Properties of Integers and Divisibility Practice Questions — GRE General Test — Page 2

Question 21Question →

If $n$ is a positive integer such that $n$ is divisible by $12$ and $n^2$ is divisible by $180$ , what is the least possible value of $n$ ?

30

60

90

180

360

Show answer & explanation

Answer: 60

Answer

60

The correct answer is 60. The prime factorization of 12 is

2^2 \times 3^1

, requiring

n

to have at least two factors of 2 and one factor of 3. The prime factorization of 180 is

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

. For

n^2

to be divisible by 180,

n

must contribute a factor of 5 (which becomes

5^2

in

n^2

). Multiplying the minimal prime factors yields

2^2 \times 3^1 \times 5^1 = 60

.

Step-by-Step Solution

1

Analyze the prime factorization required for

n

to be divisible by 12

Since

12 = 2^2 \times 3^1

, any positive integer

n

divisible by 12 must contain at least

2^2

and

3^1

in its prime factorization.

An integer must contain all prime factors of its divisor with at least equal exponents.

2

Analyze the prime factorization required for

n^2

to be divisible by 180

Since

180 = 2^2 \times 3^2 \times 5^1

,

n^2

must contain at least

2^2

,

3^2

, and

5^1

. Because

n^2

doubles all prime exponents of

n

,

n

must contribute at least

5^1

, which makes the factor of 5 in

n^2

equal to

5^2

.

The exponent of any prime factor in a perfect square

n^2

must be even.

3

Combine the minimal prime factor requirements for

n

The minimum prime factorization for

n

is

2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60

.

Taking the minimum required power of each prime factor yields the smallest positive integer fulfilling both conditions.

Key Concept

Prime Factorization and Divisibility of Powers

Question 22Question →

When the positive integer $n$ is divided by 7, the remainder is 3, and when $n$ is divided by 11, the remainder is 5. If $n$ is a three-digit integer less than 200 that is divisible by 6, what is the value of $n$ ?

Show answer & explanation

Answer: 192

Answer

The value of

n

is 192.

To find

n

, combine the remainder requirements

n \equiv 3 \pmod 7

and

n \equiv 5 \pmod{11}

. The smallest positive integer solution is

38

. Because

\text{lcm}(7, 11) = 77

, all valid integers take the form

n = 77m + 38

. Restricting

n

to three-digit numbers less than 200 gives candidates

n = 115

(

m=1

) and

n = 192

(

m=2

). Among these, only 192 is divisible by 6 (since

192 = 6 \times 32

).

Step-by-Step Solution

1

Formulate remainder congruences for the given conditions

n \equiv 3 \pmod{7}

and

n \equiv 5 \pmod{11}

Dividing

n

by 7 leaves remainder 3, and dividing by 11 leaves remainder 5.

2

Find the smallest positive integer satisfying both congruences

n = 38

Checking values

11k + 5

:

5, 16, 27, 38

.

38 \div 7 = 5

remainder 3, so 38 satisfies both conditions.

3

Determine the general solution for

n

using the Chinese Remainder Theorem logic

n = 77m + 38

for integer

m \ge 0

Since 7 and 11 are coprime, the solutions repeat every

\text{lcm}(7, 11) = 77

.

4

Evaluate candidate values for

n

such that

100 \le n < 200

For

m=1

,

n = 115

; for

m=2

,

n = 192

These are the only three-digit integers less than 200 of the form

77m + 38

.

5

Apply the final constraint that

n

must be divisible by 6

192

is divisible by 6 (

192 = 6 \times 32

)

115

is odd, so it is not divisible by 6.

192

is even and the sum of its digits (

1+9+2=12

) is a multiple of 3, so it is divisible by 6.

Key Concept

Simultaneous congruences and combined divisibility rules (Chinese Remainder Theorem)

Estimated Time:2m 0s

Question 23Question →

Let $N = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers such that $N$ is a multiple of $8$ . If $N^2$ has $105$ distinct positive divisors and $\frac{N}{6}$ has $12$ distinct positive divisors, how many distinct positive divisors does $10N$ have?

30

36

45

48

60

Show answer & explanation

Answer: 45

Answer

The number of distinct positive divisors of

10N

is 45.

The correct answer is 45. Given

N = 2^a \cdot 3^b \cdot 5^c

, the number of positive divisors of

N^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}

is

(2a+1)(2b+1)(2c+1) = 105

. The unique factor decomposition of 105 into three factors greater than 1 is

3 \times 5 \times 7

. Because

N

is a multiple of 8,

a \ge 3

, so

2a+1 \ge 7

, forcing

2a+1 = 7

, which means

a = 3

. The remaining factor set

\{2b+1, 2c+1\} = \{3, 5\}

gives two cases for

(b, c)

:

(2, 1)

or

(1, 2)

. For

\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c

, the divisor count is

a \cdot b \cdot (c+1) = 3 \cdot b \cdot (c+1) = 12

, so

b(c+1) = 4

. Substituting

(b, c) = (2, 1)

yields

2 \times 2 = 4

, which satisfies this relation. Therefore,

(a, b, c) = (3, 2, 1)

. Now

10N = 2^{a+1} \cdot 3^b \cdot 5^{c+1} = 2^4 \cdot 3^2 \cdot 5^2

. Its total number of positive divisors is

(4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45

.

Step-by-Step Solution

1

Express the divisor count formula for

N^2

d(N^2) = (2a+1)(2b+1)(2c+1) = 105

For an integer with prime factorization

p_1^{e_1} p_2^{e_2} \cdots

, the number of divisors is

(e_1+1)(e_2+1)\cdots

. Here

N^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}

.

