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New Set: Number Properties!!!

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Re: New Set: Number Properties!!!  [#permalink]

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

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Re: New Set: Number Properties!!!  [#permalink]

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New post 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.

Answer: C.
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Re: New Set: Number Properties!!!  [#permalink]

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New post 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.

Answer: C.


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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New Set: Number Properties!!!  [#permalink]

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New post 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.


Answer: D[/quote]

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
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Re: New Set: Number Properties!!!  [#permalink]

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New post 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.


Answer: D


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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