Praxis Math Quiz: Apply Number Properties

20 questions · exam conditions

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Apply Number PropertiesQuestion 1 of 20

Two gears in a machine are aligned by a mark. The first gear has 24 teeth and the second has 36 teeth. How many complete rotations must the first, smaller gear make before the two marks are aligned again for the first time?

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Praxis Math Quiz

Praxis Math Quiz: Apply Number Properties

Practice Apply Number Properties in Praxis Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Number Properties, giving you a quick way to practice the rules, question types, and explanations that matter most for Praxis Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

2
3 (correct answer)
12
72

Explanation: The gears will align again after a number of teeth have passed that is the least common multiple (LCM) of the number of teeth on each gear. First, find the LCM of 24 and 36.\nPrime factorization of 24 is

2^3 \cdot 3

.\nPrime factorization of 36 is

2^2 \cdot 3^2

.\nThe LCM is the product of the highest powers of all prime factors:

2^3 \cdot 3^2 = 8 \cdot 9 = 72

. So, the marks align after 72 teeth have passed the intersection point. The question asks for the number of rotations of the first gear. Since the first gear has 24 teeth, the number of rotations is

72 \div 24 = 3

Question 2

At a bus station, Bus A departs every 12 minutes and Bus B departs every 20 minutes. If both buses depart together at 1:00 PM, what time will it be the third time they depart together (including the 1:00 PM departure)?

2:00 PM
2:20 PM
3:00 PM (correct answer)
4:00 PM

Explanation: First, find the time interval at which the buses depart together. This is the least common multiple (LCM) of 12 and 20. \n

12 = 2^2 \cdot 3

20 = 2^2 \cdot 5

\nLCM =

2^2 \cdot 3 \cdot 5 = 60

. The buses depart together every 60 minutes, or every hour. The first departure together is at 1:00 PM. The second will be 60 minutes later at 2:00 PM. The third departure will be another 60 minutes later, at 3:00 PM.

Question 3

How many positive integers are common factors of 36 and 90?

4
5
6 (correct answer)
9

Explanation: The common factors of two numbers are the factors of their greatest common divisor (GCD). First, find the GCD of 36 and 90.\n

36 = 2^2 \cdot 3^2

90 = 2 \cdot 3^2 \cdot 5

\nThe GCD is

2^1 \cdot 3^2 = 2 \cdot 9 = 18

. Now, find how many factors the GCD (18) has. The factors of 18 are 1, 2, 3, 6, 9, and 18. There are 6 common factors.

Question 4

For a school event, a teacher wants to arrange chairs in rows. Whether the teacher places 5, 6, or 8 chairs in a row, there are always 2 chairs left over. What is the smallest possible number of chairs greater than 2 that the teacher could have?

62
82
122 (correct answer)
242

Explanation: Let

C

be the total number of chairs. The problem states that if you subtract 2 from

C

, the result is a multiple of 5, 6, and 8. So,

C-2

must be the least common multiple (LCM) of 5, 6, and 8. \n

5 = 5^1

6 = 2 \cdot 3

8 = 2^3

\nThe LCM is

2^3 \cdot 3 \cdot 5 = 8 \cdot 3 \cdot 5 = 120

. So,

C-2 = 120

. This means

C = 122

. This is the smallest number of chairs greater than 2 that satisfies the conditions.

Question 5

A florist has 90 roses and 75 tulips to create identical bouquets for a wedding. If she must use all the flowers with none left over, what is the greatest number of identical bouquets she can create?

5
6
15 (correct answer)
450

Explanation: To find the greatest number of identical bouquets, we need to find the greatest common divisor (GCD) of the number of roses (90) and tulips (75). The number of bouquets must be a factor of both 90 and 75.\nPrime factorization of 90 is

2 \cdot 3^2 \cdot 5

.\nPrime factorization of 75 is

3 \cdot 5^2

.\nThe GCD is the product of the lowest powers of common prime factors:

3^1 \cdot 5^1 = 15

. Therefore, the greatest number of identical bouquets she can create is 15.