2

Determine the value of

a

using the divisibility condition

a = 3

Since

N

is a multiple of

8 = 2^3

, we must have

a \ge 3

, which implies

2a+1 \ge 7

. The prime factorization of

105

into three odd factors greater than

1

is

3 \times 5 \times 7

. Thus,

2a+1 = 7

, giving

a = 3

.

3

Solve for exponents

b

and

c

using the divisor count of

\frac{N}{6}

b = 2

and

c = 1

Since

\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c

, its divisor count is

a \cdot b \cdot (c+1) = 3b(c+1) = 12

, so

b(c+1) = 4

. From step 2,

\{2b+1, 2c+1\} = \{3, 5\}

, so either

(b, c) = (2, 1)

or

(b, c) = (1, 2)

. Testing

(b, c) = (2, 1)

gives

2(1+1) = 4

, which satisfies the equation.

4

Calculate the prime factorization and divisor count of

10N

d(10N) = (4+1)(2+1)(2+1) = 45

Since

10N = 2 \cdot 5 \cdot (2^3 \cdot 3^2 \cdot 5^1) = 2^4 \cdot 3^2 \cdot 5^2

, applying the divisor formula yields

5 \times 3 \times 3 = 45

.

Key Concept

Divisor Count Formula & Prime Factorization Constraints

Estimated Time:2m 30s

Question 24Question →

If $x$ is any integer, which of the following expressions must be divisible by $2$ ?

x^2 + 1

x^2 + x + 4

x^3 + 2x + 1

2x^2 + x + 3

x^2 + 3x + 1

Show answer & explanation

Answer: $x^2 + x + 4$

Answer

The expression

x^2 + x + 4

must be divisible by

2

for all integer values of

x

.

The expression

x^2 + x + 4

can be rewritten as

x(x + 1) + 4

. For any integer

x

, either

x

or

x + 1

is even, so their product

x(x + 1)

is always even (a multiple of

2

). Adding

4

(which is also even) to an even number always yields an even number. Therefore,

x^2 + x + 4

is divisible by

2

for all integer values of

x

.

Step-by-Step Solution

1

Rewrite the expression by factoring out

x

from the first two terms.

x^2 + x + 4 = x(x + 1) + 4

Grouping terms isolates the product of two consecutive integers.

2

Analyze the parity of the product

x(x + 1)

.

Because

x

and

x + 1

are consecutive integers, one of them must be even, making the product

x(x + 1)

an even integer.

The product of any integer and its immediate successor is always even.

3

Add

4

to the even product

x(x + 1)

.

The sum of an even integer and an even integer (

4

) is always an even integer.

Adding an even integer to an even integer preserves divisibility by

2

.

Key Concept

Parity of consecutive integer products and integer divisibility rules

Question 25Question →

Let $k$ be a positive integer. The integer $k$ has exactly $6$ positive divisors, $3k$ has exactly $8$ positive divisors, and $5k$ has exactly $12$ positive divisors. Which of the following could be the value of $k$ ? Indicate all such values.

Select all that apply

18

45

63

81

99

Show answer & explanation

Answer: $18$ ; $63$ ; $99$

Answer

The correct values of

k

are

18

,

63

, and

99

.

The integers

18

,

63

, and

99

are all of the form

k = 3^2 \times q^1 = 9q

, where

q

is a prime number other than

3

or

5

(specifically

q = 2, 7, 11

). Each has

(2+1)(1+1) = 6

divisors,

3k = 3^3 \times q^1

has

(3+1)(1+1) = 8

divisors, and

5k = 3^2 \times 5^1 \times q^1

has

(2+1)(1+1)(1+1) = 12

divisors.

Step-by-Step Solution

1

Analyze the prime factorization form of

k

based on its divisor count

d(k) = 6

.

Since

6 = 6 \times 1

or

3 \times 2

,

k

must be of the form

p^5

or

p^2 q^1

, where

p

and

q

are distinct prime numbers.

The formula for the total number of positive divisors of an integer

n = p_1^{a_1} p_2^{a_2} \dots p_m^{a_m}

is

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

.

2

Test the prime factorization form

k = p^5

.

k = p^5

is impossible.

If

p \neq 3

, then

3k = 3^1 p^5

, which would have

(1+1)(5+1) = 12

divisors (contradicting

d(3k) = 8

). If

p = 3

,

3k = 3^6

has

7

divisors. If

p = 5

,

5k = 5^6

has

7

divisors.

3

Test the prime factorization form

k = p^2 q^1

under different prime identities.

k

must be of the form

3^2 q^1

, where

q

is a prime number distinct from

3

and

5

.

If

p = 3

and

q \neq 3, 5

, then

3k = 3^3 q^1

has

(3+1)(1+1) = 8

divisors, and

5k = 3^2 \times 5^1 \times q^1

has

(2+1)(1+1)(1+1) = 12

divisors. Any other choice for

p

or

q

fails to yield

d(3k) = 8

or

d(5k) = 12

.

4

Evaluate the given options against the required form

k = 9q

where

q

is a prime distinct from

3

and

5

.

18 = 9 \times 2

(

q=2

, valid prime),

63 = 9 \times 7

(

q=7

, valid prime), and

99 = 9 \times 11

(

q=11

, valid prime) are correct.

45 = 9 \times 5

(

q=5

, invalid because

q \neq 5

) and

81 = 3^4

(invalid form) are incorrect.

Only primes

q \notin \{3, 5\}

preserve the required divisor counts for

3k

and

5k

.

Key Concept

Divisors formula and impact of prime multiplication on prime factorization exponents

Estimated Time:2m 30s

Question 26Question →

When the positive integer $n$ is divided by $12$ , the remainder is $7$ . What is the remainder when $n^2 + 3n + 5$ is divided by $12$ ?

