New Set: Number Properties!!! : Data Sufficiency (DS)

2013-03-25T04:50:20-07:00

[size=150]Number Properties is a key topic on the GMAT and it's important to have a solid understanding of it. You'll encounter questions that test your ability to work with integers, multiples and factors, primes, odds and evens, and more. In this set of GMAT questions from GMAT Club Tests, you'll face 11 challenging problems that will put your number properties knowledge to the test. Good luck![/size] [b]1. If x is an integer, what is the value of x?[/b] (1) |23x| is a prime number (2) [m]2[square_root]x^2[/square_root][/m] is a prime number. [b]2. If a positive integer n has exactly two positive factors what is the value of n?[/b] (1) n/2 is one of the factors of n (2) The lowest common multiple of n and n + 10 is an even number. [b]3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?[/b] (1) Both x and y is have 3 positive factors. (2) Both [m][square_root]x[/square_root][/m] and [m][square_root]y[/square_root][/m] are prime numbers [b]4. If each digit of the positive three-digit integer [m]k[/m] is a positive multiple of 4, what is the value of [m]k[/m]?[/b] (1) The units digit of [m]k[/m] is the least common multiple of the tens and hundreds digits of [m]k[/m]. (2) [m]k[/m] is NOT a multiple of 3. [b]5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?[/b] (1) The median of {a!, b!, c!} is an odd number. (2) c! is a prime number [b]6. Set S consists of more than two integers. Are all the numbers in set S negative?[/b] (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. [b]7. Is x the square of an integer?[/b] (1) When x is divided by 12 the remainder is 6 (2) When x is divided by 14 the remainder is 2 [b]8. List A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the list less than 1/5?[/b] (1) Reciprocal of the median is a prime number (2) The product of any two terms of the list is a terminating decimal [b]9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?[/b] (1) ab = 2 (2) 0 < a < b < 2 [b]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?[/b] (1) 9 is NOT a factor of N (2) 125 is a factor of N BONUS QUESTION: [b]11. If x and y are positive integers, is x a prime number?[/b] (1) |x - 2| < 2 - y (2) x + y - 3 = |1-y| [size=200][highlight]The solutions can be found below on this page.[/highlight][/size] [gmatclubtest_1][/gmatclubtest_1]

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Bunuel

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

29 Mar 2013, 06:11

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11. If x and y are positive integers, is x a prime number?

(1) |x - 2| < 2 - y . The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus 0 < 2 - y, thus y < 2 (if y is more than or equal to 2, then \(2-y\leq{0}\) and it cannot be greater than |x - 2|). Next, since given that y is a positive integer, then y=1.

So, we have that: \(|x - 2| < 1\), which implies that \(-1 < x-2 < 1\), or \(1 < x < 3\), thus \(x=2=prime\). Sufficient.

(2) x + y - 3 = |1-y|. Since y is a positive integer, then \(1-y\leq{0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D.

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

29 Mar 2013, 05:06

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

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

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

29 Mar 2013, 04:43

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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]

29 Mar 2013, 04:12

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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?

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.

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

29 Mar 2013, 05:52

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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 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.

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

29 Mar 2013, 03:55

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

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

Answer: D.

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

29 Mar 2013, 06:03

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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?

\(N = 3^x*5^y\) has 12 positive factors means that (x+1)(y+1)=12=2*6=6*2=3*4=4*3. We can have the following cases:

\(N = 3^1*5^5\);
\(N = 3^5*5^1\);
\(N = 3^2*5^3\);
\(N = 3^3*5^2\).

(1) 9 is NOT a factor of N. This implies that the power of 3 is less than 2, thus N could only be \(3^1*5^5\). Sufficient.

(2) 125 is a factor of N. This implies that the power of 5 is more than or equal to 3, thus N could be \(3^1*5^5\) or \(3^2*5^3\). Not sufficient.

Answer: A.

THEORY.
Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

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

29 Mar 2013, 03:30

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SOLUTIONS:

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

(1) \(|23x|\) is a prime number.

Since 23 is a prime number, this statement implies that \(x=1\) or \(x=-1\). Not sufficient.

(2) \(2\sqrt{x^2}\) is a prime number.

We can rewrite the expression as \(2\sqrt{x^2}=2|x|\). Since 2 is a prime number, this statement implies that \(x=1\) or \(x=-1\). Not sufficient.

(1)+(2) Using both conditions, \(x\) could be either 1 or -1. The information provided is still insufficient to determine the exact value of \(x\).

Answer: E

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

29 Mar 2013, 04:29

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4. If each digit of the positive three-digit integer \(k\) is a positive 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 digits of \(k\).

Possible values for \(k\) are 444, 488, 848, or 888. Not sufficient.

(2) \(k\) is NOT a multiple of 3.

Possible values for \(k\) are 448, 484, 488, 844, 848, or 884. Not sufficient.

