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If n = 1 + x, where x is the product of four consecutive [#permalink]
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04 Jun 2005, 08:45
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If n = 1 + x, where x is the product of four consecutive natural numbers, then which of the following is /are true?
(I) n is odd
(II) n and x are coprime
(III) n is a perfect square
(IV) n is not divisible by 3

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I would say I & III is always true...
N cannot be a prime..., say X1...X4 are the natural numbers, then if X1 is 0, N=1, 1 is not a prime number!
in the above case 1/3 is not perfectly divisible by 3 but if you picked numbers you see

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II.
n and x are consecutive integers.

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Dan wrote: II.
n and x are consecutive integers.
What do you mean? are you trying to explain what "coprime" means?

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coprimes are like 6 and 7, or 120 and 121, where the greatest common factor is 1.
but since in n = 1 + x, n and x are consecutive integers, then they are coprime.
For the other choices:
I n could be odd (1*2*3*4) or even (2*3*4*5)
II True
III 1+(1*2*3*4) = 25 perfect square; or 1+(2*3*4*5) = 120 which is not
IV 25 not div by 3; 120 it is.

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Dan wrote: I n could be odd (1*2*3*4) or even (2*3*4*5) II True III 1+(1*2*3*4) = 25 perfect square; or 1+(2*3*4*5) = 120 which is not IV 25 not div by 3; 120 it is.
1+2*3*4*5=121 which makes III and IV true (in your reasoning)
For I, n couldn't be odd since at least 2 out of 4 consecutive integers are even, and multiplying whatever number by an even number, you get an even number, and adding 1 to an even number makes it always odd.

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Re: Uhm...what's a coprime? [#permalink]
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04 Jun 2005, 11:43
thearch wrote: If n = 1 + x, where x is the product of four consecutive natural numbers, then which of the following is /are true?
(I) n is odd (II) n and x are coprime (III) n is a perfect square (IV) n is not divisible by 3
(1) true  x is always even , hence n is always odd
(2) true  n and x are consecutive integers,hence are coprime
(3) true  x is always divisible by 24 ,and any number divisible by 24 is of the form p^2  1 ( where p is a prime >= 5) , so since x is of the form p^2  1 , n = p ^ 2 ( which is a sqaure , and is add , since p is prime)
(4) true = since n is of the for p^2 where p is prime, it cannot be divisible by 3.
HMTG.

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apparently 0 could be considered a natural number

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Re: Uhm...what's a coprime? [#permalink]
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04 Jun 2005, 14:29
thearch wrote: If n = 1 + x, where x is the product of four consecutive natural numbers, then which of the following is /are true?
(I) n is odd > true even +1 is odd
(II) n and x are coprime > true greatest common divisor is 1
(III) n is a perfect square > true n = 1 +(m2)(m1)m(m+1) =
= (m^2m1)^2
(IV) n is not divisible by 3 > true since x is divisible by 3 always

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Re: Uhm...what's a coprime? [#permalink]
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04 Jun 2005, 14:40
thearch wrote: If n = 1 + x, where x is the product of four consecutive natural numbers, then which of the following is /are true?
(I) n is odd (II) n and x are coprime (III) n is a perfect square (IV) n is not divisible by 3
I think I, II and III have to be true

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Dan wrote: coprimes are like 6 and 7, or 120 and 121, where the greatest common factor is 1.
but since in n = 1 + x, n and x are consecutive integers, then they are coprime.
For the other choices:
I n could be odd (1*2*3*4) or even (2*3*4*5) II True III 1+(1*2*3*4) = 25 perfect square; or 1+(2*3*4*5) = 120 which is not IV 25 not div by 3; 120 it is.
Are 6 and 7 coprime?? I would say NO
Lets take 21 as the number 3 and 7 are co primes because there are no other factors for 21 other than 3 and 7(excluding the number itself and 1) which are prime and factors of the number 21......
I would say I,III are the right answers

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Dan wrote: coprimes are like 6 and 7, or 120 and 121, where the greatest common factor is 1.
but since in n = 1 + x, n and x are consecutive integers, then they are coprime.
For the other choices:
I n could be odd (1*2*3*4) or even (2*3*4*5) II True III 1+(1*2*3*4) = 25 perfect square; or 1+(2*3*4*5) = 120 which is not IV 25 not div by 3; 120 it is.
1+(2*3*4*5) = 121 = 11*11 not 120

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gmat2me2 wrote: Are 6 and 7 coprime?? I would say NO
why not? their greatest common divisor is 1, therefore they are coprimes.

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sparky wrote: gmat2me2 wrote: Are 6 and 7 coprime?? I would say NO
why not? their greatest common divisor is 1, therefore they are coprimes.
Ooops my bad....Yes they are co prime indeed.......Their GCD is 1.....

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Re: Uhm...what's a coprime? [#permalink]
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04 Jun 2005, 21:52
sparky wrote: (I) n is odd > true even +1 is odd (II) n and x are coprime > true greatest common divisor is 1 (III) n is a perfect square > true n = 1 +(m2)(m1)m(m+1) = = (m^2m1)^2 (IV) n is not divisible by 3 > true since x is divisible by 3 always
:b: This is how one should approach the problem.
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good show guys!!
OA is:
true
true
true
true

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