1

3

4

7

15

Show answer & explanation

Answer: $3$

Answer

The remainder when

n^2 + 3n + 5

is divided by

12

is

3

.

The correct answer is obtained by replacing

n

with its remainder

7

modulo

12

in the expression

n^2 + 3n + 5

. Evaluating gives

7^2 + 3(7) + 5 = 49 + 21 + 5 = 75

. Dividing

75

by

12

yields

6

with a remainder of

3

.

Step-by-Step Solution

1

Express

n

in terms of its quotient and remainder modulo

12

.

n = 12q + 7

for some non-negative integer

q

, meaning

n \equiv 7 \pmod{12}

.

By the division algorithm, any integer

n

leaving a remainder of

7

when divided by

12

can be represented as

12q + 7

.

2

Substitute

n \equiv 7 \pmod{12}

into the polynomial expression

n^2 + 3n + 5

.

n^2 + 3n + 5 \equiv 7^2 + 3(7) + 5 \pmod{12}

.

Properties of modular arithmetic allow direct substitution of equivalent remainders into polynomial expressions.

3

Evaluate the arithmetic expression.

7^2 + 3(7) + 5 = 49 + 21 + 5 = 75

.

Compute the total numeric value before finding the final remainder modulo

12

.

4

Divide the total sum by

12

to find the final remainder.

75 = 12 \times 6 + 3

, so the remainder is

3

.

A remainder modulo

12

must be an integer in the range from

0

to

11

.

Key Concept

Properties of Remainders and Modular Substitution

Estimated Time:1m 15s

Question 27Question →

Let $n$ be a positive integer such that $n$ is divisible by $12$ and $n^2$ is divisible by $270$ . Which of the following statements MUST be true? Select all such statements.

Select all that apply

n

is divisible by

180

.

n^2

is divisible by

10,800

.

n

is divisible by

24

.

n^3

is divisible by

250,000

.

n

has at least

18

positive divisors.

Show answer & explanation

Answer: $n$ is divisible by $180$ .; $n^2$ is divisible by $10,800$ .; $n$ has at least $18$ positive divisors.

Answer

The statements asserting that

n

is divisible by

180

,

n^2

is divisible by

10,800

, and

n

has at least

18

positive divisors MUST be true.

Because

12 = 2^2 \cdot 3^1

divides

n

,

n

contains at least two factors of

2

and one factor of

3

. Because

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

divides

n^2

,

n^2

must contain at least three factors of

3

and one factor of

5

, meaning

n

itself must contain at least two factors of

3

and one factor of

5

. Therefore,

n

must be a multiple of

2^2 \cdot 3^2 \cdot 5^1 = 180

. Consequently: (1)

n

is divisible by

180

; (2)

n^2

is a multiple of

180^2 = 32,400

, which is divisible by

10,800

; (3)

n

has at least as many divisors as

180

, which has

(2+1)(2+1)(1+1) = 18

positive divisors.

Step-by-Step Solution

1

Determine the prime factorization constraints on

n

from the given divisibility conditions.

Since

12 = 2^2 \cdot 3^1

divides

n

,

v_2(n) \ge 2

and

v_3(n) \ge 1

. Since

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

divides

n^2

,

v_2(n^2) \ge 1

,

v_3(n^2) \ge 3

, and

v_5(n^2) \ge 1

.

Divisibility requires that the exponent of each prime factor in the divisor is less than or equal to the exponent in the dividend.

2

Deduce the minimal exponents of primes in

n

.

v_2(n) \ge 2

,

2 \cdot v_3(n) \ge 3 \implies v_3(n) \ge 2

, and

2 \cdot v_5(n) \ge 1 \implies v_5(n) \ge 1

. Thus,

n

must be a multiple of

2^2 \cdot 3^2 \cdot 5^1 = 180

.

Exponents of prime factors in integer factorizations must be non-negative integers.

3

Evaluate each statement against the minimum required value

n = 180 = 2^2 \cdot 3^2 \cdot 5^1

.

1.

180 \mid n

is true.
2.

n^2

is a multiple of

180^2 = 32,400

, which is divisible by

10,800

(

32,400 = 3 \cdot 10,800

). True.
3.

180

is not divisible by

24

. False.
4.

n^3

has

5^3

, whereas

250,000

requires

5^5

. False.
5. The number of positive divisors of

180

is

(2+1)(2+1)(1+1) = 18

. Any multiple of

180

has at least

18

positive divisors. True.

Testing the baseline integer

n = 180

identifies necessary conditions and counterexamples.

Key Concept

Properties of Integers and Divisibility

Question 28Question →

Let $n$ be a positive integer such that $n$ has exactly 15 positive divisors. If $n$ is divisible by 18, but $n$ is NOT divisible by 8, what is the remainder when $n$ is divided by 7?

1

2

3

4

5

Show answer & explanation

Answer: 2

Answer

The remainder when

n

is divided by 7 is 2.

The total number of divisors of a positive integer with prime factorization

p_1^{a} p_2^{b} \dots

is

(a+1)(b+1)\dots

. Given that

n

has 15 divisors, the possible exponent forms are 14 (since

14+1=15

) or 4 and 2 (since

(4+1)(2+1)=15

). Since

n

is divisible by

18 = 2 \cdot 3^2

, its prime factors must be 2 and 3. The form

p^{14}

is eliminated because it contains only one prime factor. Testing the two permutations for

p^4 \cdot q^2

:
1. If

n = 2^4 \cdot 3^2 = 144

,

n

is divisible by 8 (

144 / 8 = 18

), which contradicts the condition that

n

is NOT divisible by 8.
2. If

n = 2^2 \cdot 3^4 = 324

,

n

is divisible by 18 (

324 / 18 = 18

) and is NOT divisible by 8 (

324 / 8 = 40.5

).
Dividing 324 by 7 gives

324 = 7 \times 46 + 2

, so the remainder is 2.

