2^50(50!) ----- [1]

Now, let's go back to the examples.
If we had h(4) + 1, we would get 9 --> largest prime is 3, smallest prime is 3 as well
If we had h(6) + 1, we would get 49 --> largest prime is 7, smallest prime is 7 as well
If we had h(8) + 1, we would get 385 --> largest prime is 11, smallest prime is 5

You get the picture now:
The biggest possible prime of h(n)+1 is always going to be bigger than the biggest prime of h(n).

So back to h(100) = 2^50(50!). The largest prime of this function is 47. We know h(100)+1 must have a larger prime than 47 so the answer is e.

Note:
I know you'll be asking, why are we testing the largest possible prime. This is because when we are asked for the smallest possbile value of anything, we want to have a buffer. So if you can pass the largest value, surly your smallest value will be in the range. I can get you an example if you're not clear on this note.

Wilfred,

Pls get me an example cos I am a bit confused !
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24 Apr 2007, 19:00
Bluebird wrote:
ywilfred,

I admit that I do not understand what you mean about needing a buffer with the largest value of something. Can you provide me with the example you mention in your explanation post?

Thanks!

Bluebird

hi, sorry for the late reply...

Let's say you want to buy an apple from the store down your road but you don't know how much that apple costs. You recall that the highest price ever quoted was a hundred bucks for an apple elsewhere (I'm exaggerating the price of an apple here). So in order to get an apple within one trip, all you have to do is make sure you have at least a hundred bucks with you. That way, you cover all range of prices from lowest possible to highest possible... That's what I meant by a buffer - giving yourself room to maneuver
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24 Apr 2007, 19:04
now back to the question. we know the highest possible prime of h(100) = 47. h(100)+1 is going to have its largest prime set at greater than 47. We don't compare smallest prime because the answer we arrive at might come up short and that leaves us no room to manuever.
24 Apr 2007, 19:04
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