Number Properties and Arithmetic Practice Questions — GMAT — Page 12

Question 221Question →

The positive integer $n$ has a prime factorization of the form $2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers. If $n$ is divisible by both 12 and 15, and $n$ has exactly 24 positive integer divisors, what is the least possible value of $n$ ?

180

360

480

540

600

Show answer & explanation

Answer: 360

Answer

The least possible value of

n

is 360.

To find the least value of

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

divisible by

12 = 2^2 \times 3

and

15 = 3 \times 5

, we require

a \ge 2

,

b \ge 1

, and

c \ge 1

. The number of positive divisors is

(a+1)(b+1)(c+1) = 24

. Testing valid factor triples for 24 with

a+1 \ge 3

,

b+1 \ge 2

, and

c+1 \ge 2

gives possible values

n = 480

(from

a=5, b=1, c=1

),

n = 360

(from

a=3, b=2, c=1

),

n = 540

(from

a=2, b=3, c=1

), and

n = 600

(from

a=3, b=1, c=2

). The minimum among these valid integers is 360.

Step-by-Step Solution

1

Determine the minimum exponent constraints from divisibility conditions.

Since

n

is divisible by

12 = 2^2 \times 3^1

, we must have

a \ge 2

and

b \ge 1

. Since

n

is divisible by

15 = 3^1 \times 5^1

, we must have

b \ge 1

and

c \ge 1

. Combining these,

a \ge 2

,

b \ge 1

, and

c \ge 1

.

Divisibility requires that the prime factorization of

n

contains at least the prime powers present in the prime factorizations of 12 and 15.

2

Set up the total divisor count equation.

The total number of positive integer divisors of

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

is given by

(a+1)(b+1)(c+1) = 24

.

The divisor counting formula states that for a number with prime factorization

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

, the total number of positive divisors is

(e_1+1)(e_2+1)\dots

3

Find all integer factor triples

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

that multiply to 24 subject to

a+1 \ge 3

,

b+1 \ge 2

, and

c+1 \ge 2

.

The valid factorizations of 24 into three factors meeting the bounds are:
-

(6, 2, 2) \implies (a, b, c) = (5, 1, 1)

, giving

n = 2^5 \times 3^1 \times 5^1 = 480

-

(4, 3, 2) \implies (a, b, c) = (3, 2, 1)

, giving

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

-

(3, 4, 2) \implies (a, b, c) = (2, 3, 1)

, giving

n = 2^2 \times 3^3 \times 5^1 = 540

-

(4, 2, 3) \implies (a, b, c) = (3, 1, 2)

, giving

n = 2^3 \times 3^1 \times 5^2 = 600

To minimize

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

, we must evaluate all valid permutations of exponents consistent with the divisor product constraint.

4

Compare the resulting values of

n

to find the minimum.

Comparing 480, 360, 540, and 600, the smallest value is 360.

Assigning the larger exponent 3 to the smallest prime base 2 and exponent 2 to prime base 3 minimizes the overall product.

Key Concept

Divisor Counting Formula and Prime Factorization Constraints

Estimated Time:2m 0s

Question 222Question →

Set $S$ consists of $n$ consecutive integers listed in increasing order. The product of the least integer and the greatest integer in Set $S$ is equal to $0$ . If the sum of all the integers in Set $S$ is $21$ , how many integers are in Set $S$ ?

6

7

8

12

14

Show answer & explanation

Answer: 7

Answer

7

The given sum of the consecutive integers is positive (

21

), which requires the set of consecutive integers to start at

0

rather than end at

0

. Setting up the sum formula for consecutive integers from

0

to

k

gives

\frac{k(k+1)}{2} = 21

, solving to

k = 6

. Including

0

, the terms in the set are

0, 1, 2, 3, 4, 5, 6

, which yields

6 - 0 + 1 = 7

integers in total.

Step-by-Step Solution

1

Analyze the condition for the product of the least and greatest integers.

Since the product of the least integer

a

and greatest integer

b

is

0

, at least one of these extreme terms must be

0

.

