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! 1. If x is an integer, what is the value of x? (1) |23x| is a prime number (2) 2\sqrt{x^2} is a prime number. 2. 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. 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 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 . (2) k is NOT a multiple of 3. 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 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. 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 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? (1) Reciprocal of the median is a prime number (2) The product of any two terms of the list is a terminating decimal 9. If denotes the greatest integer less than or equal to x for any number x, is + = 1 ? (1) ab = 2 (2) 0 < a < b < 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 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| The solutions can be found below on this page. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT. DS questions in GMAT Focus encompass various types of word problems, such as: Word Problems Work Problems Distance Problems Mixture Problems Percent and Interest Problems Overlapping Sets Problems Statistics Problems Combination and Probability Problems While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..." Check GMAT Syllabus for Focus Edition You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT. So, you can ignore these questions. The topic is locked. Hope it helps.

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New Set: Number Properties!!!

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Bunuel

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27 Aug 2017, 06:17

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nishantt7

Bunuel

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 $y-2\leq{0}$ 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.

Hi Bunuel

Can you please elaborate the solution of this one?
I did it with critical point method and ended with 2 equations in case of x>2(x+y=4) and x<2(x+y=0) and 2 equations with x<1(x+2y=4) and x>1(x=2).
How to proceed further?

For (1): In $|x - 2| \lt 2 - y$, the left hand side of the inequality (|x - 2|) is an absolute value. An absolute value of a number cannot be negative, so the left hand side of the inequality (|x - 2|) is not negative, which means that the right hand side of the inequality, which is greater than the left hand side, also cannot be negative. This gives $0 \lt 2 - y$.

For (2): We know from the stem that y is a positive integer. Since y is a positive integer (1, 2, 3, ...), then 1 - y = 1 - positive = non-positive, which measn that |1-y|=-(1-y). Therefore, for (2) we'd have $x + y - 3 = -(1-y)$, which gives $x=2=prime$. Sufficient.

Hope it's clear.

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05 Jan 2020, 19:36

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Bunuel

I'm still having difficulty understanding when to consider only the positive root when GMAT gives the even square root. It seems like when the value is defined that this rule will apply (e.g. sqrt(25) is 5, not -5); when a variable is provided instead, the answer is the abs(x), and 'x' itself can be +/-.

Can you help me better understand? I only considered the pos. root in this problem and got the wrong answer as a result.

Thanks!

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05 Jan 2020, 22:13

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rwx5861

$\sqrt{...}$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($\sqrt{...}$) always means non-negative square root.

The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$\sqrt{9} = 3$, NOT +3 or -3;
$\sqrt[4]{16} = 2$, NOT +2 or -2;
Similarly $\sqrt{\frac{1}{16}} = \frac{1}{4}$, NOT +1/4 or -1/4.

Notice that in contrast, the equation $x^2 = 9$ has TWO solutions, +3 and -3. Because $x^2 = 9$ means that $x =-\sqrt{9}=-3$ or $x=\sqrt{9}=3$.

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05 Dec 2020, 06:50

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I read on gmat club that the value of square root of x is considered positive value only (We consider + or - for absolute values only). So, in statement 2 why are we considering square root of x square as +1 or -1?

Bunuel

SOLUTIONS:

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

(1) $|23x|$ is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) $2\sqrt{x^2}$ is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.

Bunuel

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05 Dec 2020, 07:30

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gayatri259

Bunuel

Both the square root sign ($\sqrt{}$) and absolute value cannot give negative result.

The value of $\sqrt{x^2}$ is still positive for both 1 and -1:

$\sqrt{x^2}=\sqrt{1^2}=\sqrt{1}=1$;
$\sqrt{x^2}=\sqrt{(-1)^2}=\sqrt{1}=1$.

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Hi Bunnel,

could you please tell how the list of 10 terms will look like with product of any two term resulting to terminating decimal ? can we have all the numbers in list as 1/2 ?

