Even-Odd Properties and Sign Rules Alıştırma Soruları — GRE General Test

Soru 1Soru →

Let $p$ and $q$ be integers such that $p < 0 < q$ , $(-1)^p + (-1)^q = 0$ , and $p^2 q + p$ is an even integer. Which of the following expressions must be negative?

(-p)^q

(-q)^p

q^p

p^q

(q - p)^p

Cevabı ve açıklamayı göster

Cevap: $p^q$

Cevap

p^q

From

(-1)^p + (-1)^q = 0

,

p

and

q

must have opposite parities. Factoring

p^2 q + p

into

p(pq + 1)

shows that if

p

were odd,

q

would be even, leading to an odd product

\text{odd} \times \text{odd} = \text{odd}

. Because

p^2 q + p

is given as even,

p

must be an even integer and

q

must be an odd integer. Given

p < 0

,

p

is a negative even integer, and

q

is a positive odd integer. The expression

p^q

represents a negative base raised to an odd exponent, which is guaranteed to be negative.

Adım Adım Çözüm

1

Determine the parities of

p

and

q

using the equation

(-1)^p + (-1)^q = 0

.

One of

p

or

q

is even and the other is odd.

For the sum

(-1)^p + (-1)^q

to equal

0

, one term must equal

1

and the other must equal

-1

, meaning

p

and

q

have opposite parities.

2

Analyze the parity of

p^2 q + p = p(pq + 1)

to determine which variable is even.

p

must be even, and

q

must be odd.

If

p

were odd,

q

would be even, making

pq

even,

pq+1

odd, and

p(pq+1)

odd. Since

p^2 q + p

is even,

p

must be even and

q

must be odd.

3

Evaluate the sign of

p^q

using sign rules for exponents.

p^q < 0

p

is negative (

p < 0

) and

q

is a positive odd integer. A negative number raised to an odd power is always negative.

Anahtar Kavram

Parity and sign rules for negative bases and integer exponents

Soru 2Soru →

Suppose $a$ , $b$ , and $c$ are integers such that $a < 0 < b < c$ . If $a(b - c)$ is an odd integer and $a^2 b + b c$ is an even integer, which of the following expressions MUST be a positive even integer?

b(c - a)

a b c

a(b - c)

(a - c) b

(a + c) b

Cevabı ve açıklamayı göster

Cevap: $b(c - a)$

Cevap

b(c - a)

is guaranteed to be a positive even integer.

The expression

b(c - a)

consists of

b

(which is positive and even) multiplied by

(c - a)

(which is positive and even, as subtracting a negative odd number from a positive odd number yields a positive even number). The product of two positive even integers is always a positive even integer.

Adım Adım Çözüm

1

Determine the signs of variables

a

,

b

, and

c

.

a < 0

(negative),

b > 0

(positive), and

c > 0

(positive).

Directly given by the inequality

a < 0 < b < c

.

2

Analyze parity from the given condition that

a(b - c)

is odd.

a

is odd and

(b - c)

is odd.

A product of two integers is odd if and only if both factors are odd.

3

Analyze parity from the second condition

a^2 b + b c = b(a^2 + c)

being even.

b

must be even, and

c

must be odd.

Since

a

is odd,

a^2

is odd. If

b

were odd, then

a^2 + c

would need to be even (making

c

odd), but if both

b

and

c

were odd,

b - c

would be even, contradicting step 2. Therefore,

b

must be even. Since

b

is even and

b - c

is odd,

c

must be odd.

4

Evaluate the sign and parity of

b(c - a)

.

c - a = \text{odd} - (\text{negative odd}) = \text{positive even}

. Since

b

is positive even,

b(c - a) = \text{positive even} \times \text{positive even} = \text{positive even}

.

Subtracting a negative number yields addition (

c - a > 0

), and subtracting an odd integer from an odd integer produces an even integer.

Anahtar Kavram

Even-Odd Parity and Integer Sign Properties under Multiplication and Subtraction

Soru 3Soru →

Let $m$ and $n$ be negative integers such that $m < n$ and $m - n$ is an odd integer. Which of the following expressions must be a positive even integer?

(m - n)^2 + 1

m + n + 1

(-1)^n (m - n)

m^2 + n^2

(m - n)^3

Cevabı ve açıklamayı göster

Cevap: $(m - n)^2 + 1$

Cevap

(m - n)^2 + 1

must be a positive even integer.

