Properties of Integers and Divisibility Practice Questions — GRE General Test

Question 1Question →

How many positive integers $n$ less than $30$ satisfy the condition that $n^2 - 1$ is divisible by $24$ ?

5

8

9

10

15

Show answer & explanation

Answer: 10

Answer

10

For

n^2 - 1

to be divisible by

24 = 8 \times 3

,

n

must be an odd integer that is not divisible by 3, which means

\gcd(n, 6) = 1

. The positive integers less than 30 coprime to 6 are 1, 5, 7, 11, 13, 17, 19, 23, 25, and 29, totaling 10 integers.

Step-by-Step Solution

1

Factor the expression and analyze divisibility requirements for 24.

n^2 - 1 = (n - 1)(n + 1)

. Since

24 = 8 \times 3

with

\gcd(8, 3) = 1

,

n^2 - 1

must be divisible by both

8

and

3

.

Decompose 24 into coprime prime-power factors to evaluate modular conditions independently.

2

Determine the condition for

n^2 - 1

to be divisible by 8.

If

n

is even,

n^2 - 1

is odd and cannot be divisible by 8. If

n

is odd, let

n = 2k + 1

; then

n^2 - 1 = 4k(k + 1)

. Since one of

k

or

k + 1

is always even,

4k(k + 1)

is divisible by 8. Thus,

n

must be odd.

Analyze parity requirements for the power of 2.

3

Determine the condition for

n^2 - 1

to be divisible by 3.

If

n

is a multiple of 3,

n^2 - 1 \equiv -1 \pmod 3

, which is not divisible by 3. If

n

is not a multiple of 3,

n \equiv 1

or

2 \pmod 3

, so

n^2 \equiv 1 \pmod 3

and

n^2 - 1

is divisible by 3. Thus,

n

cannot be a multiple of 3.

Analyze remainder properties modulo 3.

4

Combine the conditions and count valid positive integers

n < 30

.

The combined requirement is

\gcd(n, 6) = 1

(

n

is odd and not a multiple of 3). The positive integers less than 30 satisfying this are 1, 5, 7, 11, 13, 17, 19, 23, 25, and 29. Counting these yields 10 integers.

Enumerate all integers meeting the coprime condition within the specified domain.

Key Concept

Modular arithmetic properties of quadratic expressions and coprimality constraints

Question 2Question →

How many positive integers less than $100$ are divisible by both $4$ and $6$ , but are NOT divisible by $8$ ?

Show answer & explanation

Answer: 4

Answer

The correct numerical answer is 4.

To be divisible by both 4 and 6, an integer must be a multiple of

\text{LCM}(4, 6) = 12

. The positive integers less than 100 that are multiples of 12 are 12, 24, 36, 48, 60, 72, 84, and 96 (8 integers). Among these, those divisible by 8 are multiples of

\text{LCM}(12, 8) = 24

, which are 24, 48, 72, and 96 (4 integers). Subtracting the excluded integers yields

8 - 4 = 4

.

Step-by-Step Solution

1

Find the least common multiple of 4 and 6.

LCM(4, 6) = 12

An integer divisible by both 4 and 6 must be a multiple of their least common multiple.

2

Count positive integers less than 100 that are multiples of 12.

The multiples are 12, 24, 36, 48, 60, 72, 84, and 96, giving 8 integers.

The largest multiple of 12 strictly less than 100 is 96 (12 × 8).

3

Identify multiples of 12 that are also divisible by 8.

Since LCM(12, 8) = 24, these are the multiples of 24: 24, 48, 72, and 96, giving 4 integers.

Any integer divisible by both 12 and 8 must be a multiple of 24.

4

Subtract the excluded integers from the total count.

8 - 4 = 4

We exclude the multiples of 8 from the set of multiples of 12.

Key Concept

Divisibility, Least Common Multiple (LCM), and Set Exclusion

Question 3Question →

What is the least positive integer $n$ that leaves a remainder of $3$ when divided by $7$ , a remainder of $4$ when divided by $5$ , and is divisible by $9$ ?

Show answer & explanation

Answer: 234

Answer

The least positive integer satisfying all three conditions is 234.

To find the least positive integer

n

that satisfies

n \equiv 3 \pmod{7}

,

n \equiv 4 \pmod{5}

, and

n \equiv 0 \pmod{9}

, we first find a general expression for integers meeting the first two conditions. Checking values of

5m + 4

modulo 7 gives

24

as the smallest positive integer matching both. The combined condition is

n \equiv 24 \pmod{35}

, or

n = 35k + 24

. Requiring

35k + 24

to be divisible by 9 gives

8k + 6 \equiv 0 \pmod{9}

, which simplifies to

k \equiv 6 \pmod{9}

. The smallest non-negative integer value for

k

is

6

, leading to

n = 35(6) + 24 = 234

.