Step-by-Step Solution

1

Analyze the divisor count formula for

n

The prime factorization of

n

must be either

p^{14}

or

p^4 \cdot q^2

for distinct prime numbers

p

and

q

.

The total number of positive divisors of an integer with prime factorization

p_1^{a} p_2^{b} \dots

is given by

(a+1)(b+1)\dots = 15

. Since 15 factors as

15 \times 1

or

5 \times 3

, the exponent structures are 14 or 4 and 2.

2

Apply the divisibility conditions by 18 and 8

n

must equal

2^2 \cdot 3^4 = 324

.

Because

n

is divisible by

18 = 2^1 \cdot 3^2

, its prime factors must include both 2 and 3, ruling out

p^{14}

. Thus

\{p, q\} = \{2, 3\}

. If

n = 2^4 \cdot 3^2 = 144

, then

n

is divisible by 8 (

2^3

), violating the given constraint. Therefore,

n

must be

2^2 \cdot 3^4 = 324

, which is divisible by 18 (

324 / 18 = 18

) and not divisible by 8 (

324 / 8 = 40.5

).

3

Compute the remainder when 324 is divided by 7

324 divided by 7 yields a quotient of 46 and a remainder of 2.

324 = 7 \times 46 + 2

because

7 \times 46 = 322

.

Key Concept

Prime factorization, divisor count formula, and divisibility constraints

Question 29Question →

Let $x$ be a positive integer. If the expression $x^2 + 3x$ is divisible by $6$ , which of the following statements MUST be true?

x

is divisible by

3

x

is divisible by

6

x

is an even integer

x

is an odd integer

x + 1

is a prime number

Show answer & explanation

Answer: $x$ is divisible by $3$

Answer

x

is divisible by

3

Factoring

x^2 + 3x

yields

x(x+3)

. For any integer

x

, one of

x

or

x+3

is even and the other is odd, so their product

x(x+3)

is always divisible by

2

. For the product to also be divisible by

3

, either

x

or

x+3

must be divisible by

3

. Because

x

and

x+3

differ by

3

,

x+3

is a multiple of

3

if and only if

x

is a multiple of

3

. Thus,

x

must be divisible by

3

.

Step-by-Step Solution

1

Factor the given algebraic expression

x^2 + 3x = x(x + 3)

Factoring allows us to analyze the parity and prime factors of each component separately.

2

Analyze parity (divisibility by 2)

If

x

is even,

x+3

is odd

\Rightarrow

product is even. If

x

is odd,

x+3

is even

\Rightarrow

product is even.

The expression

x(x+3)

is guaranteed to be divisible by

2

for all integers

x

.

3

Analyze divisibility by 3

For

x(x+3)

to be divisible by

3

, at least one factor must be divisible by

3

. Since

x+3 \equiv x \pmod 3

,

x+3

is divisible by

3

if and only if

x

is divisible by

3

.

Since divisibility by

2

is always satisfied, divisibility by

6

requires

x(x+3)

to be divisible by

3

, which requires

x

to be a multiple of

3

.

Key Concept

Divisibility properties of factored integer expressions and modular arithmetic

Estimated Time:1m 15s

Question 30Question →

Let $n$ be a positive integer of the form $n = 2^a \cdot 3^b$ , where $a$ and $b$ are positive integers. If $n$ is divisible by $6$ and $n^2$ has exactly $35$ positive divisors, how many positive divisors does $6n$ have?

12

18

20

24

70

Show answer & explanation

Answer: 20

Answer

The integer

6n

has 20 positive divisors.

The correct answer is 20. Since

n = 2^a \cdot 3^b

, we have

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

. The number of divisors of

n^2

is

(2a+1)(2b+1) = 35

. Given

n

is divisible by 6,

a, b \ge 1

, which means

2a+1 \ge 3

and

2b+1 \ge 3

. The factorizations of 35 are

5 \times 7

and

7 \times 5

, giving exponent values of 2 and 3. Then

6n = 2^{a+1} \cdot 3^{b+1}

, which has

(a+2)(b+2) = (2+2)(3+2) = 20

positive divisors.

Step-by-Step Solution

1

Express

n^2

in terms of its prime factorization.

n^2 = (2^a \cdot 3^b)^2 = 2^{2a} \cdot 3^{2b}

Squaring an integer doubles the exponents of its prime factors.

2

Set up the equation for the number of positive divisors of

n^2

.

(2a + 1)(2b + 1) = 35

The number of positive divisors of

p_1^{e_1} p_2^{e_2}

is given by

(e_1 + 1)(e_2 + 1)

.

3

Solve for the exponents

a

and

b

given that

n

is divisible by 6.

Since

n

is divisible by 6, both

a \ge 1

and

b \ge 1

. Thus

2a + 1 > 1

and

2b + 1 > 1

. The only factor pairs of 35 into two integers greater than 1 are

5 \times 7

or

7 \times 5

. Therefore,

\{2a+1, 2b+1\} = \{5, 7\}

, which yields

\{a, b\} = \{2, 3\}

.

Divisibility by 6 guarantees both prime factors 2 and 3 are present in

n

.

4

Find the prime factorization of

6n

and calculate its number of positive divisors.