Because

m - n

is given as an odd integer and

m < n

,

m - n

is a negative odd integer. Squaring any negative odd integer gives a positive odd integer. Adding

1

to a positive odd integer yields a positive even integer, which guarantees the result is always positive and even.

Adım Adım Çözüm

1

Analyze the parity and sign of

m - n

m - n

is a negative odd integer.

Since

m < n

,

m - n < 0

. The problem specifies that

m - n

is odd.

2

Evaluate the expression

(m - n)^2

(m - n)^2

is a positive odd integer.

Squaring any non-zero real number yields a positive result. Squaring an odd integer always yields an odd integer.

3

Add 1 to

(m - n)^2

(m - n)^2 + 1

is an even integer greater than or equal to 2.

Adding 1 to a positive odd integer converts it into a positive even integer.

Anahtar Kavram

Parity rules under arithmetic operations and exponents with signed integers

Tahmini Süre:1m 15s

Soru 4Soru →

Let $x$ and $y$ be non-zero integers that satisfy all of the following conditions:
I. $(-x)^{y + 1} < 0$
II. $x^2 y + x$ is odd
III. $x < y$

Which of the following statements MUST be true? Select all that apply.

Geçerli olan tümünü seçin

y^x

is an even integer

x^y - y

is an odd integer

-x < -y

(x + y)^2

is an odd integer

(-x)^y < 0

Cevabı ve açıklamayı göster

Cevap: $y^x$ is an even integer; $x^y - y$ is an odd integer; $(x + y)^2$ is an odd integer

Cevap

The statements that must be true are '

y^x

is an even integer', '

x^y - y

is an odd integer', and '

(x + y)^2

is an odd integer'.

From the given conditions, we deduce that

x

is a positive odd integer and

y

is a positive even integer with

x < y

. Therefore:
- Raising the positive even integer

y

to the positive power

x

yields an even integer (

y^x

is even).
- Raising the odd integer

x

to the positive power

y

yields an odd integer, and subtracting the even integer

y

leaves an odd integer (

x^y - y

is odd).
- The sum of odd

x

and even

y

is odd, and squaring an odd integer yields an odd integer (

(x + y)^2

is odd).

Adım Adım Çözüm

1

Analyze Condition I:

(-x)^{y + 1} < 0

x > 0

(a positive integer) and

y

is an even integer.

For a number raised to an integer power to be strictly negative, the base must be negative and the exponent must be odd. Hence,

-x < 0 \implies x > 0

, and

y + 1

is odd

\implies y

is even.

2

Analyze Condition II:

x^2 y + x

is odd

x

is an odd integer.

Factor the expression as

x(xy + 1)

. A product of two integers is odd if and only if both factors are odd. Thus,

x

must be odd. (Additionally,

xy + 1

must be odd

\implies xy

is even, which holds since

y

is even).

3

Analyze Condition III:

x < y

y

is a positive even integer.

Since

x

is a positive odd integer (

x \ge 1

) and

x < y

,

y

must also be a positive integer (

y \ge 2

).

4

Evaluate the statements based on the derived properties (

x

is positive odd,

y

is positive even,

x < y

)

The statements '

y^x

is an even integer', '

x^y - y

is an odd integer', and '

(x + y)^2

is an odd integer' are guaranteed to be true.

1)

\text{even}^{\text{positive odd}} = \text{even}

. 2)

\text{odd}^{\text{positive even}} - \text{even} = \text{odd} - \text{even} = \text{odd}

. 3)

(\text{odd} + \text{even})^2 = \text{odd}^2 = \text{odd}

.

Anahtar Kavram

Deducing parity and signs of variables using exponent rules and arithmetic properties of even and odd numbers.

Tahmini Süre:2m 0s

Soru 5Soru →

If $k$ and $m$ are integers such that $k < 0 < m$ , $(-1)^k = -1$ , and $(-1)^m = 1$ , which of the following expressions MUST be a negative odd integer?

k - m

k + m

k m

m - k

k^2 - m

Cevabı ve açıklamayı göster

Cevap: $k - m$

Cevap

The expression

k - m

MUST be a negative odd integer.