Step-by-Step Solution

1

Set up system of modular congruences for the remainders

n \equiv 3 \pmod{7}

,

n \equiv 4 \pmod{5}

, and

n \equiv 0 \pmod{9}

Translates the remainder and divisibility conditions into mathematical equations.

2

Combine the first two congruences using the Chinese Remainder Theorem approach

n \equiv 24 \pmod{35}

, so

n = 35k + 24

for an integer

k \ge 0

Since

\text{lcm}(5, 7) = 35

, the solutions to the combined system repeat every 35 integer values.

3

Enforce the divisibility condition by 9 on

n = 35k + 24

35k + 24 \equiv 0 \pmod{9} \implies 8k + 6 \equiv 0 \pmod{9} \implies -k + 6 \equiv 0 \pmod{9} \implies k \equiv 6 \pmod{9}

Reduces coefficients modulo 9 to find the values of

k

that make

n

a multiple of 9.

4

Calculate the smallest positive integer

n

corresponding to

k = 6

n = 35(6) + 24 = 234

Choosing

k = 6

yields the smallest non-negative integer for

k

that satisfies all conditions.

Key Concept

Simultaneous congruences and divisibility constraints

Question 4Question →

If $n$ is an integer that is divisible by $15$ , which of the following statements MUST also be true? Select all such statements.

Select all that apply

n

is divisible by

3

n

is divisible by

5

n + 15

is divisible by

15

n

is divisible by

30

n

is an even integer

Show answer & explanation

Answer: $n$ is divisible by $3$ ; $n$ is divisible by $5$ ; $n + 15$ is divisible by $15$

Answer

The statements asserting that

n

is divisible by

3

,

n

is divisible by

5

, and

n + 15

is divisible by

15

must all be true.

Any integer divisible by

15

must be divisible by all factors of

15

, which includes

3

and

5

. Furthermore, adding

15

to a multiple of

15

yields another multiple of

15

, so

n + 15

is also divisible by

15

.

Step-by-Step Solution

1

Analyze the prime factor decomposition of the divisor

The prime factorization of

15

is

3 \times 5

. Therefore, any integer

n

that is a multiple of

15

can be expressed as

n = 15k = 3 \times 5 \times k

for some integer

k

.

Understanding factor relationships allows evaluation of divisibility rules.

2

Evaluate divisibility of

n

by

3

and

5

Since

n = 3(5k)

,

n

is divisible by

3

. Since

n = 5(3k)

,

n

is divisible by

5

. Both statements are always true.

Any multiple of a composite number is also a multiple of that number's factors.

3

Evaluate the expression

n + 15

n + 15 = 15k + 15 = 15(k + 1)

. Since

k + 1

is an integer,

n + 15

is a multiple of

15

. This statement is always true.

Adding a multiple of

15

to another multiple of

15

yields a multiple of

15

.

4

Test counterexamples for remaining statements

For

n = 15

:

15

is not divisible by

30

, so divisibility by

30

is not guaranteed. Additionally,

15

is odd, so being an even integer is not guaranteed.

A statement must hold for all possible values of

n

to be necessarily true.

Key Concept

Divisibility by a composite number implies divisibility by all of its factors, and multiples of a number remain multiples when another multiple of that number is added.

Question 5Question →

If $n$ is a positive integer such that $n^2$ is divisible by 72, what is the smallest possible value of $n$ ?

6

12

18

24

36

Show answer & explanation

Answer: 12

Answer

12

The prime factorization of 72 is

2^3 \times 3^2

. For

n^2

to be a multiple of 72, the prime factorization of

n^2

must contain at least three factors of 2 and two factors of 3. Since

n^2

is a perfect square, the exponents of all prime factors in

n^2

must be even. Therefore, the exponent of 2 in

n^2

must be at least 4, making the minimum value of

n^2

equal to

2^4 \times 3^2 = 144

. Taking the square root yields

n = 12

, which is the smallest positive integer.

Step-by-Step Solution

1

Find the prime factorization of 72.

72 = 2^3 \times 3^2

.

Decomposing 72 into prime factors reveals the minimum prime factor requirements for

n^2

.

2

Determine the prime factorization required for

n^2

.

For

n^2

to be divisible by

2^3 \times 3^2

, its prime factorization must contain at least three factors of 2 and two factors of 3. Because

n^2

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

n^2

must contain at least

2^4 \times 3^2 = 144

.

Exponents of prime factors in any perfect square must be even integers.

