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

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2
3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

(1) Both x and y is have 3 positive factors.
(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers

(1) Both x and y is have 3 positive factors.

Consecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 ....
Among these numbers, only 4 and 9 are consecutive perfect squares that have 3 positive factors ( for instance 16 = 4*4 = 2*2*2*2 --> 5 factors and SO ON )
Hence, y=9 and x=4 --> 9 = 4.2 +1 --> R = 1

Thus, (1) SUFFICIENT

(2) Both $$\sqrt{x}$$ and $$\sqrt{y}$$ are prime numbers
Consecutive perfect squares could be : 4 , 9 , 16 , 25 , 36 , 49 , 64 ....
Among these numbers, only 4 and 9 are consecutive perfect squares that have their square roots as prime numbers ( for example : $$\sqrt{16}$$ = 4, which is not a prime number and SO ON )
Hence, y=9 and x=4 --> 9 = 4.2 --> R = 1

Thus, (2) SUFFICIENT

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Originally posted by Rock750 on 26 Mar 2013, 03:57.
Last edited by Rock750 on 29 Mar 2013, 04:20, edited 1 time in total.
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Re: New Set: Number Properties!!!  [#permalink]

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1
4. Each digit of the three-digit integer N is a multiple of 4, what is the value of K?

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K
(2) K is NOT a multiple of 3.

Given that each digit of the three-digit integer N is a multiple of 4 , K could be : 444 or 448 or 484 or 488 ...

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K

So , K should b equal to 444 (LCM(4,4) = 4) OR equalt to 888 (LCM(8,8) = 8) OR equalt to 488 (LCM(4,8) = 8) ....
Hence, (1) NOT SUFFICIENT

(2) K is NOT a multiple of 3.

So, K could be equal to 448 or 484 ...
Hence, (2) NOT SUFFICIENT

(1) + (2)

Both 488 and 848 have their units digit as the LCM of the tens and hundreds and are not a mulitple of 3
Hence, (1) + (2) NOT SUFFICIENT

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

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

(1) The median of {a!, b!, c!} is an odd number.
(2) c! is a prime number

(1) The median of {a!, b!, c!} is an odd number.

The median of {a!, b!, c!} is an odd number --> b! is the median --> b! is odd --> b can take two values : 0 or 1

If b = 0 and b>a then a is a negative integer --> impossible
Hence b must be equal to 1 , and because b>a and The median of {a!, b!, c!} is an odd number., then a = 0 and c = 2 because the median must be equal to b! wich is 1.

(1) SUFFICIENT

(2) c! is a prime number

In this case, c can be equal to 0, 1 or 2

(2) INSUFFICIENT

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

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6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative
(2) The product of the smallest and largest integers in the list is a prime number.

(1) The product of any three integers in the list is negative
Considering S with 3 integers : the set can be constitued with + + - OR - - - , So this statement is INSUFFICIENT

(2) The product of the smallest and largest integers in the list is a prime number.

By the usual definition of prime for integers, negative integers can not be prime.
So, for the product of the smallest and largest integers in the list to be a prime number, the largest and the smallest both must be either positive or negative.
For example : Set = {13, 8 , 6 , 1} : the product of the largest and smallest = 13*1 = 13prime
And Set {-1, -5 -9 , -13} : the product of the largest and smallest = -13 * -1 = 13 prime
So, we cannot conclude if all the numbers in set S are negative or not , this statement alone is INSUFFICIENT

(1) + (2) --> Given by the second statement that ALL THE NUMBERS IN SET S MUST HAVE THE SAME SIGN, and given by the first statement that ANY THREE INTEGERS in the list is negative --> THE ANSWER IS YES : ALL THE NUMBERS IN SET S ARE NEGATIVE

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

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7. Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

(1) When x is divided by 12 the remainder is 6

when x is divided by 12 the remainder is 6 --> x = 12q + 6 = 3*2*(2q+1) ; So the answer is NO, x cannot be the square of any integer

For example :
q = 1 : x = 18
q=2 : x = 30 ...

(1) SUFFICIENT

(2) When x is divided by 14 the remainder is 2

When x is divided by 14 the remainder is 2 --> x = 14p + 2 = 2*(7p+1)

For example :
p = 1 --> x = 16 = 4^2
p = 2 --> x = 30

So, we dont know --> (2) NOT SUFFICIENT

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

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8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number
(2) The product of any two terms of the set is a terminating decimal

Set A consist of 10 terms A = { x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 , x10 } where xi >_ xi-1 ( i = 1.2....10 ) and xi = 1/Pj Pj: prime number

Median A = (x5 + x6) / 2 < 1/5 ??