Given

k < 0 < m

,

k

is negative and

m

is positive. The relation

(-1)^k = -1

shows

k

is odd, while

(-1)^m = 1

shows

m

is even. Subtracting a positive even integer

m

from a negative odd integer

k

yields

k - m

, which must be less than 0 (negative) and odd (odd minus even).

Adım Adım Çözüm

1

Determine the parity and sign of

k

k

is a negative odd integer.

k < 0

specifies that

k

is negative, and

(-1)^k = -1

implies that the exponent

k

must be odd.

2

Determine the parity and sign of

m

m

is a positive even integer.

m > 0

specifies that

m

is positive, and

(-1)^m = 1

implies that the exponent

m

must be even.

3

Evaluate the sign and parity of

k - m

k - m

is strictly negative and odd.

Since

k < 0

and

m > 0

,

k - m = k + (-m) < 0

. By parity rules,

\text{odd} - \text{even} = \text{odd}

.

Anahtar Kavram

Even-Odd Properties and Sign Rules

Soru 6Soru →

For how many integer values of $n$ in the interval $-15 \le n \le 15$ is the value of the expression $(-1)^{n^2 + n} - (-1)^{3n}$ equal to $2$ ?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The correct answer is 16.

Because

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

is the product of two consecutive integers, it is guaranteed to be even for every integer

n

. Consequently,

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

. Substituting this into the given equation yields

1 - (-1)^{3n} = 2

, which reduces to

(-1)^{3n} = -1

. A power of

-1

equals

-1

if and only if the exponent is odd, so

3n

must be odd, which requires

n

itself to be odd. In the interval

[-15, 15]

, there are 16 odd integers: 8 negative odd integers and 8 positive odd integers.

Adım Adım Çözüm

1

Determine the parity of

n^2 + n

The expression

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

represents the product of two consecutive integers. Because one of any two consecutive integers is even, their product is always even. Therefore,

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

for all integers

n

.

Simplifying the exponent with a known parity rule reduces the expression to a constant.

2

Isolate

(-1)^{3n}

in the equation

Substituting

1

into the original equation gives

1 - (-1)^{3n} = 2

, which simplifies to

(-1)^{3n} = -1

.

Isolating the exponential term reveals the sign condition required for the equality to hold.

3

Find the parity condition for

n

For

(-1)^{3n}

to equal

-1

, the exponent

3n

must be an odd integer. Since

3

is odd, the product

3n

is odd if and only if

n

is odd.

Applying the product parity rule (

\text{odd} \times \text{odd} = \text{odd}

) relates the condition on

3n

back to

n

.

4

Count the odd integers in the interval

[-15, 15]

The odd integers in the interval are

-15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15

. There are 16 such integers.

Counting all qualifying values within the specified range yields the final numeric answer.

Anahtar Kavram

Parity rules for consecutive integers and exponents of negative numbers

Soru 7Soru →

Consider three non-zero integers $x$ , $y$ , and $z$ that satisfy all of the following conditions:

I. $(-1)^{x^2 y + z} = -1$
II. $x y z^2 < 0$
III. $x + y$ is an even integer

Which of the following expressions MUST be an odd integer?

(x + z)(y + z)

x y z

x + y + z

x(y + z)

x z + y

Cevabı ve açıklamayı göster

Cevap: $(x + z)(y + z)$

Cevap

The expression

(x + z)(y + z)

MUST be an odd integer.

Condition I dictates that

x^2 y + z

is odd. Condition III states that

x + y

is even, meaning

x

and

y

share the same parity. If

x

and

y

are both even,

x^2 y

is even, forcing

z

to be odd. If

x

and

y

are both odd,

x^2 y

is odd, forcing

z

to be even. Consequently,

z

always has the opposite parity of both

x

and

y

. Therefore,

(x + z)

is always odd and

(y + z)

is always odd. The product of two odd integers,

(x + z)(y + z)

, is guaranteed to be odd.

Adım Adım Çözüm

1

Analyze Condition I for exponent parity

x^2 y + z

must be an odd integer

For

(-1)^k = -1

, the exponent

k

must be odd.

2

Analyze Condition III for shared parity of

x

and

y

x

and

y

are either both even or both odd

The sum of two integers is even if and only if they share the same parity.

3

Deduce parity relationship for

z

across cases

In Case 1 (

x, y

even),

x^2 y

is even, so

z

must be odd. In Case 2 (

x, y

odd),

x^2 y

is odd, so

z

must be even.

