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My Wrong Ones...

by goyalsau » Thu Oct 21, 2010 5:41 am
How many real n's are there such that n! is a perfect square?

(E) More than 3
(C) 2
(D) 3
(B) 1
(A) 0


The distance of any point P(x,y) from the origin is defined as d(x,y) = max{|x|,{|y|}. If d(x,y) =3 then all such possible point P will lie on the circumference of

(E) a hexagon
(A) an equilateral triangle
(D) a right traingle
(B) a square
(C) a circle

The line AB of length 6 m is a tangent to the inner one of the two concentric circles at point C and a chord to the outer circle. Find the radius of the outer circle if both the radii have integer values.

(B) 4 m
(C) 6 m
(A) 5 m
(D) 3 m
(E) 7 m

If the radius of a sphere is increased by 2 cm, its surface area increases by 352 cm2 . The radius of sphere before the change is

(A) 3 cm
(C) 6 cm
(D) 7 cm
(B) 4 cm
(E) 5 cm

In a box containing 15 apples, exactly 6 apples are rotten. Each day one apple is taken out from the box. What is the probability that after four days there are exactly 8 apples in the box that are not rotten?

(C) 2/13
(A) 12/91
(B) 1/7
(D) 2/7
(E) None of these
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by shovan85 » Thu Oct 21, 2010 5:54 am
goyalsau wrote:How many real n's are there such that n! is a perfect square?

(E) More than 3
(C) 2
(D) 3
(B) 1
(A) 0
Why your options are wrongly numbered ;)?

IMO [spoiler]C (2)[/spoiler]

I am not sure but if you say n! = 1*2*....*(n-1)*n. Thus, We will never have at least one common factor between n-1 and n (co-prime it is called I guess). When you multiply these two in the factorial sequence i believe we will definitely have an extra prime integer which will not let the n! as a perfect square.

Only 1 satisfies. 1! = 1 and 1^2 = 1.
Sorry I missed 0!

Let me know if I m correct.
Last edited by shovan85 on Thu Oct 21, 2010 6:12 am, edited 1 time in total.
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by shovan85 » Thu Oct 21, 2010 6:03 am
goyalsau wrote:
The distance of any point P(x,y) from the origin is defined as d(x,y) = max{|x|,{|y|}. If d(x,y) =3 then all such possible point P will lie on the circumference of

(E) a hexagon
(A) an equilateral triangle
(D) a right traingle
(B) a square
(C) a circle
IMO B
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by goyalsau » Thu Oct 21, 2010 6:04 am
shovan85 wrote: Why your options are wrongly numbered ;)?
I am sorry for that, I was so upset after the test , I just didn't notice that . It may be because of software . I don't know . I hope it does not make much a difference

Well Even i Thought that way and answered 1, But its 2.
I doubt , It has to do something with 0. I am not sure about 0 so may be.
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by goyalsau » Thu Oct 21, 2010 6:05 am
shovan85 wrote:
goyalsau wrote:
The distance of any point P(x,y) from the origin is defined as d(x,y) = max{|x|,{|y|}. If d(x,y) =3 then all such possible point P will lie on the circumference of

(E) a hexagon
(A) an equilateral triangle
(D) a right traingle
(B) a square
(C) a circle
IMO B
Buddy Explain .
I already have the official answers .
So please try to explain your reasoning as well.
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by msbinu » Thu Oct 21, 2010 6:05 am
isn't the answer 2??

0! = 1 (which is a perfect square )
1! = 1 which is also a perfect square

so total 2


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by shovan85 » Thu Oct 21, 2010 6:13 am
msbinu wrote:isn't the answer 2??

0! = 1 (which is a perfect square )
1! = 1 which is also a perfect square

so total 2


Regards,
Binu
True Boss.... I missed that as I was looking foe a reason why it happens. Thanks for updating :)
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by novel » Thu Oct 21, 2010 6:15 am
shovan85 wrote:
goyalsau wrote:
The distance of any point P(x,y) from the origin is defined as d(x,y) = max{|x|,{|y|}. If d(x,y) =3 then all such possible point P will lie on the circumference of

(E) a hexagon
(A) an equilateral triangle
(D) a right traingle
(B) a square
(C) a circle
IMO B
Please explain.I thought the ans is C
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by shovan85 » Thu Oct 21, 2010 6:17 am
goyalsau wrote:
shovan85 wrote:
goyalsau wrote:
The distance of any point P(x,y) from the origin is defined as d(x,y) = max{|x|,{|y|}. If d(x,y) =3 then all such possible point P will lie on the circumference of

(E) a hexagon
(A) an equilateral triangle
(D) a right traingle
(B) a square
(C) a circle
IMO B
Buddy Explain .
I already have the official answers .
So please try to explain your reasoning as well.
Sure man but be patient. I always give reasoning to my answer. I solved it but it requires a diagram so took time ;)


See the picture below. Consider only Quadrant I.

When u see the line x = 3 that means all the points on line will have X co-ordinate as 3 and y is LESS THAN 3. But after point B the Y will be greater than X.
Now consider from B to C. all y = 3 and x < 3.

As we are taking the MAX of absolute values they will not matter in the Quadrants. Let me know if I m correct :)
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by fskilnik@GMATH » Thu Oct 21, 2010 6:27 am
shovan85 wrote:
msbinu wrote:isn't the answer 2??