1) can my list look like : 1/2,1/2, 1/2.... as per me this also satisfy the statement 2 and median is equal to 0.5 or can list have all the term as 1/5 , in that median will be 0.2?

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?

(2) The product of any two terms of the list is a terminating decimal. This statement implies that the list must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

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06 Dec 2020, 10:28

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abhinavsodha800

For (2) the set could be any combination of 1/2's and 1//5:
{1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2}
{1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5, 1/5}
{1/5, 1/5, 1/5, 1/5, 1/5, 1/2, 1/2, 1/2, 1/2, 1/2}
...

The median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. So, we have a NO answer to the question. Sufficient.

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18 May 2021, 00:06

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Bunuel

Hey Bunuel,

Had a doubt here. For case 2, since the question stem says that "$2\sqrt{x^2}$ is a prime number" the answer to that has been to be a positive number+2 right ? So out of the two cases +1 and -1, we can't take the value -1 . So Case 2 is sufficient right? Please correct me if I am wrong.

Bunuel

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18 May 2021, 00:16

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ash431994

Bunuel

Both 1 and -1 worrk there.

$2\sqrt{x^2}=2\sqrt{1^2}=2\sqrt{1}=2$;
$2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2$.

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18 May 2021, 01:40

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Bunuel

ash431994

Bunuel

Both 1 and -1 worrk there.

$2\sqrt{x^2}=2\sqrt{1^2}=2\sqrt{1}=2$;
$2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2$.

But sqrt(1) can be both +1 or -1 right? and here since it says 2 is prime, it has to be a positive value.

Bunuel

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18 May 2021, 02:05

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ash431994

Bunuel

ash431994

Both 1 and -1 worrk there.

$2\sqrt{x^2}=2\sqrt{1^2}=2\sqrt{1}=2$;
$2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2$.

But sqrt(1) can be both +1 or -1 right? and here since it says 2 is prime, it has to be a positive value.

No. This has already been addressed in this thread couple of times.

$\sqrt{...}$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($\sqrt{...}$) always means non-negative square root. Check this post: https://gmatclub.com/forum/new-set-numb ... l#p2434955

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07 Mar 2024, 09:35

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Bunuel

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

The question says x and y are consecutive perfect squares, not that x and y are squares of consecutive numbers. Is it right to assume that they are squares of consecutive numbers?

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Haffun

Bunuel

The question says x and y are consecutive perfect squares, not that x and y are squares of consecutive numbers. Is it right to assume that they are squares of consecutive numbers?

x and y being consecutive perfect squares implies that they are squares of consecutive integers. For example, 9 and 16 are consecutive perfect squares, and they are squares of consecutive integers 3 and 4, respectively.

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For (2), it is given that it is prime number, so it should be 1 only as -2 is not a prime number. Please explain.

Bunuel

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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Aashimabhatia

For (2), it is given that it is prime number, so it should be 1 only as -2 is not a prime number. Please explain.

Bunuel

No, all is correct there. x can be 1 or -1. BOTH result $2\sqrt{x^2}$ to be a prime number 2.

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26 Dec 2025, 06:12

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Bunuel - if I may quote a rule, I picked on this forum - When the gmat provides square root sign for an even root, then the only accepted answer is the positive root. i.e \sqrt{}25 is 5 and not +5 or -5. So, why in statement 2, do we have -1 or 1?

Bunuel

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Your doubt has been addressed in the thread. Please review.

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Bunuel, how do we jump from 1−y≤0 to |1−y|=−(1−y)?

Bunuel

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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Bunuel, how do we jump from 1−y≤0 to |1−y|=−(1−y)?

Because when x <= 0, |x| = -x.

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Updated on: 26 Dec 2025, 07:14

Originally posted by Bunuel on 26 Dec 2025, 07:13.
Last edited by Bunuel on 26 Dec 2025, 07:14, edited 1 time in total.

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Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

Word Problems
Work Problems
Distance Problems
Mixture Problems
Percent and Interest Problems
Overlapping Sets Problems
Statistics Problems
Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore these questions. The topic is locked.

Hope it helps.