(1) Reciprocal of the median is a prime number

we don't know anything about x5 and x6 so this statement is INSUFFICIENT

(2) The product of any two terms of the set is a terminating decimal

Given that statement --> the set A is composed only by 1/2 or 1/5 or both many times --> the mean will always be greater than or equalt to 1/5 but never less than 1/5 --> the answer to the question is NO --> this statement ALONE is SUFFICIENT

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

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1. If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2)2\sqrt{x2}is a prime number

Hellooz,

St 1, we have |23x| is a prime no.
Possible values of x are 1 and -1 and hence not sufficient

St 2 Now 2\sqrt{x2} is equal 2 |x| is equal 2 x.
Since in GMAT even root is considered postive only therefore we have 2\sqrt{x2} = 2x and 2x is a prime no and hence value of x is 1

Ans should be B
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Re: New Set: Number Properties!!!  [#permalink]

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If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n
(2) The lowest common multiple of n and n + 10 is an even number.

We need to find the value of n and n has exactly 2 postive factors

This implies that n is of the form n= a^1 where a is prime factor of n.

From St 1, we have n/2 is one of the factors of n.
Now any prime no has only 1 and the no itself as its factor therefore n can be any of the prime nos ie. 2,3,5,7.... and so on. However when n = 2, n/2 = 1 and 1 is factor of n and hence the value n will be 2
Note that any other prime no when divided by 2 will not give an integer as 2 is the only even prime no

So B,C and E ruled out

St 2. We have multiple options here
The lowest common multiple of n and n + 10 is an even number.

n =2 and n+ 10 =12, LCM is 12( Even no) value of n =2-----> n is prime
n=4 and n+10 = 14 , LCM is 28 (even no)value of n =4 -----> n is not prime

Hence D ruled out,

Ans should be A
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Re: New Set: Number Properties!!!  [#permalink]

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3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

(1) Both x and y is have 3 positive factors.
(2) Both \sqrt{x}and \sqrt{y}are prime numbers

From the Q stem we get x and y are consecutive perfect squares ie. x=4, y=9 or x=9 then y=16 and so on
We need to find y/x

St 1. Both x and y have 3 postive factors
Consider X , Since X has 3 psotive factors and x is a perfect square x is of the form x = a^2 where is "a" is a prime no. Then possible factors of x are 1,a, and a2
Similarly for y= b^2 where b is prime no, factors will be 1,b,and b^2

Clearly we can multiple remainders

for ex x=4 , y =9 , R (y/x)= 1
x= 25, y=36, R (y/x)= 11

Hence not sufficient and therefore A and D ruled out

St 2. \sqrt{x} and \sqrt{y} are prime no which means x=a^2 and y= b^2 where a and b are prime no and x and y are sqaure of prime nos.

This is same information as St1 and hence B and C ruled out

Ans should be E
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Re: New Set: Number Properties!!!  [#permalink]

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4. Each digit of the three-digit integer N is a multiple of 4, what is the value of K?

(1) The units digit of K is the least common multiple of the tens and hundreds digit of K
(2) K is NOT a multiple of 3.

I think K should be N is statement choices

Lets look at the possible 3 digit no options based on the given Question stem. We have
444, 888,484, 448,844 and so on....the digit integer N will have only 4' s and 8's

St 1 implies that no is of the form 488------> where 8 in unit place is LCM of 4 and 8 (tens place)
or it could be 888-----> Again 8 in unit place is LCM of 8 and 8 in tens and hundreds place

We have 2 options hence A alone not sufficient therefore A and ruled out

2. St N is not a multiple of 3----> means N can be 448,844,884,848 and so on.. B alone is not sufficent combining we get

following choices 484,848...Again we have 2 possible values of N and therefore

Ans should be E
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Re: New Set: Number Properties!!!  [#permalink]

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

(1) The median of {a!, b!, c!} is an odd number.
(2) c! is a prime number

We need to find if a,b and c are consecutive integers or not

st 1 says The median of {a!, b!, c!} is an odd number.

We know that 2! =2 and 3!=6 and 4!=24 -----> for any factorial greater then equal to 2! will always be even no as it will have 2

This implies possible options of a,b and c are 0,1 and 2
0!=1 and 1!=1 and 2!= 2 so our set is (0,1 and 2) and median is 1 which is odd. This implies that a,b and c are consecutive integers
Now also we can have any other no in place of 2!, ----> we can have 3! or 4! our median will be odd...a,b and c are not CI

So A and D ruled out

St 2 says c! is a prime number

c can only be 2 as factorial of other nos greater then 2 will have some factor and a and b can be any nos
a=-3 , b=-2 and c= 2----> a,b and c are not CI
but again if c=2, b=1 and a=0 then they are CI

So B alone not sufficient

Combining we get only case of c=2,b=1 and a=0 and therefore
Ans should be C
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Re: New Set: Number Properties!!!  [#permalink]

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6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative
(2) The product of the smallest and largest integers in the list is a prime number.