3

Take the square root to find the minimum value of

n

.

n = \sqrt{2^4 \times 3^2} = 2^2 \times 3^1 = 12

.

The smallest positive integer

n

must equal the square root of the minimal perfect square multiple of 72.

Key Concept

Prime Factorization and Divisibility of Perfect Squares

Question 6Question →

Let $p$ and $q$ be positive integers such that $18$ is a factor of $p$ and $45$ is a factor of $q$ . Which of the following statements MUST be true? Select all such statements.

Select all that apply

The sum

p + q

is divisible by

9

.

The product

p \cdot q

is divisible by

270

.

The sum

p + q

is divisible by

18

.

The product

p \cdot q

is divisible by

1620

.

The sum

p^2 + q^2

is divisible by

81

.

Show answer & explanation

Answer: The sum $p + q$ is divisible by $9$ .; The product $p \cdot q$ is divisible by $270$ .; The sum $p^2 + q^2$ is divisible by $81$ .

Answer

The statements asserting that the sum

p + q

is divisible by

9

, the product

p \cdot q

is divisible by

270

, and the sum

p^2 + q^2

is divisible by

81

MUST be true.

The correct options are those stating that the sum

p + q

is divisible by

9

, the product

p \cdot q

is divisible by

270

, and the sum

p^2 + q^2

is divisible by

81

. Each of these statements can be proven algebraically by substituting

p = 18a

and

q = 45b

into the expressions and factoring out the required divisor.

Step-by-Step Solution

1

Express

p

and

q

in terms of their prime factorizations and integer multipliers.

p = 18a = (2 \cdot 3^2)a

and

q = 45b = (3^2 \cdot 5)b

for positive integers

a

and

b

.

Establishing explicit algebraic representations allows verification of divisibility for any combination of

p

and

q

.

2

Evaluate the divisibility of the sum

p + q

.

p + q = 18a + 45b = 9(2a + 5b)

.

Factoring out

9

proves that

p + q

is always divisible by

9

. However,

2a + 5b

is not necessarily even (e.g., if

a=1, b=1

,

2a+5b=7

), so

p+q

is not guaranteed to be divisible by

18

.

3

Evaluate the divisibility of the product

p \cdot q

.

p \cdot q = (18a)(45b) = 810ab = 270(3ab)

.

Since

810ab

is a multiple of

270

, the product is always divisible by

270

. However, it is not necessarily a multiple of

1620

unless

ab

contains an additional factor of

2

.

4

Evaluate the divisibility of the sum of squares

p^2 + q^2

.

p^2 + q^2 = (18a)^2 + (45b)^2 = 324a^2 + 2025b^2 = 81(4a^2 + 25b^2)

.

Factoring out

81

proves that

p^2 + q^2

is always divisible by

81

.

Key Concept

Divisibility rules and algebraic factoring of linear and quadratic integer expressions

Question 7Question →

What is the smallest positive integer $n$ such that $n$ is a multiple of $180$ , the only prime factors of $n$ are $2$ , $3$ , and $5$ , and $n$ has exactly $36$ positive integer divisors?

Show answer & explanation

Answer: 1440

Answer

1440

To minimize

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

subject to

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

with

a \ge 2

,

b \ge 2

, and

c \ge 1

, we evaluate all valid factor partitions of

36

. The partition

(a+1, b+1, c+1) = (6, 3, 2)

yields

(a, b, c) = (5, 2, 1)

, giving

n = 2^5 \cdot 3^2 \cdot 5^1 = 1440

, which is the smallest possible integer satisfying all criteria.

Step-by-Step Solution

1

Express

n

in terms of prime factorization and establish exponent inequalities.

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

with

a \ge 2

,

b \ge 2

, and

c \ge 1

.

Since

n

is a multiple of

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

and contains no other prime factors, its prime powers must at least match those of

180

.

2

Set up the divisor count equation.

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

, where

a+1 \ge 3

,

b+1 \ge 3

, and

c+1 \ge 2

.

The total number of positive divisors of

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

is given by

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

.

3

Determine all valid factorizations of

36

into three factors

(x, y, z) = (a+1, b+1, c+1)

.

The valid factor sets

\{x, y, z\}

satisfying

x, y \ge 3

and

z \ge 2

are

\{4, 3, 3\}

and

\{6, 3, 2\}

.

Factor sets containing a factor of

2

for

x

or

y

(such as

\{9, 2, 2\}

) are invalid because

a+1 \ge 3

and

b+1 \ge 3

.

4

Calculate the value of

n

for all valid assignments of exponents.

From

\{4, 3, 3\}

:

(3, 2, 2) \implies 1800

,

(2, 3, 2) \implies 2700

,

(2, 2, 3) \implies 4500

.
From

\{6, 3, 2\}

:

(5, 2, 1) \implies 1440

,

(2, 5, 1) \implies 4860

.