By the zero-product property,

a \cdot b = 0 \implies a = 0

or

b = 0

.

2

Determine which extreme integer is equal to 0.

The least integer

a

must be

0

.

If

b = 0

, all integers in the set would be

\leq 0

, so their sum would be

\leq 0

. Since the given sum is

21 > 0

, the set must consist of non-negative integers starting at

0

.

3

Set up the sum equation for consecutive integers starting at 0.

\frac{k(k+1)}{2} = 21 \implies k(k+1) = 42 \implies k = 6

.

The sum of integers from

0

to

k

is

0 + \frac{k(k+1)}{2} = 21

.

4

Count the total number of terms in the set.

n = 6 - 0 + 1 = 7

.

The number of integers in an inclusive range from

a

to

b

is given by

b - a + 1

.

Key Concept

Consecutive Integers, Zero Property, and Inclusive Term Counting

Estimated Time:1m 30s

Question 223Question →

Let $N = 2^4 \times 3^2 \times 5$ . How many distinct prime factors does the integer $N + 9$ have?

1

2

3

4

6

Show answer & explanation

Answer: 1

Answer

The integer

N + 9

has 1 distinct prime factor.

Evaluating

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

gives

720

. Adding 9 yields

729

, which prime factorizes as

3^6

. Because 3 is the only unique prime base in this expression,

N + 9

has exactly 1 distinct prime factor.

Step-by-Step Solution

1

Calculate the numerical value of

N

from its prime factorization.

N = 2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = 720

Evaluating

N

is necessary to compute

N + 9

.

2

Add 9 to

N

to find the value of

N + 9

.

N + 9 = 720 + 9 = 729

The question asks for the distinct prime factors of the sum

N + 9

.

3

Find the prime factorization of 729.

729 = 3^6

Prime factorization reveals the prime bases of the integer.

4

Count the number of distinct prime factors of 729.

The only distinct prime factor is 3, so the count is 1.

Distinct prime factors are the unique prime numbers appearing in the prime factorization.

Key Concept

Prime Factorization of an Expression Sum

Estimated Time:1m 30s

Question 224Question →

Set $S$ consists of consecutive integers. The sum of all positive integers in set $S$ is $300$ , and the median of set $S$ is $-4.5$ . How many negative integers are in set $S$ ?

Show answer & explanation

Answer: 33

Answer

The set S contains 33 negative integers.

To find the number of negative integers in set S, first determine the maximum positive integer N in the set using the sum formula for consecutive positive integers: \frac{N(N+1)}{2} = 300, which yields N = 24. Next, use the property that for an evenly spaced set, the median is the average of the smallest and largest numbers: -4.5 = \frac{\text{Smallest} + 24}{2}. Solving for the smallest integer gives -33. Finally, count the negative integers in set S, which run from -33 to -1 inclusive: -1 - (-33) + 1 = 33.

Step-by-Step Solution

1

Find the largest integer in Set S

The largest integer is 24.

The positive integers in set S must form a sequence from 1 to N. The sum of positive integers is given by \frac{N(N + 1)}{2} = 300, which simplifies to N(N + 1) = 600. Since 24 \times 25 = 600, N = 24.

2

Find the smallest integer in Set S using the median formula

The smallest integer is -33.

In any set of consecutive integers, the median is equal to the arithmetic mean of the smallest and largest terms: \text{Median} = \frac{\text{Smallest} + \text{Largest}}{2}. Substituting -4.5 for the median and 24 for the largest term gives -4.5 = \frac{\text{Smallest} + 24}{2} \implies \text{Smallest} + 24 = -9 \implies \text{Smallest} = -33.

3

Count the number of negative integers in Set S

There are 33 negative integers.

The negative integers in set S range from -33 to -1, inclusive. The count is calculated as -1 - (-33) + 1 = 33.