0! = 1 (which is a perfect square )
1! = 1 which is also a perfect square

so total 2


Regards,
Binu
True Boss.... I missed that as I was looking foe a reason why it happens. Thanks for updating :)
Ok... we have seen that when n=0 and n=1 we have two solutions. But why there are no others for n at least 2? It seems to me that the "bullet-proof" answer depends on higher-Math...
Bertrand´s Postulate wrote: There is always a prime between x and 2x, where x is any positive integer greater than 1.
Let us assume this is true (sure...) and use it for our purposes...

Let p be the greatest prime present at the prime decomposition of n! (n is at least 2, do not forget), then I state that this prime appears just once, and that is enough because if we had n! a perfect square, every prime factor of n! should appear an even number of times!

Proof of my statement:

Let n! be equal to p. (other primes less than p with their corresponding exponents, please note that "2" is there) ; if n! had a second p factor, that means that n! would have the prime number that is between p and 2p (Bertrand´s Postulate), meaning that p was not the greatest prime number present at n! decomposition, what is absurd. Done!

Hope you like it!

Regards,
Fabio.
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by shovan85 » Thu Oct 21, 2010 6:44 am
goyalsau wrote: In a box containing 15 apples, exactly 6 apples are rotten. Each day one apple is taken out from the box. What is the probability that after four days there are exactly 8 apples in the box that are not rotten?

(C) 2/13
(A) 12/91
(B) 1/7
(D) 2/7
(E) None of these
End of the 4th day we should have 8 not rotten apple.
So in the 4 days only one NOT ROTTEN apple has to be taken.

Let us denote Rotten Apple as R and Not Rotten as N

_ _ _ _ let be the 4 days

Thus total possibilities are

R R R N
R R N R
R N R R
N R R R

thus answer will be = 4 * (6/15 * 5/14 * 4/13 * 9/12) = 12/91

IMO A

If you have any doubts please go through the below post I have just explained my way of approach to a such kind of problem. Still issue let me know :)

https://www.beatthegmat.com/probabilty-q ... tml#310592
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by shovan85 » Thu Oct 21, 2010 6:50 am
fskilnik wrote: Ok... we have seen that when n=0 and n=1 we have two solutions. But why there are no others for n at least 2? It seems to me that the "bullet-proof" answer depends on higher-Math...
Bertrand´s Postulate wrote: There is always a prime between x and 2x, where x is any positive integer greater than 1.
Let us assume this is true (sure...) and use it for our purposes...

Let p be the greatest prime present at the prime decomposition of n! (n is at least 2, do not forget), then I state that this prime appears just once, and that is enough because if we had n! a perfect square, every prime factor of n! should appear an even number of times!
Excellent!! I realized the same while taking my chance up to 11! but I did not know such postulates are available for this.

Thanks a lot.

What about my Co-prime funda?
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by fskilnik@GMATH » Thu Oct 21, 2010 7:21 am
shovan85 wrote:We will never have at least one common factor between n-1 and n (co-prime it is called I guess).
Thanks for asking me if this statement of yours is correct, Shovan.

(1) Two integers are said coprimes (or, more commonly used, relatively prime) if their greater common divisor is 1, that is, if their only common positive factor is 1.

(2) Yes, two consecutive integers are always relatively prime, and the reason is simple:

> 2 divides just one of them (sure...)
> each prime greater than 2 divides at most one of them (I am sure you will guess why...)

Hint: [spoiler] in m consecutive integers (m at least 2, sure), exactly one of them is divisible by m, therefore...[/spoiler]

> if x is an integer (greater than 1) that is not prime but divides two consecutive integers, then it has a prime in its prime decomposition that does the same, so apply the former to verify that you must have been drunk to believe that... ;)


Hope you like it all.

Regards,
Fabio.

P.S.: I don´t know if we should write co-primes or coprimes.
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by shovan85 » Thu Oct 21, 2010 7:25 am
goyalsau wrote:
The line AB of length 6 m is a tangent to the inner one of the two concentric circles at point C and a chord to the outer circle. Find the radius of the outer circle if both the radii have integer values.

(B) 4 m
(C) 6 m
(A) 5 m
(D) 3 m
(E) 7 m
See the diagram below. Let OC be r and OB be R.

Pythagoras Theo. r^2 + 3^2 = R^2

Now all should be integers (r and R) so we need to see the triplet that satisfy this.
Some example of triplets are 3-4-5, 6-8-10, 5-12-13

Now we have 3 and R>r so from the triplet R = 5cm

IMO A 5cm
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by shovan85 » Thu Oct 21, 2010 7:28 am
fskilnik wrote:
shovan85 wrote:We will never have at least one common factor between n-1 and n (co-prime it is called I guess).
Thanks for asking me if this statement of yours is correct, Shovan.

(1) Two integers are said coprimes (or, more commonly used, relatively prime) if their greater common divisor is 1, that is, if their only common positive factor is 1.

(2) Yes, two consecutive integers are always relatively prime, and the reason is simple:

> 2 divides just one of them (sure...)
> each prime greater than 2 divides at most one of them (I am sure you will guess why...)
Yes Very true. Thanks :)
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