Let Set S has n integers so n>2

From St1,we have 2 possibilities for
All 3 nos are negative ie. -*-*-= Negative no
Or 2 nos are positive and one negative ---> +*-*+

So A and ruled out

St2 Since n> 2, Lets take n=4 and the nos be n1<n2<n3<n4
Let all nos be positive.
Since the product of smallest and largest is a prime no then smallest no will have to be 1 and largest not can be any prime no greater than n2 and n3.
n2 and n3 can be any nos

Another case will be n4<n3<n2<n1....in this case let us assume all nos are negative and n1 is the largest nos and n4 is the smallest in this case as well
n4*n1=n1n4----> which can be a prime no
Ex -7*-1=7 is a prime no

Since we have 2 cases so Option is ruled out.

Now Combining we get a case where all nos are negative the 2 statement conditions are met and hence should be C

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

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7. Is x the square of an integer?

(1) When x is divided by 12 the remainder is 6
(2) When x is divided by 14 the remainder is 2

From Question stem, we have whether x=I^2 where I is some Integer

From St1, Possible values of x will be 6,18,30,42...

In General we have x= 12a+ R where R=6 and a is some integer

I^2= 12a+6-----> I^2= 2^2*3a+6

Now the RHS expression cannot be a square of any Integer I and therefore X is not a square of any integer.

So option B,C and E ruled out

St 2 When x is divided by 14 the remainder is 2

x= 14b+2 ---->Possible values of x are 2,16, 30,44,58...
We have 2 cases ie. X=16 a perfect sqaure and X = 2 or 30 or 44 not a perfect square
Hence B alone is not sufficient

So Ans should be A
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Re: New Set: Number Properties!!!  [#permalink]

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9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

(1) ab = 2
(2) 0 < a < b < 2

From St 1, we have ab=2 (Note that a and b are of same sign)
Possible values are

a= 2, b= 1, Is exp [a] + [b] = 1 ---Clearly no as  +  = 3=1
a=-1, b=-2, Is exp [a] + [b] = 1 No as -----> Expression equals -3 ([-1]+[-2] = -3)
a=Sqt 2 , b= Sqt 2, Exp [Sqt2] + [sqt 2] = 1+1 =2 hence no
For any value of a and b the expression does not seem to be true and hence [a]+[b]=1 is not true for all cases (I hope this doesn't come back to bite me)

Hence ans should be A so B,C and E ruled out

St2 0 < a < b < 2

Let a = 0.3, b =1.3 so we have [0.3]+[1.3]= 0+1 and hence expression is true
but if a= 1.1 and b = 1.6 then we have [1.1]+[1.6] = 1+1 =2 not equal to 1

Since 2 cases are possible with st2 therefore ans should be A
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Re: New Set: Number Properties!!!  [#permalink]

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10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

(1) 9 is NOT a factor of N
(2) 125 is a factor of N

Since x and y are postive and N has 12 postive factors which means we can have more than 1 value for N for different combination of prime factors

N = 3^x*5^y ------ >Now No. factors is given by (x+1)(y+1)= 12
Possible combinations
(x+1)(y+1)
1 12
2 6
3 4

and vice versa

St1 : 9 is not a factor of N,
Now if X =2 or 3 then 9 becomes a factor of N therefore possible values of x are 0 and 1
For x =0 the number N will be 5^11
For x =1, then the number will be 3^1 *5^5 (Factors will be 3, 5, 3*5, 5^5, 5^4, 5^3, 5^2, 3*5^5 , 3*5^4, 3*5^3,3*5^2 and 1 )

So 2 values of N possible hence not sufficient. Option A & D ruled out

St 2 :125 is a factor of N

Now if y= 3 then 125 becomes factor of N therefore possible values of y are 3 or 6
When y=3, x= 2 then no N will be = 3^2*5^3
or when y=5, x=1 then no N will be = 3^1*5^5

2 values of N possible so option B ruled out

Combining 2 statements we get N = 3^1*5^5

Hence Ans should be C
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Re: New Set: Number Properties!!!  [#permalink]

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1
BONUS QUESTION:
11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