Key Concept

Properties of consecutive integer sets, median-mean equivalence, and inclusive range counting

Estimated Time:1m 30s

Question 225Question →

Set $S$ consists of consecutive integers from $a$ to $b$ , inclusive, such that $a < 0 < b$ . The number of negative integers in set $S$ is equal to three times the number of positive integers in set $S$ . If the sum of all elements in set $S$ is $-68$ , how many integers are in set $S$ ?

15

16

17

21

13

Show answer & explanation

Answer: 17

Answer

17

Set

S

consists of

12

negative integers (

-12

through

-1

),

4

positive integers (

1

through

4

), and the integer

0

. The sum of positive integers is

10

and the sum of negative integers is

-78

, giving a total sum of

-68

. The total number of integers in the set is

12 + 4 + 1 = 17

.

Step-by-Step Solution

1

Define the number of terms and bounds of set

S

in terms of positive integers

b

.

The positive integers are

1, 2, \dots, b

, so there are

b

positive integers. The negative integers are

-1, -2, \dots, a

. Since the number of negative integers is

3

times the number of positive integers, there are

3b

negative integers, which means

a = -3b

. Set

S

also includes

0

.

Expressing all elements in terms of

b

allows setting up a single variable equation.

2

Calculate the sum of all elements in set

S

as a function of

b

.

The sum of the positive integers is

\frac{b(b+1)}{2}

. The sum of the negative integers is

-\frac{3b(3b+1)}{2}

. The total sum is

\frac{b(b+1)}{2} - \frac{3b(3b+1)}{2} = \frac{b^2 + b - 9b^2 - 3b}{2} = \frac{-8b^2 - 2b}{2} = -4b^2 - b

.

The sum of consecutive integers from

1

to

n

is

\frac{n(n+1)}{2}

.

3

Solve for

b

using the given sum of

-68

.

-4b^2 - b = -68 \implies 4b^2 + b - 68 = 0

. Factoring gives

(4b + 17)(b - 4) = 0

. Since

b

must be a positive integer,

b = 4

. Therefore,

a = -3(4) = -12

.

Solving the quadratic equation identifies the exact boundaries of the set.

4

Calculate the total number of integers in set

S

.

The set contains integers from

-12

to

4

, inclusive. The total count of integers is

4 - (-12) + 1 = 17

.

The number of integers in an inclusive range

[a, b]

is given by

b - a + 1

.

Key Concept

Consecutive Integers and Number Sets

Question 226Question →

If $x$ and $y$ are integers such that $-8 \le x \le 4$ and $-2 \le y \le 5$ , $x y^2 > 0$ , and $x^3 y < 0$ , how many distinct integer values are possible for the product $x y$ ?

Show answer & explanation

Answer: 6

Answer

The total number of distinct integer values for the product

x y

is 6.

Analyzing the given inequalities reveals sign constraints on both variables:

x y^2 > 0

forces

x > 0

(since

y^2 > 0

for non-zero

y

), limiting

x

to integer values

1, 2, 3, 4

. Then

x^3 y < 0

forces

y < 0

(since

x^3 > 0

), limiting

y

to integer values

-2, -1

. Multiplying each value of

x

by each value of

y

gives the products

-1, -2, -3, -4, -6, -8

, which constitutes exactly 6 distinct values.

Step-by-Step Solution

1

Determine the sign and allowable integer values of

x

x \in \{1, 2, 3, 4\}

Since

y^2

is strictly positive for any non-zero real number

y

,

x y^2 > 0

implies

x > 0

and

y \neq 0

.

2

Determine the sign and allowable integer values of

y

y \in \{-2, -1\}

Because

x > 0

,

x^3

is also positive. For

x^3 y < 0

,

y

must be negative.

3

Calculate products for all valid pairs of

(x, y)

and remove duplicates

The distinct product values are

-1, -2, -3, -4, -6, -8

, giving 6 values in total.

Evaluating

x y

for

x \in \{1, 2, 3, 4\}

and

y \in \{-2, -1\}

yields

-1, -2, -3, -4

when

y = -1

and

-2, -4, -6, -8

when

y = -2

.

Key Concept

Even powers of non-zero numbers are always positive, while odd powers preserve the sign of the base number.

Question 227Question →

If $K = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers, such that $15K$ has 15 more positive factors than $K$ , and $6K$ has 12 more positive factors than $K$ , what is the total number of positive factors of $K$ ?

10

12

15

18

27

Show answer & explanation

Answer: 12

Answer

The total number of positive factors of

K

is 12.

By setting up factor count equations from prime exponents, we find

a=2, b=1, c=1

. Applying the formula

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

yields

(3)(2)(2) = 12

.

Step-by-Step Solution

1

Set up the formula for the number of positive factors of

K

,

15K

, and

6K

.

For

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

, the number of factors is

d(K) = (a+1)(b+1)(c+1)

. For

15K = 2^a \cdot 3^{b+1} \cdot 5^{c+1}

,

d(15K) = (a+1)(b+2)(c+2)

. For

6K = 2^{a+1} \cdot 3^{b+1} \cdot 5^c

,

d(6K) = (a+2)(b+2)(c+1)

.

Prime factorizations determine the number of positive factors via the product of (exponent + 1).

2

Express the given factor differences algebraically.

d(15K) - d(K) = (a+1)[(b+2)(c+2) - (b+1)(c+1)] = (a+1)(b+c+3) = 15

. Also,

d(6K) - d(K) = (c+1)[(a+2)(b+2) - (a+1)(b+1)] = (c+1)(a+b+3) = 12

.

Factor out common terms to simplify the system of equations.

3

Solve for the positive integers

a

,

b

, and

c

.

Since

b \ge 1

and

c \ge 1

,

b+c+3 \ge 5

. The only factor of 15 greater than or equal to 5 is 5 (since

b+c+3=15

would require

a=0

). Thus

b+c+3=5 \implies b=1, c=1

and

a+1=3 \implies a=2

. Checking in the second equation:

(1+1)(2+1+3) = 2 \cdot 6 = 12

, which holds.

Integer constraints limit the potential factor pairs of 15.

4

Calculate

d(K)

.

d(K) = (2+1)(1+1)(1+1) = 3 \cdot 2 \cdot 2 = 12

.

Substitute

a=2, b=1, c=1

into the factor counting formula.

Key Concept

Prime Factorization and Number of Divisors

Estimated Time:2m 0s

Question 228Question →

Let $m$ and $n$ be integers such that $-4 \le m \le 6$ and $-8 \le n \le 3$ . If $m^2 n < 0$ and $m + n > 0$ , what is the minimum possible value of the product $m \cdot n$ ?

Show answer & explanation

Answer: -30

Answer

The minimum possible value of

m \cdot n

is

-30

.

The correct answer is -30. Analyzing

m^2 n < 0

shows that

n

must be negative and

m

cannot be zero. The condition

m + n > 0

implies

m > -n = |n|

, making

m

positive. Since

m \le 6

, the maximum possible value for

|n|

is 5, which corresponds to

n = -5

. When

n = -5

,

m

must be strictly greater than 5, leaving

m = 6

as the only valid value. The product is

6 \times (-5) = -30

, which is the minimum value achievable under all constraints.

Step-by-Step Solution

1

Determine the signs of

m

and

n

using the given inequalities.

n < 0

and

m > 0

with

m \ne 0

.

Since

m^2

is strictly positive for any non-zero integer

m

,

m^2 n < 0

forces

n < 0

. Then

m + n > 0

requires

m > -n > 0

.

2

Find the range of valid integer values for

n

.

n \in \{-5, -4, -3, -2, -1\}

.

Since

m \le 6

and

m > -n

, we must have

-n \le 5

, which gives

n \ge -5

.

3

Evaluate the minimum product

m \cdot n

across all allowed values of

n

.

The minimum product is

-30

, occurring when

n = -5

and

m = 6

.

To minimize a negative product, maximize the absolute product

m \cdot |n|

. When

n = -5

,

m

must be

6

, yielding

6 \cdot (-5) = -30

.

Key Concept

Positive and Negative Number Properties with Inequalities

Question 229Question →

If $x$ and $y$ are non-zero real numbers such that $\frac{x - y}{x} > 2$ and $x y^2 < 0$ , which of the following inequalities must be true?

x + y < 0

x y > 0

x + y > 0

x - y > 0

\frac{y}{x} > -1

Show answer & explanation

Answer: $x + y > 0$

Answer

x + y > 0

From

x y^2 < 0

, since

y \neq 0

,

y^2

is strictly positive, which forces

x < 0

. Next, expanding

\frac{x - y}{x} > 2

by multiplying both sides by negative

x

reverses the inequality direction to

x - y < 2x

. Subtracting

x

from both sides gives

-y < x

, which is equivalent to

x + y > 0

. Thus,

x + y > 0

must always be true.

Step-by-Step Solution

1

Determine the sign of

x

using

x y^2 < 0

.

Since

y \neq 0

,

y^2 > 0

. For the product

x y^2

to be negative,

x

must be negative (

x < 0

).

A positive quantity multiplied by a negative quantity yields a negative product.

2

Simplify the inequality

\frac{x - y}{x} > 2

taking the sign of

x

into account.

Multiplying both sides by

x

(and reversing the inequality because

x < 0

) gives

x - y < 2x

.

Multiplying or dividing an inequality by a negative value reverses the inequality sign.

3

Rearrange the terms of

x - y < 2x

to isolate the relationship between

x

and

y

.

Subtracting

x

from both sides yields

-y < x

, which rearranges to

x + y > 0

(or

y > -x

).

Adding

y

to both sides converts

-y < x

directly into

x + y > 0

.

Key Concept

Deducing signs and algebraic bounds in inequalities involving negative variables

Estimated Time:1m 30s

Question 230Question →

If $a$ , $b$ , and $c$ are non-zero real numbers such that $a b^2 c^3 < 0$ and $\frac{a}{b} > 0$ , which of the following expressions MUST be negative?

a^2 b c

a b c

a b c^2

\frac{a}{b c^2}

\frac{a c}{b}

Show answer & explanation

Answer: $a^2 b c$

Answer

The expression

a^2 b c

must be negative.

For any non-zero real number

a

, the term

a^2

is strictly positive. The given conditions show that

b^2 > 0

, which simplifies

a b^2 c^3 < 0

to

a c < 0

, proving

a

and

c

have opposite signs. Furthermore,

\frac{a}{b} > 0

means

a

and

b

have the same sign. Combining these facts shows that

b

and

c

must always have opposite signs, so

b c < 0

. Therefore,

a^2 b c = (a^2)(b c) = (+) \times (-) < 0

in all valid cases.

Step-by-Step Solution

1

Analyze the sign implication of

a b^2 c^3 < 0

.

a c < 0

, meaning

a

and

c

have opposite signs.

Because

b \neq 0

,

b^2

is strictly positive (

b^2 > 0

). Dividing

a b^2 c^3 < 0

by

b^2

yields

a c^3 < 0

. Since

c^3

has the exact same sign as

c

, it follows that

a c < 0

.

2

Analyze the sign implication of

\frac{a}{b} > 0

.

a

and

b

must have the same sign (

a b > 0

).

A quotient of two non-zero numbers is positive if and only if both numbers have the same sign.

3

Combine sign conditions to evaluate the possible cases for

a, b, c

.

Either Case 1:

a > 0, b > 0, c < 0

, or Case 2:

a < 0, b < 0, c > 0

.

If

a > 0

, then

b > 0

(from step 2) and

c < 0

(from step 1). If

a < 0

, then

b < 0

(from step 2) and

c > 0

(from step 1).

4

Test the expression

a^2 b c

under both cases.

In Case 1:

(+)^2 \cdot (+) \cdot (-) = (-)

. In Case 2:

(-)^2 \cdot (-) \cdot (+) = (-)

. Thus

a^2 b c < 0

always.

Since

a^2 > 0

for all non-zero real numbers

a

, and

b

and

c

have opposite signs in every valid scenario (

b c < 0

), the product

a^2 (b c)

is always strictly negative.

Key Concept

Deducing signs of variables using exponent rules and inequalities

Estimated Time:1m 30s

Question 231Question →

Let $a$ , $b$ , and $c$ be integers such that $-5 \le a \le 3$ , $-4 \le b \le 6$ , and $-3 \le c \le 4$ . If $a b < 0$ and $b c^2 > 0$ , what is the maximum possible value of the expression $\frac{a - c}{b}$ ?

Show answer & explanation

Answer: 2

Answer

The maximum possible value of the expression is 2.

From

b c^2 > 0

, since

c^2 > 0

for any non-zero real number

c

, it follows that

b > 0

and

c \neq 0

. Given

a b < 0

and

b > 0

,

a

must be strictly negative (

a < 0

). To maximize

\frac{a - c}{b}

where

b > 0

, we maximize the numerator

(a - c)

and minimize the denominator

b

. The maximum negative integer for

a

is

-1

, and the minimum integer for

c

is

-3

, giving a maximum numerator of

-1 - (-3) = 2

. The minimum positive integer for

b

is

1

. Hence, the maximum value is

\frac{2}{1} = 2

.

Step-by-Step Solution

1

Analyze the sign of

b c^2 > 0

b > 0

and

c \neq 0

The square of any non-zero real number is strictly positive (

c^2 > 0

), so

b

must be positive for the product

b c^2

to be positive.

2

Analyze the sign of

a b < 0

a < 0

Since

b > 0

,

a

must be negative for their product to be less than zero.

3

Determine candidate integer sets

a \in \{-5, -4, -3, -2, -1\}

,

b \in \{1, 2, 3, 4, 5, 6\}

,

c \in \{-3, -2, -1, 1, 2, 3, 4\}

Filter the given intervals by the sign restrictions (

a < 0

,

b > 0

,

c \neq 0

).

4

Optimize numerator

(a - c)

and denominator

b

a = -1

,

c = -3

,

b = 1

To maximize a quotient with a positive denominator, make the numerator as large as possible (

a - c = -1 - (-3) = 2

) and the denominator as small as possible (

b = 1

).

5

Evaluate the maximum value

2

\frac{-1 - (-3)}{1} = \frac{2}{1} = 2

.

Key Concept

Deducing sign constraints from products and powers of variables and optimizing algebraic quotients under sign and domain restrictions.

Estimated Time:1m 30s

Question 232Question →

If $m$ and $n$ are real numbers such that $m < 0 < n$ and $m^2 - 3mn - 4n^2 = 0$ , what is the value of $\frac{m + 2n}{m - n}$ ?

Show answer & explanation

Answer: -0.5

Answer

The value of the expression is -0.5.

Factoring

m^2 - 3mn - 4n^2 = 0

yields

(m - 4n)(m + n) = 0

, giving solutions

m = 4n

or

m = -n

. Given

m < 0 < n

,

m

is negative and

n

is positive, so

m

and

n

must have opposite signs. The solution

m = 4n

implies that

m

and

n

have the same sign (both positive), which violates

m < 0

. Therefore,

m

must equal

-n

. Substituting

m = -n

into

\frac{m + 2n}{m - n}

results in

\frac{-n + 2n}{-n - n} = \frac{n}{-2n} = -0.5

.

Step-by-Step Solution

1

Factor the quadratic equation

m^2 - 3mn - 4n^2 = 0

.

(m - 4n)(m + n) = 0

Factoring allows determination of the possible linear relationships between

m

and

n

.

2

Apply the sign condition

m < 0 < n

to eliminate the invalid case.

m = -n

Since

n > 0

,

4n

must be positive. Setting

m = 4n

would make

m > 0

, contradicting

m < 0

. Thus,

m + n = 0

, giving

m = -n

.

3

Substitute

m = -n

into

\frac{m + 2n}{m - n}

.

\frac{-n + 2n}{-n - n} = \frac{n}{-2n} = -0.5

Replacing

m

with

-n

simplifies the ratio to a constant value.

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