6n = (2^1 \cdot 3^1) \cdot (2^a \cdot 3^b) = 2^{a+1} \cdot 3^{b+1}

. Using

\{a, b\} = \{2, 3\}

, the exponents are

a+1

and

b+1

, which equal 3 and 4 (in either order). The number of divisors is

(3+1)(4+1) = 4 \times 5 = 20

.

Adding 1 to each exponent in

n

gives the exponents for

6n

, leading to

(a+2)(b+2) = (2+2)(3+2) = 20

.

Key Concept

Divisor Count Formula and Prime Factorization Constraints

Estimated Time:2m 0s

Question 31Question →

If $k$ is a positive integer such that $k$ is divisible by $15$ and $k^2$ is divisible by $360$ , what is the minimum possible number of positive divisors of $k$ ?

6

8

12

18

24

Show answer & explanation

Answer: 12

Answer

The minimum possible number of positive divisors of

k

is 12.

The minimum possible number of positive divisors occurs when

k

has the smallest possible prime exponents satisfying all divisibility conditions. Since

15 = 3 \times 5

divides

k

,

k

must have prime factors

3^1

and

5^1

. For

k^2

to be divisible by

360 = 2^3 \times 3^2 \times 5^1

,

k^2

must have at least

2^3

, which requires

k

to have at least

2^2 = 4

. Thus, the minimal prime factorization of

k

is

k = 2^2 \times 3^1 \times 5^1 = 60

. Using the divisor count formula

(e_1 + 1)(e_2 + 1)(e_3 + 1)

, we get

(2+1)(1+1)(1+1) = 12

positive divisors.

Step-by-Step Solution

1

Find the prime factorizations of 15 and 360.

15 = 3^1 \times 5^1

and

360 = 2^3 \times 3^2 \times 5^1

.

Expressing numbers in terms of prime factors reveals the required prime exponents for divisibility.

2

Determine the minimum exponents of the prime factors required for

k

.

Since

15

divides

k

,

k

must contain at least

3^1

and

5^1

. For

k^2

to be divisible by

360 = 2^3 \times 3^2 \times 5^1

,

k^2

must contain at least

2^3

, which means

k

must contain at least

2^2

. The required factors

3^2

and

5^1

in

k^2

are automatically provided since

(3^1)^2 = 3^2

and

(5^1)^2 = 5^2

. Thus, the minimal

k

is

2^2 \times 3^1 \times 5^1 = 60

.

Taking the minimum required exponent for each prime factor minimizes the total number of divisors of

k

.

3

Calculate the number of positive divisors of the minimal

k

.

The number of divisors for

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

is

(2 + 1)(1 + 1)(1 + 1) = 3 \times 2 \times 2 = 12

.

The divisor count formula adds 1 to each prime exponent and multiplies the results.

Key Concept

Properties of Integer Divisibility and Divisor Counting

Estimated Time:1m 30s

Question 32Question →

What is the largest two-digit positive integer $n$ such that when $n$ is divided by $4$ , the remainder is $3$ , and when $n$ is divided by $5$ , the remainder is $2$ ?

Show answer & explanation

Answer: 87

Answer

The largest two-digit positive integer satisfying both remainder conditions is 87.

Any integer satisfying both remainder requirements must be of the form

n = 20m + 7

for an integer

m

. Testing values for

m

shows that

m = 4

yields

n = 87

, which is the largest two-digit integer fitting this rule.

Step-by-Step Solution

1

Set up modular arithmetic equations for the given remainder conditions.

n \equiv 3 \pmod 4

and

n \equiv 2 \pmod 5

Translate the verbal description of remainders into mathematical congruence relations.

2

Combine the congruence relations to find the general form of

n

.

n = 20m + 7

for non-negative integers

m

The least common multiple of

4

and

5

is

20

, meaning solutions repeat every

20

units.

3

Find the maximum integer

m

that produces a two-digit integer.

For

m = 4

,

n = 87

. For

m = 5

,

n = 107

.

Two-digit integers are strictly less than

100

.

Key Concept

Simultaneous Remainders and Divisibility Cycles

Estimated Time:1m 30s

Question 33Question →

Let $n$ be a positive integer that is a factor of $180$ . If $n$ is divisible by $6$ but is not divisible by $4$ , what is the maximum possible number of positive divisors of $n$ ?

6

8

12

18

24

Show answer & explanation

Answer: 12

Answer

12

The prime factorization of

180

is

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

. Any factor

n

of

180

is of the form

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

. For

n

to be divisible by

6

, we must have

a \ge 1

and

b \ge 1

. For

n

not to be divisible by

4

, we must have

a < 2

, which forces

a = 1

. To maximize the total number of divisors

(a+1)(b+1)(c+1)

, we select the largest possible values for

b

and

c

, which are

b = 2

and

c = 1

. This yields

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

, giving

(1+1)(2+1)(1+1) = 12

positive divisors.

Step-by-Step Solution

1

Find the prime factorization of 180.

180 = 2^2 \times 3^2 \times 5^1

Any positive factor

n

of

180

must have the form

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

, where

0 \le a \le 2

,

0 \le b \le 2

, and

0 \le c \le 1

.

2

Apply the divisibility conditions to determine the possible values of the exponent

a

.

a = 1

Since

n

is divisible by

6 = 2 \times 3

,

a \ge 1

and

b \ge 1

. Since

n

is not divisible by

4 = 2^2

,

a < 2

. Thus,

a

must equal

1

.

3

Maximize the divisor count formula

(a+1)(b+1)(c+1)

using the available ranges for

b

and

c

.

b = 2

and

c = 1

To maximize the number of positive divisors, choose the maximum allowable values for

b

(

b=2

) and

c

(

c=1

).