Question 6

How many distinct positive integers are factors of the number 180?

9
12
18 (correct answer)
24

Explanation: To find the number of factors of an integer, first find its prime factorization.

180 = 18 \cdot 10 = (2 \cdot 3^2) \cdot (2 \cdot 5) = 2^2 \cdot 3^2 \cdot 5^1

. The number of factors is found by taking each exponent in the prime factorization, adding 1 to it, and then multiplying these new numbers together. The exponents are 2, 2, and 1. So, the number of factors is

(2+1)(2+1)(1+1) = 3 \cdot 3 \cdot 2 = 18

Question 7

The integer $k$ is a multiple of 6 and a factor of 120. Which of the following is NOT a possible value for $k$ ?

12
30
48 (correct answer)
60

Explanation: We need to check which of the given options satisfies both conditions. The conditions are: (1)

k

is a multiple of 6, and (2)

k

is a factor of 120. Let's test each option.\nA) 12 is a multiple of 6 (

6 \cdot 2 = 12

) and a factor of 120 (

120 \div 12 = 10

). So, 12 is a possible value.\nB) 30 is a multiple of 6 (

6 \cdot 5 = 30

) and a factor of 120 (

120 \div 30 = 4

). So, 30 is a possible value.\nC) 48 is a multiple of 6 (

6 \cdot 8 = 48

), but it is not a factor of 120 (

120 \div 48 = 2.5

). So, 48 is NOT a possible value.\nD) 60 is a multiple of 6 (

6 \cdot 10 = 60

) and a factor of 120 (

120 \div 60 = 2

). So, 60 is a possible value.

Question 8

The least common multiple of an integer $x$ and 18 is 72. The greatest common divisor of $x$ and 18 is 6. What is the value of $x$ ?

12
16
24 (correct answer)
36

Explanation: We can use the property that for any two positive integers

a

and

b

a \cdot b = \text{GCD}(a, b) \cdot \text{LCM}(a, b)

. In this case,

a=x

and

b=18

. So,

x \cdot 18 = \text{GCD}(x, 18) \cdot \text{LCM}(x, 18)

. We are given

\text{LCM}(x, 18) = 72

and

\text{GCD}(x, 18) = 6

. Plugging in the values:

x \cdot 18 = 6 \cdot 72

, which simplifies to

x \cdot 18 = 432

. Solving for

x

x = 432 \div 18 = 24

Question 9

If the greatest common divisor of two distinct positive integers $x$ and $y$ is $x$ , which of the following statements must be true?

$y$ is a factor of $x$ .
$y$ is a multiple of $x$ . (correct answer)
$x$ and $y$ are both odd.
$x$ is greater than $y$ .

Explanation: The greatest common divisor (GCD) of two integers is the largest integer that divides both of them without a remainder. If the GCD of

x

and

y

x

itself, it means that

x

must divide both

x

and

y

. Since

x

always divides

x

, the condition implies that

x

must also be a factor of

y

. If

x

is a factor of

y

, then

y

is a multiple of

x

. For example, GCD(4, 12) = 4; here, 12 is a multiple of 4.

Question 10

A positive integer $n$ has exactly three positive factors. Which of the following must be true about $n$ ?

$n$ is a prime number.
$n$ is an odd number.
$n$ is the square of a prime number. (correct answer)
$n$ is the cube of a prime number.

Explanation: The number of factors of an integer is determined by the exponents in its prime factorization. If the factorization is

p_1^{a_1} p_2^{a_2} \dots

, the number of factors is

(a_1+1)(a_2+1)\dots

. For a number to have exactly 3 factors, this product must equal 3. Since 3 is a prime number, there can only be one term in the product, and it must be 3. This means

a_1+1=3

, so

a_1=2

. The prime factorization must be of the form

p^2

for some prime number

p

. Therefore,

n

must be the square of a prime number. For example,

9=3^2

has factors 1, 3, 9.

4=2^2

has factors 1, 2, 4.

Question 11

Let $M$ be the least common multiple of 18, 24, and 30. Which of the following is a prime factor of $M$ ?

7
5 (correct answer)
11
13

Explanation: To find the least common multiple (LCM), we first find the prime factorization of each number. \n

18 = 2 \cdot 9 = 2 \cdot 3^2

24 = 8 \cdot 3 = 2^3 \cdot 3

30 = 3 \cdot 10 = 2 \cdot 3 \cdot 5

\nThe LCM is the product of the highest powers of all prime factors involved (2, 3, and 5).

M = \text{LCM} = 2^3 \cdot 3^2 \cdot 5^1 = 8 \cdot 9 \cdot 5 = 360

. The prime factors of

M

are 2, 3, and 5. Of the choices given, only 5 is a prime factor of

M

Question 12

Two positive integers have a least common multiple of 84 and a greatest common divisor of 2. If one of the integers is 12, what is the other integer?

7
14 (correct answer)
28
42

Explanation: For any two positive integers

a

and

b

a \cdot b = \text{GCD}(a, b) \cdot \text{LCM}(a, b)

. Let the two integers be 12 and

x

. We are given LCM = 84 and GCD = 2. So,

12 \cdot x = 2 \cdot 84

, which is

12x = 168

. Dividing both sides by 12 gives

x = 14

. We can check our answer: GCD(12, 14) = 2, and LCM(12, 14) = LCM(

2^2 \cdot 3

2 \cdot 7

) =

2^2 \cdot 3 \cdot 7 = 84

Question 13

When a positive integer $k$ is divided into 150, the remainder is 6. When $k$ is divided into 205, the remainder is 5. What is the greatest possible value of $k$ ?

8 (correct answer)
12
16
24

Explanation: If dividing 150 by

k

gives a remainder of 6, it means that

150 - 6 = 144

is perfectly divisible by

k

. If dividing 205 by

k

gives a remainder of 5, it means that

205 - 5 = 200

is perfectly divisible by

k

. So,

k

must be a common divisor of 144 and 200. To find the greatest possible value of

k

, we find the greatest common divisor (GCD) of 144 and 200. \n

144 = 12^2 = (2^2 \cdot 3)^2 = 2^4 \cdot 3^2

200 = 2 \cdot 100 = 2 \cdot 10^2 = 2 \cdot (2 \cdot 5)^2 = 2^3 \cdot 5^2

\nThe GCD is

2^3 = 8

. A divisor must be larger than the remainder. Since

k=8

is larger than both remainders (6 and 5), it is a valid answer.

Question 14

If an integer is divisible by both 12 and 15, then it must also be divisible by which of the following?

18
24
30 (correct answer)
180

Explanation: An integer divisible by both 12 and 15 must be a multiple of their least common multiple (LCM). Let's find the LCM of 12 and 15. \n

12 = 2^2 \cdot 3

15 = 3 \cdot 5

\nThe LCM is

2^2 \cdot 3 \cdot 5 = 60

. Any number that is a multiple of 60 must be divisible by any factor of 60. Let's check the options to see which one is a factor of 60. \nA) 18 is not a factor of 60. \nB) 24 is not a factor of 60. \nC) 30 is a factor of 60. Therefore, any multiple of 60 must also be a multiple of 30. \nD) 180 is a multiple of 60, not a factor, so not every multiple of 60 is a multiple of 180.

Question 15

Let $G$ be the greatest common divisor of 48 and 72, and let $L$ be the least common multiple of 10 and 12. What is the value of $L - G$ ?

12
24
36 (correct answer)
48

Explanation: This problem requires two separate calculations. First, find

G

, the GCD of 48 and 72. \n

48 = 16 \cdot 3 = 2^4 \cdot 3

72 = 8 \cdot 9 = 2^3 \cdot 3^2

\nGCD(48, 72) =

2^3 \cdot 3^1 = 8 \cdot 3 = 24

. So,

G = 24

. Second, find

L

, the LCM of 10 and 12. \n

10 = 2 \cdot 5

12 = 2^2 \cdot 3

\nLCM(10, 12) =

2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5 = 60

. So,

L = 60

. Finally, calculate

L - G = 60 - 24 = 36

Question 16

The number $K$ is the product of three distinct prime numbers. Which of the following must be the number of positive factors of $K^2$ ?

8
9
12
27 (correct answer)

Explanation: Let the three distinct prime numbers be

p

q

, and

r

. Then

K = p \cdot q \cdot r = p^1 \cdot q^1 \cdot r^1

. Now, we find

K^2

K^2 = (p^1 \cdot q^1 \cdot r^1)^2 = p^2 \cdot q^2 \cdot r^2

. To find the number of factors of

K^2

, we add 1 to each exponent in its prime factorization and multiply the results. The exponents are 2, 2, and 2. So, the number of factors is

(2+1)(2+1)(2+1) = 3 \cdot 3 \cdot 3 = 27

Question 17

The product of two positive integers is 600, and their least common multiple is 120. What is their greatest common divisor?

5 (correct answer)
10
20
50

Explanation: For any two positive integers

a

and

b

, the product of the numbers is equal to the product of their greatest common divisor (GCD) and least common multiple (LCM). The formula is

a \cdot b = \text{GCD}(a, b) \cdot \text{LCM}(a, b)

. We are given that

a \cdot b = 600

and

\text{LCM}(a, b) = 120

. Plugging these values into the formula gives

600 = \text{GCD}(a, b) \cdot 120

. To find the GCD, divide the product by the LCM:

\text{GCD} = 600 \div 120 = 5

Question 18

The number $N$ is defined by its prime factorization, $N = 2^3 \cdot 5^2 \cdot 11$ . Which of the following integers is NOT a factor of $N$ ?

40
55
75 (correct answer)
100

Explanation: A number is a factor of

N

if its prime factorization is a 'subset' of

N

's; that is, it only contains prime factors of

N

raised to powers less than or equal to their powers in

N

. The prime factors of

N

are 2, 5, and 11. \nA)

40 = 8 \cdot 5 = 2^3 \cdot 5^1

. This is a factor of

N

. \nB)

55 = 5^1 \cdot 11^1

. This is a factor of

N

. \nC)

75 = 3 \cdot 25 = 3^1 \cdot 5^2

. This is NOT a factor of

N

because it contains the prime factor 3, which is not a prime factor of

N

. \nD)

100 = 4 \cdot 25 = 2^2 \cdot 5^2

. This is a factor of

N

Question 19

If $x$ and $y$ are positive integers such that $x$ is a multiple of 4 and $y$ is a multiple of 5, then the product $xy$ must be a multiple of which of the following?

10
15
20 (correct answer)
25

Explanation: We are given that

x

is a multiple of 4, so we can write

x = 4a

for some integer

a

. We are also given that

y

is a multiple of 5, so we can write

y = 5b

for some integer

b

. The product

xy

(4a)(5b) = 20ab

. Since

a

and

b

are integers,

ab

is an integer. Therefore,

xy

is always 20 times some integer, which means

xy

must be a multiple of 20.

Question 20

Two positive integers have a sum of 28 and a greatest common divisor of 4. If one of the integers is 12, what is their least common multiple?

48 (correct answer)
56
64
72

Explanation: Let the two integers be 12 and

b

. We are given their sum is 28, so

12 + b = 28

, which gives us

b = 16

. We need to verify that GCD(12, 16) = 4. Finding the prime factorizations:

12 = 2^2 \cdot 3

and

16 = 2^4

. The GCD is

2^2 = 4

, which matches the given condition. Now we find the LCM using the formula: LCM(12, 16) =

\frac{12 \times 16}{\text{GCD}(12, 16)} = \frac{192}{4} = 48