Assigning larger exponents to smaller prime bases minimizes the overall product.

5

Select the minimum integer value among all candidates.

The smallest value is

1440

.

Comparing all valid candidates

1440, 1800, 2700, 4500, 4860

, the minimum is

1440

.

Key Concept

Prime Factorization and Divisor Count Constraints

Estimated Time:2m 30s

Question 8Question →

When the positive integer $n$ is divided by $12$ , the remainder is $7$ . Which of the following could be the remainder when $n^2 + 5n + 2$ is divided by $24$ ?

1

2

5

7

11

Show answer & explanation

Answer: 2

Answer

2

Because

n \equiv 7 \pmod{12}

,

n

can be expressed as

12k + 7

. Testing the two cases for the integer

k

(even vs. odd) shows that

n \equiv 7 \pmod{24}

or

n \equiv 19 \pmod{24}

. Substituting

n = 19

into

n^2 + 5n + 2

yields

458

, and

458 = 19 \times 24 + 2

, which leaves a remainder of

2

. Therefore, the value

2

is a valid possible remainder.

Step-by-Step Solution

1

Express

n

algebraically based on the given remainder condition.

Since

n

leaves a remainder of

7

when divided by

12

,

n

can be written as

n = 12k + 7

for some non-negative integer

k

.

By the division algorithm, any integer leaving remainder

r

when divided by

m

is of the form

mk + r

.

2

Analyze

n

modulo

24

by considering the parity of

k

.

If

k

is even (

k = 2m

), then

n = 24m + 7 \equiv 7 \pmod{24}

. If

k

is odd (

k = 2m + 1

), then

n = 24m + 19 \equiv 19 \pmod{24}

.

To evaluate an expression modulo

24

when given information modulo

12

, split the integer into even and odd multiplier cases.

3

Evaluate

n^2 + 5n + 2 \pmod{24}

for both possible residue classes of

n \pmod{24}

.

Case 1 (

n \equiv 7 \pmod{24}

):

7^2 + 5(7) + 2 = 49 + 35 + 2 = 86

. Since

86 = 3 \times 24 + 14

, the remainder is

14

. Case 2 (

n \equiv 19 \pmod{24}

):

19^2 + 5(19) + 2 = 361 + 95 + 2 = 458

. Since

458 = 19 \times 24 + 2

, the remainder is

2

.

Substituting representative values of

n

into

n^2 + 5n + 2

determines all possible remainder outputs modulo

24

.

4

Identify which of the options matches a valid remainder.

The only possible remainders are

14

and

2

. Among the given choices,

2

is present.

Matching the derived valid remainder set

\{2, 14\}

against the options identifies the correct answer.

Key Concept

Modular Arithmetic and Remainder Properties under Quadratic Transformations

Estimated Time:2m 0s

Question 9Question →

When the positive integer $n$ is divided by $9$ , the remainder is $5$ . What is the remainder when $n^2 + 4n + 7$ is divided by $9$ ?

Show answer & explanation

Answer: 7

Answer

The remainder when

n^2 + 4n + 7

is divided by

9

is

7

.

Any positive integer

n

that leaves a remainder of

5

when divided by

9

can be represented as

n = 9k + 5

for some non-negative integer

k

. Substituting this into

n^2 + 4n + 7

yields

(9k + 5)^2 + 4(9k + 5) + 7 = 81k^2 + 90k + 25 + 36k + 20 + 7 = 81k^2 + 126k + 52

. Since

81k^2

and

126k

are both divisible by

9

, the remainder of the entire expression when divided by

9

is determined entirely by

52

. Dividing

52

by

9

gives

52 = 9 \times 5 + 7

, so the final remainder is

7

.

Step-by-Step Solution

1

Express the integer

n

in terms of its remainder upon division by

9

.

n = 9k + 5

for some non-negative integer

k

, or equivalently

n \equiv 5 \pmod{9}

.

By the division algorithm, any integer divided by

9

can be written as a multiple of

9

plus the remainder.

2

Substitute

n \equiv 5 \pmod{9}

into the polynomial expression

n^2 + 4n + 7

.

n^2 + 4n + 7 \equiv 5^2 + 4(5) + 7 = 25 + 20 + 7 = 52 \pmod{9}

.

Properties of modular arithmetic allow substitution of remainder values into polynomial expressions.

3

Find the remainder of

52

when divided by

9

.

52 = 9 \times 5 + 7

, which gives a remainder of

7

.

Dividing

52

by

9

yields a quotient of

5

and a remainder of

7

, which is strictly between

0

and

8

.