To satisfy

x^2 y + z = \text{odd}

,

x^2 y

and

z

must have opposite parities.

4

Evaluate the parity of

(x + z)(y + z)

(x + z)

is odd and

(y + z)

is odd, so their product is odd

In both cases,

z

has opposite parity to both

x

and

y

. Adding two integers of opposite parity always yields an odd integer, and the product of two odd integers is always odd.

Anahtar Kavram

Parity rules under exponentiation and algebraic combination

Tahmini Süre:2m 0s

Soru 8Soru →

If $x$ is an odd integer and $y$ is an even integer, which of the following expressions must result in an even integer? Select all that apply.

Geçerli olan tümünü seçin

xy + x

xy + y

(x + 1)(y + 1)

x^2 + y

x(y + 3)

Cevabı ve açıklamayı göster

Cevap: $xy + y$ ; $(x + 1)(y + 1)$

Cevap

The expressions that must be even are

xy + y

and

(x + 1)(y + 1)

.

The expression

xy + y

is guaranteed to be even because

xy

is even (product of odd and even) and adding another even integer

y

produces an even sum. The expression

(x + 1)(y + 1)

is also guaranteed to be even because adding 1 to an odd integer

x

creates an even integer

x + 1

, and multiplying an even number by any integer always yields an even product.

Adım Adım Çözüm

1

Analyze fundamental parity rules for addition and multiplication of integers.

Recall that

\text{odd} \times \text{even} = \text{even}

,

\text{odd} \times \text{odd} = \text{odd}

,

\text{even} + \text{even} = \text{even}

, and

\text{odd} + \text{even} = \text{odd}

.

Parity rules govern the even/odd behavior of combined expressions.

2

Evaluate the expression

xy + y

.

Since

x

is odd and

y

is even,

xy

is even. Then

\text{even} + y\,(\text{even}) = \text{even}

.

Adding two even terms always results in an even number.

3

Evaluate the expression

(x + 1)(y + 1)

.

Since

x

is odd,

x + 1

is even. The product of an even integer and any integer (

y + 1

) is always even.

An even factor guarantees an even product.

Anahtar Kavram

Even and Odd Parity Rules under Arithmetic Operations

Soru 9Soru →

If $n$ is a negative odd integer, which of the following expressions MUST be a positive even integer?

n^2 + 1

-n^2 + 1

(n + 1)^2

2n - 1

n^3 + 1

Cevabı ve açıklamayı göster

Cevap: $n^2 + 1$

Cevap

The expression

n^2 + 1

must be a positive even integer.

Squaring any negative odd integer

n

produces a positive odd integer (

n^2 \ge 1

). Adding

1

to a positive odd integer always yields a positive even integer (

n^2 + 1 \ge 2

).

Adım Adım Çözüm

1

Analyze the parity and sign of

n^2

Since

n

is negative and odd, any even power of

n

(such as

n^2

) produces a positive odd integer.

Negative times negative yields positive, and odd times odd yields odd.

2

Analyze the effect of adding

1

to

n^2

n^2 + 1

is the sum of a positive odd integer and

1

, which results in a positive even integer.

Adding

1

to any odd integer results in an even integer, and adding

1

to a positive integer maintains positivity.

Anahtar Kavram

Parity and sign rules for exponents and addition

Soru 10Soru →

If $m$ is an even integer and $n$ is an odd integer, what is the value of $(-1)^m + (-1)^n$ ?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

0

Since

m

is an even integer,

(-1)^m = 1

. Since

n

is an odd integer,

(-1)^n = -1

. Adding

1

and

-1

yields

0

.

Adım Adım Çözüm

1

Evaluate

(-1)^m

for an even integer

m

(-1)^m = 1

Raising

-1

to an even integer power always results in

1

.

2

Evaluate

(-1)^n

for an odd integer

n

(-1)^n = -1

Raising

-1

to an odd integer power always results in

-1

.

3

Sum the two evaluated expressions

1 + (-1) = 0

Combining

1

and

-1

equals

0

.

Anahtar Kavram

Exponent sign rules for even and odd powers

Soru 11Soru →

Let $n$ be an integer such that $-15 \le n \le 15$ . How many integer values of $n$ satisfy both of the following conditions?