4

Calculate the maximum number of positive divisors.

(1+1)(2+1)(1+1) = 2 \times 3 \times 2 = 12

For

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

, the total number of positive divisors is

12

.

Key Concept

Divisor count formula and prime factor constraints

Estimated Time:1m 30s

Question 34Question →

Let $m$ and $n$ be positive integers such that $72^2 \cdot m = n^3$ . Which of the following statements MUST be true? Select all that apply.

Select all that apply

m

is divisible by

9

n

is divisible by

36

m

must be a perfect square

n

is divisible by

54

The product

m \cdot n

is divisible by

108

Show answer & explanation

Answer: $m$ is divisible by $9$ ; $n$ is divisible by $36$ ; The product $m \cdot n$ is divisible by $108$

Answer

The statements that

m

is divisible by

9

,

n

is divisible by

36

, and the product

m \cdot n

is divisible by

108

MUST be true.

Analyzing prime factorizations shows that

n^3 = 2^6 \cdot 3^4 \cdot m

. For the right-hand side to be a perfect cube, the power of

3

must be elevated to at least

6

, requiring

m

to contain

3^2 = 9

as a factor. Consequently,

n^3

contains at least

2^6 \cdot 3^6 = 36^3

, forcing

n

to be a multiple of

36

. Finally, multiplying

m = 9k^3

by

n = 36k

yields

324k^4

, which is a multiple of

108

for all integer values of

k

.

Step-by-Step Solution

1

Express

72^2

in terms of its prime factorization.

72 = 2^3 \cdot 3^2 \implies 72^2 = (2^3 \cdot 3^2)^2 = 2^6 \cdot 3^4

Prime factorization allows us to analyze the exponents required for

n^3

to be a perfect cube.

2

Determine the minimum prime factor requirements for

m

and

n

.

2^6 \cdot 3^4 \cdot m = n^3 \implies

exponent of

2

in

m

must be

\ge 0

(a multiple of 3), exponent of

3

in

m

must be

\ge 2

(since

4 + 2 = 6

is a multiple of 3).

Every prime factor in a perfect cube must have an exponent divisible by 3.

3

Establish general algebraic expressions for

m

and

n

.

m = 9k^3

and

n = 36k

for any positive integer

k

.

This captures all possible integer solutions for

m

and

n

.

4

Test each statement using the general forms

m = 9k^3

and

n = 36k

.

m

is divisible by

9

(

9k^3

is a multiple of

9

);

n

is divisible by

36

(

36k

is a multiple of

36

);

m \cdot n = 324k^4 = 108(3k^4)

, which is divisible by

108

. Counterexamples show

m

need not be a square (

m=72

) and

n

need not be divisible by

54

(

n=36

).

Verifies which properties hold universally versus which can fail.

Key Concept

Prime Factor Exponents in Perfect Powers and Divisibility Rules

Question 35Question →

Let $N = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. The integer $N$ is divisible by $24$ , has exactly $36$ positive integer divisors, and $\frac{N}{5}$ is not divisible by $25$ . What is the least possible value of $a + b + c$ ?

6

7

8

9

10

Show answer & explanation

Answer: $7$

Answer

7

The correct answer is

7

. Since

24 = 2^3 \cdot 3^1

divides

N

,

a \ge 3

and

b \ge 1

. The condition that

\frac{N}{5}

is not divisible by

25

restricts

c

to either

1

or

2

. For

c=2

, the divisor equation becomes

(a+1)(b+1)(3) = 36

, so

(a+1)(b+1) = 12

. Taking

a+1=4

(

a=3

) and

b+1=3

(

b=2

) satisfies

a \ge 3

and gives a minimal sum of

a+b+c = 3+2+2 = 7

.

Step-by-Step Solution

1

Analyze divisibility conditions on exponents

Since

N

is divisible by

24 = 2^3 \cdot 3^1

, we must have

a \ge 3

and

b \ge 1

. Since

\frac{N}{5} = 2^a \cdot 3^b \cdot 5^{c-1}

is not divisible by

25 = 5^2

,

c-1 < 2 \implies c \le 2

. Because

c

is a positive integer,

c

can be

1

or

2

.

Establish structural constraints on prime exponents from the divisibility statements.

2

Set up the total divisor formula

The total number of positive integer divisors of

N

is

(a+1)(b+1)(c+1) = 36

.

Apply the standard formula for counting divisors using prime factorization.

3

Evaluate candidate values for

c = 1

and

c = 2

If

c = 1

, then

c+1 = 2

, so

(a+1)(b+1) = 18

. Since

a \ge 3 \implies a+1 \ge 4

, possible integer pairs for

(a+1, b+1)

are

(6, 3)

giving

a=5, b=2 \implies a+b+c = 8

, or

(9, 2)

giving

a=8, b=1 \implies a+b+c = 10

.
If

c = 2

, then

c+1 = 3

, so

(a+1)(b+1) = 12

. With

a+1 \ge 4

, possible pairs for

(a+1, b+1)

are

(4, 3)

giving

a=3, b=2 \implies a+b+c = 3+2+2 = 7

, or

(6, 2)

giving

a=5, b=1 \implies a+b+c = 8

.

Test allowable values of

c

to find all valid exponent triples

(a, b, c)

and minimize their sum.

4

Identify the minimum sum

The minimum sum of

a + b + c

is

7

, achieved when

a = 3

,

b = 2

, and

c = 2

.

Compare sums across all valid exponent combinations.

Key Concept

Prime Factorization, Divisibility Rules, and Number of Divisors Formula

Estimated Time:2m 0s

Question 36Question →

Let $a$ and $b$ be positive integers such that $15a = 28b$ . Which of the following statements MUST be true? Select all that apply.