Key Concept

Properties of Integer Remainders and Modular Substitution

Estimated Time:1m 15s

Question 10Question →

How many positive integer factors of $36$ are prime numbers?

2

3

4

9

1

Show answer & explanation

Answer: $2$

Answer

There are

2

positive integer factors of

36

that are prime numbers:

2

and

3

.

Listing all positive integer factors of

36

yields

1, 2, 3, 4, 6, 9, 12, 18,

and

36

. Evaluating each factor against the definition of a prime number (an integer greater than

1

divisible only by

1

and itself) shows that only

2

and

3

are prime. Therefore, there are exactly

2

prime factors.

Step-by-Step Solution

1

Find all positive integer factors of

36

.

The positive integer factors of

36

are

1, 2, 3, 4, 6, 9, 12, 18,

and

36

.

Dividing

36

by integers from

1

to

36

yields integer quotients for these nine numbers.

2

Identify which of these factors meet the definition of a prime number.

The prime factors are

2

and

3

.

A prime number is an integer greater than

1

with exactly two distinct positive divisors:

1

and itself. The integer

1

is not prime, while

4, 6, 9, 12, 18,

and

36

are composite.

3

Count the identified prime factors.

The total count is

2

.

There are exactly two distinct prime numbers (

2

and

3

) in the list of factors.

Key Concept

Prime Factors and Definition of Prime Numbers

Estimated Time:45s

Question 11Question →

When the positive integer $n$ is divided by $7$ , the remainder is $4$ . What is the remainder when $n + 15$ is divided by $7$ ?

Show answer & explanation

Answer: 5

Answer

5

Adding 15 to

n

increases the remainder by 15. The original remainder is 4, so the new sum of remainders is

4 + 15 = 19

. Dividing 19 by 7 gives a quotient of 2 and a remainder of 5.

Step-by-Step Solution

1

Express

n

algebraically based on the given remainder rule.

n = 7k + 4

for some non-negative integer

k

By the division algorithm, any integer divided by 7 with remainder 4 can be written as a multiple of 7 plus 4.

2

Substitute

n

into the expression

n + 15

.

n + 15 = 7k + 4 + 15 = 7k + 19

We need to find the remainder of this new quantity when divided by 7.

3

Extract the largest multiple of 7 from 19.

7k + 19 = 7k + 14 + 5 = 7(k + 2) + 5

Grouping multiples of 7 isolates the remaining constant term.

4

Identify the final remainder.

The remainder is 5.

7(k + 2)

is completely divisible by 7, leaving 5 as the remainder.

Key Concept

Remainder arithmetic and divisibility properties

Question 12Question →

Which of the following integers is divisible by both $3$ and $4$ ?

42

54

64

72

86

Show answer & explanation

Answer: $72$

Answer

72

For an integer to be divisible by both

3

and

4

, it must satisfy the rules for both divisors. For

72

, the sum of the digits is

7 + 2 = 9

, which is divisible by

3

. Additionally,

72 \div 4 = 18

, which is an integer. Thus,

72

is divisible by both

3

and

4

.

Step-by-Step Solution

1

Recall the divisibility rule for 3

An integer is divisible by 3 if the sum of its digits is divisible by 3.

This allows quick evaluation of divisibility by 3.

2

Recall the divisibility rule for 4

An integer is divisible by 4 if its last two digits form a number divisible by 4.

This allows quick evaluation of divisibility by 4.

3

Test

72

against both divisibility rules

Sum of digits of

72

is

7 + 2 = 9

, which is divisible by

3

. Also,

72 \div 4 = 18

, which is an integer.

72

meets both required criteria.

Key Concept

Divisibility Rules for 3 and 4

Estimated Time:45s

Question 13Question →

If $x$ and $y$ are positive integers such that $3x + 5y = 101$ and $x$ is a prime number, what is the maximum possible value of the product $xy$ ?

38

108

112

170

180

Show answer & explanation

Answer: 170

Answer

170

The equation

3x + 5y = 101

requires

101 - 3x

to be a multiple of 5. For positive integers

x

and

y

,

x

must be less than 34 and leave a remainder of 2 when divided by 5 (meaning

x

ends in 2 or 7). Testing the primes meeting this condition yields three valid pairs:

(2, 19)

with product 38,

(7, 16)

with product 112, and

(17, 10)

with product 170. Thus, 170 is the maximum product.

Step-by-Step Solution

1

Isolate

5y

and analyze modular divisibility by 5.

5y = 101 - 3x

. For

y

to be an integer,

101 - 3x

must be divisible by 5. Since

101 \equiv 1 \pmod 5

, we require

3x \equiv 1 \pmod 5

, which simplifies to

x \equiv 2 \pmod 5

.

