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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.

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.

Re: Uhm...what's a co-prime? [#permalink]
04 Jun 2005, 10:43

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 co-prime (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 co-prime
(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.

Re: Uhm...what's a co-prime? [#permalink]
04 Jun 2005, 13:29

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 co-prime -> true greatest common divisor is 1
(III) n is a perfect square -> true n = 1 +(m-2)(m-1)m(m+1) =
= (m^2-m-1)^2
(IV) n is not divisible by 3 -> true since x is divisible by 3 always

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 co-prime?? 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......

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.

Re: Uhm...what's a co-prime? [#permalink]
04 Jun 2005, 20:52

sparky wrote:

(I) n is odd -> true even +1 is odd (II) n and x are co-prime -> true greatest common divisor is 1 (III) n is a perfect square -> true n = 1 +(m-2)(m-1)m(m+1) = = (m^2-m-1)^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. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
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