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# New Set: Number Properties!!!

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Intern
Joined: 22 Dec 2018
Posts: 2
Re: New Set: Number Properties!!!  [#permalink]

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18 Jul 2019, 07:49
Oh I can see the mistake now. Thanks.

Posted from my mobile device
Intern
Joined: 16 Sep 2019
Posts: 1
Re: New Set: Number Properties!!!  [#permalink]

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27 Sep 2019, 00:13
Hi Bunuel,

Why Can't a be 1 in this case? This will imply that they are not consecutive integers.

There is no condition defined for a in these 2 statements.

The median of 1, 1 and 2 is also 1.

Bunuel wrote:
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

Math Expert
Joined: 02 Sep 2009
Posts: 59182
Re: New Set: Number Properties!!!  [#permalink]

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27 Sep 2019, 01:04
1
saisrp wrote:
Hi Bunuel,

Why Can't a be 1 in this case? This will imply that they are not consecutive integers.

There is no condition defined for a in these 2 statements.

The median of 1, 1 and 2 is also 1.

Bunuel wrote:
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

From (1) we goth that b = 1, and if a = 1, then the condition given in the stem which says that a < b < c won't be satisfied.

Hope it's clear.
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Manager
Joined: 05 Oct 2014
Posts: 106
Location: India
Concentration: General Management, Strategy
GMAT 1: 580 Q41 V28
GPA: 3.8
WE: Project Management (Energy and Utilities)

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18 Oct 2019, 05:28
If $$0 \lt x \lt y$$ and $$x$$ and $$y$$ are consecutive perfect squares, what is the remainder when $$y$$ is divided by $$x$$?

Notice that since $$x$$ and $$y$$ are consecutive perfect squares, then $$\sqrt{x}$$ and $$\sqrt{y}$$ are consecutive integers.

(1) Both $$x$$ and $$y$$ have 3 positive factors. This statement implies that $$x=(prime_1)^2$$ and $$y=(prime_2)^2$$. From above we have that $$\sqrt{x}=prime_1$$ and $$\sqrt{y}=prime_2$$ are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, $$x=(prime_1)^2=4$$ and $$y=(prime_2)^2=9$$. The remainder when 9 is divided by 4 is 1. Sufficient.

(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers. The same here: $$\sqrt{x}=2$$ and $$\sqrt{y}=3$$. Sufficient.

Bunuel I could not understand this. (1) states, both are having 3 positive factors. The consecutive prime numbers which are 2 & 3, both are having 2factors only, i.e 1 and the number itself. Moreover, Prime Numbers does not have 3 factors. Pls tell, where am I getting wrong
Math Expert
Joined: 02 Sep 2009
Posts: 59182
Re: New Set: Number Properties!!!  [#permalink]

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18 Oct 2019, 05:49
merajul wrote:
If $$0 \lt x \lt y$$ and $$x$$ and $$y$$ are consecutive perfect squares, what is the remainder when $$y$$ is divided by $$x$$?

Notice that since $$x$$ and $$y$$ are consecutive perfect squares, then $$\sqrt{x}$$ and $$\sqrt{y}$$ are consecutive integers.

(1) Both $$x$$ and $$y$$ have 3 positive factors. This statement implies that $$x=(prime_1)^2$$ and $$y=(prime_2)^2$$. From above we have that $$\sqrt{x}=prime_1$$ and $$\sqrt{y}=prime_2$$ are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, $$x=(prime_1)^2=4$$ and $$y=(prime_2)^2=9$$. The remainder when 9 is divided by 4 is 1. Sufficient.

(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers. The same here: $$\sqrt{x}=2$$ and $$\sqrt{y}=3$$. Sufficient.

Bunuel I could not understand this. (1) states, both are having 3 positive factors. The consecutive prime numbers which are 2 & 3, both are having 2factors only, i.e 1 and the number itself. Moreover, Prime Numbers does not have 3 factors. Pls tell, where am I getting wrong[/quote]

Primes have 2 factors, yes, but the square of primes have 3 factors. So, from (1) $$x=(prime_1)^2$$ and $$y=(prime_2)^2$$.
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Re: New Set: Number Properties!!!   [#permalink] 18 Oct 2019, 05:49

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