Select all that apply

a

is divisible by

7

b

is divisible by

15

a

must be an odd integer

b

must be a prime number

a + b

must be divisible by

30

Show answer & explanation

Answer: $a$ is divisible by $7$ ; $b$ is divisible by $15$

Answer

The statements asserting that

a

is divisible by

7

and that

b

is divisible by

15

must be true.

Equating prime factorizations gives

(3 \cdot 5) \cdot a = (2^2 \cdot 7) \cdot b

. Because

15

and

28

share no common prime factors,

a

must contain

2^2 \cdot 7 = 28

as a factor, and

b

must contain

3 \cdot 5 = 15

as a factor. Therefore,

a

is divisible by

7

(since

7

divides

28

) and

b

is divisible by

15

.

Step-by-Step Solution

1

Analyze the prime factorization of both sides of the equation.

The given equation

15a = 28b

can be rewritten as

(3 \cdot 5) \cdot a = (2^2 \cdot 7) \cdot b

.

Prime factorizations allow us to identify fundamental divisibility relationships between variables.

2

Determine the prime factors required for

a

and

b

.

a

must be divisible by

28

(i.e.,

a = 28k

for some integer

k \ge 1

), and

b

must be divisible by

15

(i.e.,

b = 15k

for the same integer

k

).

Since

\gcd(15, 28) = 1

, all prime factors of

28

(

2^2

and

7

) must come from

a

, and all prime factors of

15

(

3

and

5

) must come from

b

.

3

Evaluate each choice against these general forms.

Since

a = 28k

,

a

is a multiple of

7

, making the first statement true. Since

b = 15k

,

b

is a multiple of

15

, making the second statement true.

Any factor of

28

is automatically a factor of

a

, and any factor of

15

is automatically a factor of

b

.

Key Concept

Fundamental Theorem of Arithmetic and Coprime Divisibility Rules

Question 37Question →

Let $N$ be a positive integer with exactly $12$ positive divisors. If the sum of the distinct prime factors of $N$ is $12$ and $N$ is not divisible by $4$ , what is the least possible value of $N$ ?

Show answer & explanation

Answer: 126

Answer

126

The least possible value of

N

is 126. The only sets of distinct prime factors summing to 12 are

\{5, 7\}

and

\{2, 3, 7\}

. Using

\{2, 3, 7\}

produces smaller candidates. Because

N

is not divisible by 4, the exponent of 2 must be 1. The total divisor condition

(1+1)(b+1)(c+1) = 12

requires

(b+1)(c+1) = 6

, giving exponent pairs

(1, 2)

or

(2, 1)

for bases 3 and 7. Assigning the exponent 2 to 3 and 1 to 7 minimizes

N

, giving

2^1 \cdot 3^2 \cdot 7^1 = 126

.

Step-by-Step Solution

1

Find all sets of distinct prime factors that sum to 12.

The possible sets of distinct prime factors are

\{5, 7\}

and

\{2, 3, 7\}

.

Testing combinations of prime numbers (

2, 3, 5, 7, 11, \dots

):

5 + 7 = 12

and

2 + 3 + 7 = 12

are the only valid sets of distinct primes summing to 12.

2

Analyze the set of prime factors

\{2, 3, 7\}

under the condition that

N

is not divisible by 4.

The prime factorization is

N = 2^a \cdot 3^b \cdot 7^c

, where

a = 1

.

Since 2 is a prime factor of

N

,

a \ge 1

. Because

N

is not divisible by 4 (

2^2

),

a

must be strictly less than 2. Thus,

a = 1

.

3

Determine the exponents

b

and

c

using the total number of positive divisors.

The number of divisors is

(1+1)(b+1)(c+1) = 12

, which simplifies to

(b+1)(c+1) = 6

. The possible integer pairs

(b, c)

for

b, c \ge 1

are

(1, 2)

and

(2, 1)

.

The divisor count formula for

N = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}

is

(e_1+1)(e_2+1)\dots(e_k+1) = 12

.

4

Calculate values of

N

for these exponent pairs.

For

(b, c) = (1, 2)

,

N = 2^1 \cdot 3^1 \cdot 7^2 = 294

. For

(b, c) = (2, 1)

,

N = 2^1 \cdot 3^2 \cdot 7^1 = 126

.

To make

N

as small as possible, assign the larger exponent to the smaller prime base (

3^2 \cdot 7^1 < 3^1 \cdot 7^2

).

5

Compare with candidate values from the alternative set of prime factors

\{5, 7\}

.

For

\{5, 7\}

,

N = 5^3 \cdot 7^2 = 6125

or

5^5 \cdot 7^1 = 21875

, both of which are much larger than 126.

The combination

(a+1)(b+1) = 12

yields exponents of 3 and 2 (or 5 and 1), resulting in much higher prime powers.

Key Concept

Divisor count formula combined with prime factorization constraints

Question 38Question →

If $x$ is a positive integer such that $120$ is a factor of $x^2$ , what is the least possible value of $x$ ?

60

30

120

15

360

Show answer & explanation

Answer: 60

Answer

60

The prime factorization of 120 is

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

. For

x^2

to be divisible by 120,

x^2

must contain at least three 2s, one 3, and one 5. Because

x^2

is a perfect square, all exponents in its prime factorization must be even numbers. Thus, the minimum prime factorization for

x^2

is

2^4 \times 3^2 \times 5^2

, which means

x

must be at least

2^2 \times 3^1 \times 5^1 = 60

.

Step-by-Step Solution

1

Find the prime factorization of 120.

