Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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.

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.

(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;
keep on knocking, and it will be opened to you.