1. $(-1)^{n^2 + n + 1} < 0$
2. $(n - 3)(n + 4)^2 < 0$

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

17

Condition 1 is satisfied by every integer because

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

is always even (as the product of two consecutive integers), which makes

n^2 + n + 1

always odd and

(-1)^{\text{odd}} = -1 < 0

. Condition 2 requires

(n - 3)(n + 4)^2 < 0

. Since

(n+4)^2 > 0

for all

n \neq -4

, this reduces to

n < 3

while excluding

n = -4

. Within the range

-15 \le n \le 15

, there are 18 integers strictly less than 3, and removing

n = -4

leaves 17 valid values.

Adım Adım Çözüm

1

Determine the parity of the exponent

n^2 + n + 1

n^2 + n + 1

is always odd for any integer

n

, making

(-1)^{n^2 + n + 1} = -1 < 0

unconditionally true.

The product of consecutive integers

n(n+1)

is always even, so adding 1 results in an odd number.

2

Solve the sign inequality

(n - 3)(n + 4)^2 < 0

n < 3

with

n \neq -4

.

A squared expression is strictly positive except when its base is zero. At

n = -4

, the product becomes 0, violating the strict inequality.

3

Count the integer solutions in the range

-15 \le n \le 15

17 integers satisfy both conditions.

There are 18 integers less than 3 in the interval

[-15, 15]

, and excluding

n = -4

gives

18 - 1 = 17

.

Anahtar Kavram

Parity of consecutive integer products and sign rules for squared terms in inequalities

Tahmini Süre:2m 0s

Soru 12Soru →

Let $x$ , $y$ , and $z$ be non-zero integers that satisfy the following three conditions:

I. $(-1)^{x y + z} = -1$
II. $(-1)^{x z + y} = 1$
III. $x^3 y z^2 < 0$

Which of the following statements MUST be true?

x

and

y

are both even integers, and

x y < 0

x

is an odd integer, and

z

is an even integer

y

and

z

are both odd integers

x + z

is an even integer

x y z > 0

Cevabı ve açıklamayı göster

Cevap: $x$ and $y$ are both even integers, and $x y < 0$

Cevap

x

and

y

are both even integers, and

x y < 0

From Condition I,

(-1)^{xy+z} = -1

implies

xy + z

is odd, so

xy

and

z

have opposite parity. From Condition II,

(-1)^{xz+y} = 1

implies

xz + y

is even, so

xz

and

y

have the same parity. If

x

were odd,

xy

would share the parity of

y

, and

xz

would share the parity of

z

. This would mean

y

and

z

must have opposite parity (from Condition I) and the same parity (from Condition II), which is impossible. Thus,

x

must be even. With

x

even,

xy

is even, so

z

must be odd for

xy + z

to be odd. Likewise,

xz

is even, so

y

must be even for

xz + y

to be even. Finally, from Condition III, since

z^2 > 0

for non-zero

z

,

x^3 y z^2 < 0

reduces to

x^3 y < 0

, which means

xy < 0

. Thus,

x

and

y

are both even integers, and

xy < 0

.

Adım Adım Çözüm

1

Analyze Condition I for parity requirements

x y + z

must be an odd integer

Since

(-1)^k = -1

if and only if

k

is an odd integer,

x y + z

is odd. Thus,

x y

and

z

must have opposite parity.

2

Analyze Condition II for parity requirements

x z + y

must be an even integer

Since

(-1)^k = 1

if and only if

k

is an even integer,

x z + y

is even. Thus,

x z

and

y

must have the same parity.

3

Determine the parity of

x

,

y

, and

z

using proof by contradiction

x

is even,

y

is even, and

z

is odd

If

x

were odd, then

x y

would have the same parity as

y

, and

x z

would have the same parity as

z

. Condition I would require

y

and

z

to have opposite parity, while Condition II would require

y

and

z

to have the same parity, a contradiction. Therefore,

x

must be even. Since

x

is even,

x y

is even, making

z

odd (from Condition I). Similarly,

x z

is even, making

y

even (from Condition II).

4

Analyze Condition III for sign requirements

x y < 0

Since

z

is a non-zero integer,

z^2 > 0

. The inequality

x^3 y z^2 < 0

simplifies to

x^3 y < 0

. Since

x^3

has the same sign as

x

,

x^3 y < 0

implies

x y < 0

.