120 = 2^3 \times 3^1 \times 5^1

Decomposing 120 into prime factors reveals the minimal prime factor exponents required for

x^2

.

2

Determine the prime factors required for

x^2

to be divisible by 120.

The prime factorization of

x^2

must contain even exponents for all its prime factors and must be divisible by

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

. Therefore,

x^2

must contain at least

2^4 \times 3^2 \times 5^2

.

Because

x^2

is a perfect square, every prime factor in its prime factorization must have an even exponent.

3

Calculate the minimum value of

x

.

x = \sqrt{2^4 \times 3^2 \times 5^2} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60

Taking the square root of

2^4 \times 3^2 \times 5^2

gives the least positive integer

x

satisfying the condition.

Key Concept

Prime Factorization and Perfect Squares

Question 39Question →

How many positive integers $n$ less than $1,000$ are divisible by $12$ , leave a remainder of $4$ when divided by $5$ , and are not divisible by $9$ ?

11

12

13

15

17

Show answer & explanation

Answer: 12

Answer

12

The problem asks for positive integers

n < 1,000

that satisfy three conditions:

n = 12k

,

n \equiv 4 \pmod 5

, and

9 \nmid n

. Expressing

n = 12k

and substituting into the modular equation yields

2k \equiv 4 \pmod 5

, which simplifies to

k \equiv 2 \pmod 5

. For

k \le 83

, setting

k = 5m + 2

yields

17

values for

m \in \{0, 1, \dots, 16\}

. Furthermore,

12k

is divisible by

9

if and only if

k

is a multiple of

3

, which corresponds to

m \equiv 2 \pmod 3

. The values

m = 2, 5, 8, 11, 14

(

5

values) must be excluded. Subtracting

5

from

17

leaves exactly

12

valid integers.

Step-by-Step Solution

1

Express

n

in terms of its divisibility by

12

and set the upper bound.

n = 12k

for a positive integer

k

. Since

n < 1,000

, we have

12k < 1,000

, which implies

k \le 83

.

Any positive integer divisible by

12

must be a positive integer multiple of

12

.

2

Apply the modular remainder condition modulo

5

.

12k \equiv 4 \pmod 5 \implies 2k \equiv 4 \pmod 5 \implies k \equiv 2 \pmod 5

.

Since

12 \equiv 2 \pmod 5

, simplifying the congruence gives

k \equiv 2 \pmod 5

.

3

Determine the candidate set of values for

k

.

k

can be written as

k = 5m + 2

. For

1 \le k \le 83

, the non-negative integer

m

ranges from

0

to

16

inclusive, yielding

17

possible values.

The maximum value

5m + 2 \le 83

gives

5m \le 81 \implies m \le 16

.

4

Apply the condition that

n

is not divisible by

9

.

12k = 3 \times 4 \times k

is divisible by

9

if and only if

k

is divisible by

3

.

5m + 2 \equiv 2m + 2 \pmod 3

, so

2m + 2 \equiv 0 \pmod 3 \implies m \equiv 2 \pmod 3

.

Since

12

already contains one factor of

3

,

12k

needs

k

to supply another factor of

3

to be divisible by

9

.

5

Filter out the excluded values of

m

and calculate the final count.

Values of

m \in \{0, 1, \dots, 16\}

with

m \equiv 2 \pmod 3

are

m \in \{2, 5, 8, 11, 14\}

(

5

values). The number of valid integers is

17 - 5 = 12

.

Subtracting the excluded multiples of

9

from the candidate set leaves the exact number of valid integers.

Key Concept

Modular arithmetic, linear congruences, and counting integer multiples with divisibility constraints

Estimated Time:2m 0s

Question 40Question →

If $k$ is a positive integer such that $k$ is divisible by $6$ and $k + 1$ is divisible by $5$ , what is the remainder when $k^2 + 5k$ is divided by $30$ ?

0

4

6

14

24

Show answer & explanation

Answer: $6$

Answer

The remainder when

k^2 + 5k

is divided by

30

is

6

.

The correct answer is

6

. Since

k

is a multiple of

6

and leaves a remainder of

4

when divided by

5

, the general form of

k

modulo

30

is

24

. Evaluating

k^2 + 5k

modulo

30

yields

24^2 + 5(24) = 576 + 120 = 696

, which gives a remainder of

6

when divided by

30

.

Step-by-Step Solution

1

Determine the congruence class of

k

modulo 6 and modulo 5.

k \equiv 0 \pmod 6

and

k \equiv 4 \pmod 5

.

Given that

k

is divisible by 6, its remainder modulo 6 is 0. Since

k + 1

is divisible by 5,

k + 1 \equiv 0 \pmod 5

, which implies

k \equiv 4 \pmod 5

.

2

Find the smallest positive integer value of

k

modulo 30 satisfying both conditions.

k \equiv 24 \pmod{30}

.

The multiples of 6 are 0, 6, 12, 18, 24, 30, ... Among these, 24 gives a remainder of 4 when divided by 5. Since 5 and 6 are coprime,

k \equiv 24 \pmod{30}

.

3

Substitute

k \equiv 24 \pmod{30}

into the expression

k^2 + 5k

and compute the remainder modulo 30.

The remainder is

6

.

k^2 + 5k = k(k+5) \equiv 24(24+5) = 24(29) \pmod{30}

. Using modular arithmetic,

24 \equiv -6 \pmod{30}

and

29 \equiv -1 \pmod{30}

, so

(-6)(-1) = 6 \pmod{30}

.

Key Concept

Chinese Remainder Theorem and Modular Arithmetic Properties

Estimated Time:1m 30s

Properties of Integers and Divisibility

Answer

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Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept