Properties of Integers and Divisibility Practice Questions — GRE General Test — Page 3

Question 41Question →

If $p$ , $p + 2$ , and $p + 4$ are all prime numbers, what is the value of $p^2 + 5$ ?

6

9

14

19

30

Show answer & explanation

Answer: 14

Answer

The value of

p^2 + 5

is 14.

For any three consecutive odd integers

p

,

p + 2

, and

p + 4

, exactly one of them must be a multiple of 3. Because all three expressions represent prime numbers, the term divisible by 3 must be equal to 3 (since 3 is the only prime divisible by 3). Setting

p = 3

gives

p + 2 = 5

and

p + 4 = 7

, both of which are prime. Any choice of

p > 3

forces either

p + 2

or

p + 4

to be a multiple of 3 greater than 3, making it composite. Thus

p = 3

is uniquely determined, and evaluating

p^2 + 5

gives

3^2 + 5 = 14

.

Step-by-Step Solution

1

Analyze the possible remainders when the prime

p

is divided by 3.

Any positive integer

p

can be written in one of three forms:

3k

,

3k + 1

, or

3k + 2

for some integer

k

.

Dividing any integer by 3 leaves a remainder of 0, 1, or 2.

2

Test each remainder case for the expressions

p

,

p + 2

, and

p + 4

.

If

p = 3k + 1

, then

p + 2 = 3k + 3 = 3(k + 1)

, which is divisible by 3. If

p = 3k + 2

, then

p + 4 = 3k + 6 = 3(k + 2)

, which is divisible by 3.

Among any three consecutive odd numbers of the form

p, p+2, p+4

, exactly one of them must be a multiple of 3.

3

Deduce the unique value of

p

.

The only way all three numbers

p

,

p + 2

, and

p + 4

can be prime is if the one divisible by 3 is equal to 3 itself, which forces

p = 3

.

The only prime number divisible by 3 is 3 itself; any larger multiple of 3 is composite.

4

Evaluate the target expression

p^2 + 5

using

p = 3

.

3^2 + 5 = 9 + 5 = 14

.

Substitute the uniquely determined value

p = 3

into the given algebraic expression.

Key Concept

Divisibility properties of consecutive odd integers and prime number definitions

Estimated Time:1m 30s

Question 42Question →

If $n$ is a positive integer that is divisible by $6$ but not divisible by $4$ , and $n^2$ has exactly $15$ positive divisors, what is the value of $n$ ?

6

12

18

24

36

Show answer & explanation

Answer: $18$

Answer

The value of

n

is

18

.

Because

n

is divisible by

6

, it must have prime factors

2

and

3

. Since

n

is not divisible by

4

, the exponent of

2

in

n

is exactly

1

, meaning

n^2

has

2^2

as a factor. The divisor count formula for

n^2

requires

(2+1)(2b+1) = 15

, which gives

2b+1 = 5

, so

b=2

. Thus,

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

.

Step-by-Step Solution

1

Analyze the prime factorization structure of

n

based on given divisibility conditions.

Since

n

is divisible by

6 = 2 \cdot 3

, its prime factorization must contain at least one factor of

2

and one factor of

3

. Since

n

is not divisible by

4 = 2^2

, the exponent of

2

in the prime factorization of

n

must be exactly

1

. Thus,

n = 2^1 \cdot 3^b \cdot k

, where

k

contains no factors of

2

or

3

.

Establishing the exponent of

2

narrows down the search space for prime factor exponents.

2

Express

n^2

in terms of its prime factors and write the formula for its number of positive divisors.

n^2 = 2^2 \cdot 3^{2b} \cdot k^2

. The number of positive divisors of

n^2

is given by

d(n^2) = (2 + 1)(2b + 1) \cdot d(k^2) = 3(2b + 1) \cdot d(k^2) = 15

.

The total number of positive divisors of a number

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

is

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

.

3

Solve for

b

and determine if

k

has any additional prime factors.

Dividing

15

by

3

gives

(2b + 1) \cdot d(k^2) = 5

. Since

5

is prime, we must have

d(k^2) = 1

(meaning

k = 1

) and

2b + 1 = 5

, which yields

2b = 4 \implies b = 2

.

Determining the exponent of

3

fixes the exact prime factorization of

n

.

4

Calculate

n

.

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

.

Multiplying the prime factors together yields the target integer.

Key Concept

Prime Factorization and Divisor Count Formula

Estimated Time:1m 30s

Question 43Question →

If $N = 2^4 \times 3^3 \times 5^2$ , how many positive integer factors of $N$ are divisible by $6$ but not divisible by $12$ ?

Show answer & explanation

Answer: 9

Answer

The number of positive integer factors of

N

divisible by 6 but not by 12 is 9.

Any factor of

N = 2^4 \times 3^3 \times 5^2

is of the form

2^a \times 3^b \times 5^c

. For the factor to be divisible by

6

, we must have

a \ge 1

and

b \ge 1

. For it to not be divisible by

12 = 2^2 \times 3^1

, we must have

a < 2

. Therefore,

a

must equal

1

. The possible choices for

a

are

1

value (

1

), for

b

are

3

values (

1, 2, 3

), and for

c

are

3

values (

0, 1, 2

). Multiplying these options gives

1 \times 3 \times 3 = 9

.