Anahtar Kavram

Parity rules for exponentiation and basic arithmetic operations, combined with sign rules for products

Soru 13Soru →

If $a$ , $b$ , and $c$ are non-zero integers such that $a^3 b^2 c < 0$ , $a + b$ is even, and $b + c$ is odd, which of the following expressions MUST be a negative even integer?

a c (b^2 + 1)

a b c

(a + c) b^2

a^2 c (b + 1)

a c + b

Cevabı ve açıklamayı göster

Cevap: $a c (b^2 + 1)$

Cevap

The expression

a c (b^2 + 1)

must be a negative even integer.

The expression

a c (b^2 + 1)

is guaranteed to be negative because

a c < 0

(derived from

a^3 b^2 c < 0

) and

b^2 + 1 > 0

for all non-zero integers

b

. It is guaranteed to be even because

a

and

c

have opposite parity, meaning one of them must be even, rendering

a c

(and thus any integer multiple of

a c

) even.

Adım Adım Çözüm

1

Analyze the sign constraint

a^3 b^2 c < 0

.

a c < 0

, which means

a

and

c

have opposite signs.

Since

b \neq 0

,

b^2

is strictly positive. Dividing

a^3 b^2 c < 0

by

b^2

yields

a^3 c < 0

. Since

a^3

has the same sign as

a

,

a c < 0

.

2

Analyze the parity constraints

a + b

is even and

b + c

is odd.

a

and

c

have opposite parity (one is even, the other is odd).

If

a + b

is even,

a

and

b

share the same parity. If

b + c

is odd,

b

and

c

have opposite parity. Substituting the parity of

a

for

b

shows that

a

and

c

must have opposite parity.

3

Determine the sign and parity of

a c (b^2 + 1)

.

a c (b^2 + 1)

is strictly negative and even.

Since

a

and

c

have opposite parity, at least one of them is even, making the product

a c

an even integer. Since

a c < 0

and

b^2 + 1 \ge 1 > 0

, the product of negative

a c

and positive

(b^2 + 1)

is negative and even.

Anahtar Kavram

Combining sign rules (

a c < 0

) with even/odd parity logic across multiple variables.

Soru 14Soru →

Let $p$ , $q$ , and $r$ be integers such that $p < 0 < q < r$ . If $p$ is an odd integer, $q$ is an even integer, and $r$ is an odd integer, which of the following expressions must be negative? Select all that apply.

Geçerli olan tümünü seçin

p^q(q - r)

p(p - q)^r

(p - r)(-p)^r

p^r(r - p)^q

(-p)^q(q - p)

Cevabı ve açıklamayı göster

Cevap: $p^q(q - r)$ ; $(p - r)(-p)^r$ ; $p^r(r - p)^q$

Cevap

The expressions that must be negative are

p^q(q - r)

,

(p - r)(-p)^r

, and

p^r(r - p)^q

.

Expressions

p^q(q - r)

,

(p - r)(-p)^r

, and

p^r(r - p)^q

evaluate to the product of a positive factor and a negative factor in every case, making their values strictly negative.

Adım Adım Çözüm

1

Analyze the given signs and parities of variables

p < 0

(negative, odd),

q > 0

(positive, even),

r > 0

(positive, odd), with

q < r

.

Establishing the domain and sign/parity properties of each variable is essential before evaluating exponential and subtractive terms.

2

Evaluate the sign of

p^q(q - r)

p^q > 0

because an even exponent yields a positive result for non-zero bases. Since

q < r

,

(q - r) < 0

. Thus,

\text{positive} \times \text{negative} = \text{negative}

.

Demonstrates that this expression is guaranteed to be negative.

3

Evaluate the sign of

p(p - q)^r

p < 0

.

p - q < 0

, and raising a negative number to an odd exponent

r

gives a negative result. Thus,

p(p - q)^r = \text{negative} \times \text{negative} = \text{positive}

.

Shows that this expression is positive, so it cannot be negative.

4

Evaluate the sign of

(p - r)(-p)^r

p - r < 0

because subtracting a positive number from a negative number is negative.

-p > 0

, so

(-p)^r > 0

. Thus,

\text{negative} \times \text{positive} = \text{negative}

.

Demonstrates that this expression is guaranteed to be negative.

5

Evaluate the sign of

p^r(r - p)^q

p^r < 0

because a negative number raised to an odd exponent is negative.

r - p > 0

, so

(r - p)^q > 0

. Thus,

\text{negative} \times \text{positive} = \text{negative}

.