X and Y are +Integers, Is X a prime no

St1 |x - 2| < 2 - y

Can be re-written as |x - 2| +y < 2

Solving for Modulus we get

Case 1
x-2 +y <2 ----> x+y< 4
Lets say x=2 , y= 1 then is X prime----> Yes
But if X=1 and Y=2 then is X prime ----> no (However if we put these values in the given expression we get 3< 2 which is not the case ) and hence this case is not considered as it goes against the given condition

Case 2 Taking value in Mod as negative we get

-(x-2) +y < 4-----> -x+2 +y < 2, we get x> y.
and therefore for the expression we will have x=2, y = 1. Thus X is prime

SO option B,C and E ruled out

St 2 : x + y - 3 = |1-y|
x+y = |1-y| +3
Solving for mod cases we get

Case 1 : Term in modulus is positive, we get

x+y= (1-y) +3
x+y =1 -y + 3 -----> x+2y= 4
Again x=2, y= 2, Is X prime yes

Case 2 Term in Modulus is negative, we get

x+y= -(1-y) + 3
x+y = -1 +y +3
We get x =2

Therefore St2 alone is also sufficient
Hence Ans should be D
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Re: New Set: Number Properties!!!  [#permalink]

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1
9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + = 1 ?

(1) ab = 2
(2) 0 < a < b < 2

[b](1) ab = 2

ab = 2 --> b = 2/a

If a<0 --> b<0 --> [ a ] + [ b ] <0 --> Answer to the question is NO

If 0 < a _< 1 --> b >_ 2 --> [ a ] + [ b ] > 1 --> Answer to the question is NO

If a>1 --> [ a ] + [ b ] > 1 --> Answer to the question is NO

(1) SUFFICIENT

(2) 0 < a < b < 2

If a=0.3 and b=0.5 , therefore [a] = 0 and [b] = 0, and it follows that [a] + [b] = 0 ; the answer is NO
If a = 0.2 and b = 1.1 therefore [a] = 0 and [b] = 1, and if follows that [a] + [b] = 1 , the answer is YES

So , the second statement is INSUFFICIENT

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

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2
10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

(1) 9 is NOT a factor of N
(2) 125 is a factor of N

(x,y) two positive integers and N has 12 postive factors --> (x+1)(y+1)= 12

x + 1 = 2 and y + 1 = 6 --> (x,y) = (1,5)
x + 1 = 6 and y + 1 = 2 --> (x,y) = (5,1)
x + 1 = 3 and y + 1 = 4 --> (x,y) = (2,3)
x + 1 = 4 and y + 1 = 3 --> (x,y) = (3,2)

(1) 9 is NOT a factor of N

Since x is a positive integer and 9 is not a factor of N, then dicard x=2 and x=3 and x=5 --> x=1

x=1 --> y=5 --> N = 3^x*5^y = 3^1 * 5^5 --> Statement 1 ALONE SUFFICENT

(2) 125 is a factor of N

we know that : 5^3 = 125 , hence discard all possible values less than 3 for y --> two possibilties remain (x,y) = (1,5) and (x,y) = (2,3) --> Statement 2 ALONE INSUFFICIENT

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Originally posted by Rock750 on 27 Mar 2013, 07:58.
Last edited by Rock750 on 28 Mar 2013, 17:24, edited 1 time in total.
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Re: New Set: Number Properties!!!  [#permalink]

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2
BONUS QUESTION:
11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y
(2) x + y - 3 = |1-y|

(1) |x - 2| < 2 - y

If x<2 --> x=1 because x is a positive integer then x is not prime
If x=2 --> x is prime

If x>2 --> x - 2 < 2 - y
In this case :
if y<2 --> y = 1 -> x < 3 and because x>2 so this can't be true --> y must be greater than or equalt to 2
if y = 2 -> x - 2 < 0 -> x<2 so x = 1 and thus x is not prime
if y>2 hence 2 - y < 0 -> x - 2 < 2 - y < 0 -> x - 2 < 0 -> x < 2 -> x = 1 -> thus x is not prime

(1) ALONE IS SUFFICIENT

2) x + y - 3 = |1-y|

y is a positive integer
--> if y = 1 --> x = 2 thus x is prime
--> if y>1 --> |1-y| = y - 1 --> x + y - 3 = y - 1 --> x = 2 and thus x is prime

(2) ALONE IS SUFFICIENT

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Originally posted by Rock750 on 28 Mar 2013, 02:43.
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Re: New Set: Number Properties!!!  [#permalink]

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As carcass suggested, next set word problems Bunnel i wonder if you could add geometry besides so that we can have two sets for the next week.

Thanks !
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"If you don't change your life, your life will change you" Re: New Set: Number Properties!!!   [#permalink] 28 Mar 2013, 03:08

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