Properties of modular arithmetic determine which values of

x

yield integer values for

y

.

2

Determine the boundary constraints and prime condition for

x

.

Since

y

is a positive integer (

y \ge 1

),

3x < 101 \implies x \le 33

. The prime numbers

x \le 33

that satisfy

x \equiv 2 \pmod 5

(or end in 2 or 7) are

x = 2

,

x = 7

, and

x = 17

.

Only primes satisfying both the inequality bound and divisibility criteria need to be tested.

3

Calculate the corresponding values of

y

and products

xy

for each prime candidate.

For

x = 2

:

5y = 101 - 6 = 95 \implies y = 19

, product

xy = 2 \times 19 = 38

.
For

x = 7

:

5y = 101 - 21 = 80 \implies y = 16

, product

xy = 7 \times 16 = 112

.
For

x = 17

:

5y = 101 - 51 = 50 \implies y = 10

, product

xy = 17 \times 10 = 170

.

Evaluating all valid candidate pairs allows identifying the absolute maximum product.

4

Identify the maximum product.

The maximum possible value of

xy

is 170.

Comparing 38, 112, and 170 confirms that 170 is the greatest value.

Key Concept

Linear Diophantine equations with prime integer constraints

Question 14Question →

If $k$ is a positive integer less than $100$ such that $k$ is divisible by $6$ but not divisible by $8$ , what is the greatest possible value of $k$ ?

Show answer & explanation

Answer: 90

Answer

90

The positive multiples of 6 less than 100 in descending order are 96, 90, 84, etc. The largest candidate, 96, is divisible by 8 (

96 = 8 \times 12

), so it is excluded by the conditions. The next largest candidate, 90, is divisible by 6 (

90 = 6 \times 15

) but leaves a remainder of 2 when divided by 8 (

90 = 8 \times 11 + 2

). Thus, 90 is the greatest integer satisfying all conditions.

Step-by-Step Solution

1

Identify the largest multiples of 6 that are less than 100.

The multiples of 6 less than 100 in descending order are 96, 90, 84, 78, ...

Finding the greatest possible value requires testing candidates starting from the largest possible multiple of 6 below 100.

2

Test 96 against the constraint of not being divisible by 8.

96 / 8 = 12, so 96 is divisible by 8.

The question specifies that k must NOT be divisible by 8, ruling out 96.

3

Test the next candidate, 90, against the constraint.

90 / 8 = 11 R 2, so 90 is not divisible by 8.

90 meets all specified criteria: it is a positive integer less than 100, divisible by 6, and not divisible by 8.

Key Concept

Properties of Multiples and Divisibility Constraints

Question 15Question →

Let $m$ and $n$ be positive integers such that $m^2 n$ is divisible by $72$ and $m n^2$ is divisible by $108$ . What is the minimum possible value of the product $mn$ ?

12

18

24

36

72

Show answer & explanation

Answer: 36

Answer

36

The product

(m^2 n)(m n^2) = (mn)^3

must be divisible by

72 \times 108 = 2^5 \times 3^5

. For

(mn)^3

to be a valid cube of an integer, the exponents of 2 and 3 in its prime factorization must be multiples of 3 greater than or equal to 5. The smallest such multiples of 3 are 6, meaning

(mn)^3 \ge 2^6 \times 3^6 = (36)^3

, so

mn \ge 36

. Setting

m = 6

and

n = 6

confirms that

36

satisfies all requirements.

Step-by-Step Solution

1

Express the given divisibility conditions in terms of prime factorizations.

72 = 2^3 \times 3^2

and

108 = 2^2 \times 3^3

. Therefore,

m^2 n

is a multiple of

2^3 \times 3^2

, and

m n^2

is a multiple of

2^2 \times 3^3

.

Prime factorization allows us to analyze the minimum exponent requirements for 2 and 3.

2

Multiply the two expressions to find a lower bound on

(mn)^3

.

(m^2 n)(m n^2) = m^3 n^3 = (mn)^3

must be divisible by

(2^3 \times 3^2)(2^2 \times 3^3) = 2^5 \times 3^5

.

Combining the expressions yields a perfect cube

(mn)^3

.

3

Determine the minimum prime powers needed for

mn

.

Since

(mn)^3

is a perfect cube divisible by

2^5 \times 3^5

, the exponents of 2 and 3 in

(mn)^3

must be multiples of 3 that are at least 5. The smallest such exponents are 6 for both 2 and 3. Thus,

(mn)^3 \ge 2^6 \times 3^6 = (2^2 \times 3^2)^3 = 36^3

, so

mn \ge 36

.