(1)+(2) From the above, \(k\) could be 488 or 848. Still, this is not sufficient.

Answer: E

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

07 Apr 2013, 02:09

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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

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

25 Mar 2013, 07:03

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

(1) |23x| is a prime number
Since 23 si prime,\(x\)can be \(+1\) or \(-1\)
not sufficient

(2) 2\sqrt{x^2} is a prime number.
once again \(x\) can be \(+1\) or \(-1\)
not sufficient

And since 1)+2) provides no new info IMO E

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

25 Mar 2013, 07:24

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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
Not sufficient
Example: \(S = {-1,3,5}\)
the product is always <0 but 2 numbers are positive
Example: \(S = {-1,-3,-5}\)
the product is always <0 and all numbers are negative

(2) The product of the smallest and largest integers in the list is a prime number.
Not sufficient
Example: \(S = {1,3,5}\)
\(1*5=5\) prime but all positive
Example: S = \({-1,-3,-5}\)
\(-1*-5=5\) prime but all negative

(1)+(2) Sufficient IMO C
Using statement 2 we know that all are positive or all are negative, using statement 1 we know that "at least" 1 is negative=> so all are negative

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

25 Mar 2013, 07:40

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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

The question can be seen as (given statement 2):
\([a] + [b] = 1\)
case 1:\(0<(=)a<1\) => \([a]=0\) and \(1(=)<b<2\) => \([b]=1\) so \([a] + [b] = 1\)
or the opposite
case 2: \(0<(=)b<1\) => \([b]=0\) and \(1(=)<a<2\) => \([a]=1\) so \([a] + [b] = 1\)

But as ab=2 we know that one term is \(\frac{1}{2}\) and the other is \(2\)
So we are in one of the two senarios above, IMO C

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

Updated on: 25 Mar 2013, 08:57

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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?

Number of factors of \(N = (x+1)(y+1) = 12\)
the combinations are
\(3*4\) with \(x= 2\) and \(y=3\)
\(6*2\) with \(x=5\) and\(y = 1\)
and the "other way round" of each one

(1) 9 is NOT a factor of N
So x must be 1, \(x=1\)
because \((1+1)(y+1)=12\)
\(y=5\)
Sufficient

(2) 125 is a factor of N
So \(y>=3\), y can be 3 or 5, NOT sufficient
\(y=3, (3+1)(x+1)=12, x=2\)
\(y=5, (5+1)(x+1)=12, x=1\)

IMO A

Originally posted by Zarrolou on 25 Mar 2013, 07:49.
Last edited by Zarrolou on 25 Mar 2013, 08:57, edited 1 time in total.

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

25 Mar 2013, 08:34

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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
Not sufficient

(2) The product of any two terms of the set is a terminating decimal
Because \(\frac{1}{prime}\) is not a terminating decimal, with the only exception of 1/4, 1/1, 1/5 and 1/2
any set made by these three CANNOT have a median < 1/5, it can be = 1/5 but NEVER <
Some examples:
A={1,1,1,1,1/5,1/5,1/5,1/5,1/5,1/5} the median is 1/5 = 1/5
A={1,1,1,1,1/2,1/2,1/2,1/2,1/2,1/2} the median is 1/2 > 1/5
SUFFICIENT

IMO B

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

25 Mar 2013, 09:35

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11. If x and y are positive integers, is x a prime number?

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

We know that x >0 and y>0 and they are integers.

From F.S 1, we have 2-y>=0 or y<=2. Thus y can only be 2 or 1. Now if y=2, we would have 0>some thing positive or 0>0(when x also equal to 2). Either case is not possible. Thus, y can only be 1. For y=1, we can only have x = 2. Which is prime. Sufficient.

From F.S 2, we have either y>1 or y<1. Now as y is a positive integer, y can't be less than 1.For y=1, we anyways have x=2(prime).In the first case, we have y>1--> x+y-3 = y-1 or x=2(prime). Thus, Sufficient.

D.

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

25 Mar 2013, 10:06

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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 F.S 1, we have x = 12q+6 --> 6(2q+1). For x to be a square of an integer, we should have 2q+1 of the form 6^pk^2, where both q,p and k are integers and p is odd. Now we know that 2q+1 is an odd number and 6^pk^2 is even. Thus they can never be equal and hence x can never be the square of an integer. Sufficient.

From F.S 2, we have x = 14q+2 --> 2(7q+1).Just as above, we should have 7q+1 = 2^pk^2. Now for q=1, k=2 and p=1, we have 8=8, thus x is the square of an integer. But for q=0, x is not. Insufficient.

Basically for the second fact statement, we can plug in easily. No need for the elaborate theory.

A.

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

25 Mar 2013, 10:06

GMAT Data Sufficiency (DS) Questions

New Set: Number Properties!!!