Step-by-Step Solution

1

Express the prime factorization structure of a factor of

N

.

Any factor of

N

takes the form

2^a \times 3^b \times 5^c

with bounds

0 \le a \le 4

,

0 \le b \le 3

, and

0 \le c \le 2

.

Divisors of a number are formed by taking prime factors with exponents between zero and their maximum powers in the original number.

2

Determine exponent constraints for divisibility by 6 and non-divisibility by 12.

Divisibility by

6

requires

a \ge 1

and

b \ge 1

. Non-divisibility by

12

requires

a < 2

. Thus,

a = 1

exactly.

A factor must contain at least one factor of 2 and one factor of 3 to be a multiple of 6, but containing two or more factors of 2 makes it a multiple of 12.

3

Count combinations of choices for the exponents.

Exponent

a

has

1

option (

a = 1

),

b

has

3

options (

b \in \{1, 2, 3\}

), and

c

has

3

options (

c \in \{0, 1, 2\}

). Total

= 1 \times 3 \times 3 = 9

.

Applying the fundamental counting principle by multiplying the number of choices for independent prime factor exponents.

Key Concept

Counting Divisors with Prime Factorization and Divisibility Constraints

Question 44Question →

A positive integer $n$ yields a remainder of $7$ when divided by $12$ . What is the remainder when the expression $5n + 3$ is divided by $6$ ?

0

1

2

3

5

Show answer & explanation

Answer: $2$

Answer

The remainder when

5n + 3

is divided by

6

is

2

.

Since

n

leaves a remainder of

7

when divided by

12

, we can write

n = 12k + 7

for some integer

k \ge 0

. Substituting this into

5n + 3

yields

5(12k + 7) + 3 = 60k + 38

. Factoring out

6

gives

6(10k + 6) + 2

. Because

6(10k + 6)

is divisible by

6

, the remainder of the expression when divided by

6

is

2

.

Step-by-Step Solution

1

Express

n

in terms of the division algorithm for divisor

12

.

n = 12k + 7

for some non-negative integer

k

An integer that leaves a remainder of

7

when divided by

12

can be represented as

12k + 7

.

2

Substitute the expression for

n

into

5n + 3

and simplify.

5(12k + 7) + 3 = 60k + 35 + 3 = 60k + 38

Algebraic expansion allows us to analyze the entire expression modulo

6

.

3

Determine the remainder of

60k + 38

when divided by

6

.

60k + 38 = 6(10k + 6) + 2

Since

60k + 36

is an exact multiple of

6

, the leftover term

2

is the remainder.

Key Concept

Properties of Integers and Remainder Arithmetic

Estimated Time:1m 30s

Question 45Question →

Let $m$ and $n$ be positive integers such that $m$ is divisible by $18$ and $n$ is divisible by $15$ . Which of the following integers MUST be a divisor of the product $m \cdot n$ ? Select all such values.

Select all that apply

54

90

135

36

60

Show answer & explanation

Answer: $54$ ; $90$ ; $135$

Answer

The integers

54

,

90

, and

135

must be divisors of the product

m \cdot n

.

Since

m

is a multiple of

18

(

2^1 \times 3^2

) and

n

is a multiple of

15

(

3^1 \times 5^1

), their product

m \cdot n

must be a multiple of

18 \times 15 = 270

. The prime factorization of

270

is

2^1 \times 3^3 \times 5^1

. Any integer that divides

270

is guaranteed to divide

m \cdot n

for all valid values of

m

and

n

. The numbers

54

(

2^1 \times 3^3

),

90

(

2^1 \times 3^2 \times 5^1

), and

135

(

3^3 \times 5^1

) are all divisors of

270

.

Step-by-Step Solution

1

Express

m

and

n

in terms of their minimal prime factorizations.

m = 18a = 2^1 \times 3^2 \times a

and

n = 15b = 3^1 \times 5^1 \times b

for positive integers

a

and

b

.

Divisibility conditions specify the minimum prime factors that

m

and

n

must contain.

2

Find the minimal guaranteed prime factorization of the product

m \cdot n

.

m \cdot n = (2^1 \times 3^2 \times a) \times (3^1 \times 5^1 \times b) = 2^1 \times 3^3 \times 5^1 \times (ab) = 270 \times ab

.

Multiplying

m

and

n

combines their guaranteed prime factor powers.

3

Determine which choices divide

270 = 2^1 \times 3^3 \times 5^1

without requiring additional factors of

a

or

b

.

54 = 2^1 \times 3^3

,

90 = 2^1 \times 3^2 \times 5^1

, and

135 = 3^3 \times 5^1

all divide

270

. Numbers requiring

2^2

(

36

and

60

) do not necessarily divide

270

.

A number MUST divide

m \cdot n

if its prime factor powers do not exceed the minimum guaranteed powers in

m \cdot n

.

Key Concept

Divisibility of Integer Products via Prime Factorization

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