Exponents in a perfect cube factorization must be multiples of 3.

4

Verify that

mn = 36

is achievable with integer values of

m

and

n

.

Setting

m = 6

and

n = 6

gives

mn = 36

. Then

m^2 n = 6^3 = 216 = 3 \times 72

(divisible by 72) and

m n^2 = 6^3 = 216 = 2 \times 108

(divisible by 108).

Constructing valid integers

m

and

n

confirms that 36 is achievable.

Key Concept

Properties of Integers and Prime Factorization

Question 16Question →

If $k$ is an integer greater than $1$ such that $k^3 - k$ is a multiple of $240$ , what is the least possible value of $k$ ?

Show answer & explanation

Answer: 9

Answer

The least possible value of

k

is 9.

Factoring

k^3 - k

gives

(k-1)k(k+1)

, representing the product of three consecutive integers. The prime factorization of

240

is

2^4 \times 3 \times 5

. To be divisible by

240

, the product

(k-1)k(k+1)

must contain at least four factors of 2, one factor of 3, and one factor of 5. Testing integers

k > 1

: for

k=9

, the product is

(8)(9)(10) = 720

, which is

3 \times 240

. Checking all integers

1 < k < 9

confirms that no smaller integer satisfies the condition.

Step-by-Step Solution

1

Factor the algebraic expression

k^3 - k

.

k^3 - k = k(k^2 - 1) = (k-1)k(k+1)

.

This expresses the polynomial as a product of three consecutive integers.

2

Determine the prime factorization of 240.

240 = 2^4 \times 3 \times 5 = 16 \times 3 \times 5

.

For

(k-1)k(k+1)

to be a multiple of 240, the product must be divisible by 16, 3, and 5 simultaneously.

3

Test candidate values of

k > 1

to find the smallest valid integer.

For

k=9

,

(k-1)k(k+1) = (8)(9)(10) = 720

, which equals

3 \times 240

.

Testing smaller values:

k=5

yields

120

,

k=6

yields

210

,

k=7

yields

336

, and

k=8

yields

504

. None of these are divisible by 240.

Key Concept

Divisibility rules and prime factor distribution across consecutive integers

Estimated Time:2m 0s

Question 17Question →

When the positive integer $m$ is divided by $8$ , the remainder is $5$ . What is the remainder when $3m + 7$ is divided by $8$ ?

2

4

5

6

7

Show answer & explanation

Answer: $6$

Answer

The remainder is

6

.

Since

m

leaves a remainder of

5

when divided by

8

, we can evaluate

3m + 7

modulo

8

by replacing

m

with

5

. Calculating

3(5) + 7

gives

22

. Dividing

22

by

8

gives a quotient of

2

with a remainder of

6

because

22 = 8 \times 2 + 6

.

Step-by-Step Solution

1

Express integer

m

using the division algorithm.

m = 8k + 5

for some non-negative integer

k

.

Dividing

m

by

8

yields a remainder of

5

.

2

Substitute the expression for

m

into

3m + 7

.

3m + 7 = 3(8k + 5) + 7 = 24k + 15 + 7 = 24k + 22

.

This expresses the new quantity algebraically in terms of

k

.

3

Determine the remainder when

24k + 22

is divided by

8

.

24k + 22 = 8(3k + 2) + 6

, so the remainder is

6

.

The term

8(3k + 2)

is a multiple of

8

, leaving

6

as the non-negative remainder less than

8

.

Key Concept

Remainder properties of linear combinations of integers

Estimated Time:45s

Question 18Question →

For all positive integers $n$ , what is the greatest integer that MUST divide $n^5 - n$ ?

6

10

15

30

60

Show answer & explanation

Answer: 30

Answer

The greatest integer that must divide

n^5 - n

for all positive integers

n

is 30.

Factoring the expression yields

n^5 - n = (n-1)n(n+1)(n^2+1)

. The factor

(n-1)n(n+1)

consists of three consecutive integers, ensuring divisibility by both 2 and 3 (and thus by 6). Furthermore, the units digit of

n^5

is always identical to the units digit of

n

for any positive integer

n

, which guarantees that

n^5 - n

is divisible by 5. Since 6 and 5 share no common factors,

n^5 - n

must be divisible by

6 \times 5 = 30

. Testing

n = 2

gives

2^5 - 2 = 30

, proving that 30 is the largest integer that divides the expression for all positive integers

n

.

Step-by-Step Solution

1

Factor the algebraic expression

n^5 - n

.

n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = (n - 1)n(n + 1)(n^2 + 1)

.

Factoring helps isolate product terms with known divisibility rules.

2

Analyze divisibility by 2 and 3.

The sub-expression

(n - 1)n(n + 1)

is the product of three consecutive integers.

Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3. Therefore,

(n - 1)n(n + 1)

is always divisible by

2 \times 3 = 6

.

3

Analyze divisibility by 5.

By Fermat's Little Theorem (or analyzing last-digit repeating cycles of powers), for any integer

n

,

n^5 \equiv n \pmod 5

, which implies

5

divides

n^5 - n

.

The units digit of

n^5

is always equal to the units digit of

n

, so

n^5 - n

always ends in 0 or 5, making it a multiple of 5.

4

Combine the common prime factors and evaluate the maximum lower bound.

Since 2, 3, and 5 are pairwise coprime, any number divisible by 2, 3, and 5 must be divisible by

2 \times 3 \times 5 = 30

. Evaluating at

n = 2

yields

2^5 - 2 = 30

, showing no integer greater than 30 can divide

n^5 - n

for all

n

.

The greatest common divisor across all generated values of

n^5 - n

is 30.

Key Concept

Divisibility properties of consecutive integers and power mod rules

Estimated Time:1m 30s

Question 19Question →

If $a$ and $b$ are positive integers such that $15a$ is divisible by $24$ and $10b$ is divisible by $14$ , which of the following statements MUST be true? Select all such statements.

Select all that apply

a

is divisible by

8

b

is divisible by

14

The product

ab

is divisible by

56

The sum

a + b

must be an even integer

a

is divisible by

24

Show answer & explanation

Answer: $a$ is divisible by $8$ ; The product $ab$ is divisible by $56$

Answer

The statements asserting that

a

is divisible by

8

and that the product

ab

is divisible by

56

must be true.

Because

24 = 2^3 \times 3

and

15 = 3 \times 5

,

a

must provide three factors of

2

, meaning

a

is divisible by

8

. Similarly,

14 = 2 \times 7

and

10 = 2 \times 5

, so

b

must provide a factor of

7

, meaning

b

is divisible by

7

. Multiplying the minimum properties (

a

is a multiple of

8

and

b

is a multiple of

7

) guarantees that

ab

is a multiple of

56

.

Step-by-Step Solution

1

Analyze the condition

15a

is divisible by

24

Prime factorizations:

15 = 3 \times 5

and

24 = 2^3 \times 3

. For

24

to divide

15a

,

a

must supply

2^3 = 8

, so

a

is a multiple of

8

.

Determines the prime factors required for

a

.

2

Analyze the condition

10b

is divisible by

14

Prime factorizations:

10 = 2 \times 5

and

14 = 2 \times 7

. For

14

to divide

10b

,

b

must supply

7

, so

b

is a multiple of

7

.

Determines the prime factors required for

b

.

3

Evaluate each statement against the deduced minimum properties of

a

and

b

Since

a

is a multiple of

8

, the statement regarding

a

being divisible by

8

is always true. Since

a

is a multiple of

8

and

b

is a multiple of

7

,

ab

is a multiple of

8 \times 7 = 56

. Counterexamples disprove the remaining choices (

b=7

disproves divisibility of

b

by

14

and evenness of

a+b

;

a=8

disproves divisibility of

a

by

24

).

Identifies which statements MUST be true for all valid values of

a

and

b

.

Key Concept

Divisibility and Prime Factorization Constraints

Question 20Question →

When the positive integer $n$ is divided by $7$ , the remainder is $4$ . What is the remainder when $n + 18$ is divided by $7$ ?

Show answer & explanation

Answer: 1

Answer

The remainder when

n + 18

is divided by

7

is

1

.

Because

n

leaves a remainder of

4

when divided by

7

,

n

can be represented as

7k + 4

. Adding

18

gives

n + 18 = 7k + 22

. Since

7k

is a multiple of

7

, the remainder of

(7k + 22) \div 7

depends entirely on

22 \div 7

. Dividing

22

by

7

gives a quotient of

3

with a remainder of

1

.

Step-by-Step Solution

1

Represent the positive integer

n

algebraically based on the given remainder.

n = 7k + 4

for some non-negative integer

k

.

By the division algorithm, any integer

n

divided by

7

with remainder

4

can be written as

7k + 4

.

2

Add

18

to

n

.

n + 18 = (7k + 4) + 18 = 7k + 22

.

Substitute

7k + 4

for

n

in the expression

n + 18

.

3

Determine the remainder when

7k + 22

is divided by

7

.

Since

7k

is divisible by

7

, the remainder is

22 \pmod 7 = 1

.

Dividing

22

by

7

yields a quotient of

3

and a remainder of

1

(

22 = 7 \times 3 + 1

).

Key Concept

Remainder Properties under Addition

Properties of Integers and Divisibility

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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