Demonstrates that this expression is guaranteed to be negative.

6

Evaluate the sign of

(-p)^q(q - p)

-p > 0 \implies (-p)^q > 0

.

q - p > 0

. Thus,

\text{positive} \times \text{positive} = \text{positive}

.

Shows that this expression is always positive.

Anahtar Kavram

Sign rules for bases raised to even vs. odd powers, and order of operations with signed quantities.

Soru 15Soru →

Let $x$ and $y$ be integers such that $x < 0 < y$ . If $x$ and $y$ satisfy all of the following conditions:

1. $(-1)^{x y + x} = -1$
2. $(-1)^{x^2 y + y} = 1$
3. $y^2 - x^2 = 19$

What is the value of $x + y$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The value of

x + y

is 1.

Condition 1 dictates that

(-1)^{x y + x} = -1

, meaning

x y + x = x(y + 1)

is odd. For the product

x(y+1)

to be odd, both

x

and

y+1

must be odd, which means

x

is odd and

y

is even. Condition 3 factors as

(y - x)(y + x) = 19

. Given

x < 0 < y

, we know

y - x > y + x

. Since 19 is prime, its unique positive factor pair requires

y - x = 19

and

y + x = 1

. Solving this system yields

y = 10

and

x = -9

, which satisfies all sign and parity constraints. Thus,

x + y = -9 + 10 = 1

.

Adım Adım Çözüm

1

Analyze parity requirements from Condition 1

x

is odd and

y

is even

Since

(-1)^{x(y+1)} = -1

, the exponent

x(y+1)

must be odd, requiring both

x

and

y+1

to be odd.

2

Check consistency with Condition 2

Condition 2 is satisfied

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

is always even when

y

is even, making

(-1)^{x^2 y + y} = 1

true.

3

Factor difference of squares and set up system using sign rules

y - x = 19

and

y + x = 1

Since 19 is prime and

x < 0 < y

,

y - x > y + x > 0

, forcing the factor pair to be 19 and 1.

4

Solve for

x

and

y

and sum them

x = -9

,

y = 10

, giving

x + y = 1

Adding the system yields

2y = 20 \implies y = 10

, and substituting into

y + x = 1

yields

x = -9

.

Anahtar Kavram

Even-Odd Exponent Rules and Sign Properties of Integers

Soru 16Soru →

Let $p$ , $q$ , and $r$ be non-zero integers satisfying the following three conditions:

I. $p^3 q r^2 < 0$
II. $(-p)^q < 0$
III. $(-1)^{p + r} = 1$

Which of the following expressions MUST be negative?

q \cdot p^r

p \cdot (-q)^r

(-p)^r \cdot q

(p - q)^r

q \cdot (r - p)^q

Cevabı ve açıklamayı göster

Cevap: $q \cdot p^r$

Cevap

The expression

q \cdot p^r

MUST be negative.

Condition II establishes that

p > 0

and

q

is odd. Condition I establishes that

q < 0

. Since

p > 0

, raising

p

to any integer exponent

r

results in a strictly positive value (

p^r > 0

). Multiplying this positive value by negative

q

guarantees a negative outcome regardless of the value or sign of

r

.

Adım Adım Çözüm

1

Analyze Condition II:

(-p)^q < 0

Base

-p < 0 \implies p > 0

, and exponent

q

must be an odd integer.

A negative number raised to an integer power is negative if and only if the exponent is odd.

2

Analyze Condition I:

p^3 q r^2 < 0

q < 0

(negative odd integer).

Since

p > 0

,

p^3 > 0

. Also

r \neq 0 \implies r^2 > 0

. For the overall product

p^3 q r^2

to be negative,

q

must be negative.

3

Analyze Condition III:

(-1)^{p + r} = 1

p + r

is an even integer, so

p

and

r

have the same parity.

(-1)^k = 1

requires

k

to be an even integer.

4

Evaluate the sign of

q \cdot p^r

q \cdot p^r < 0

for all valid values.

Since

p > 0

, any integer power

p^r > 0

. Multiplying positive

p^r

by negative

q

yields a negative result.

Anahtar Kavram

Even-odd exponent rules and negative base sign determination

Soru 17Soru →

If $k$ and $m$ are integers such that $(-k)^5 m < 0$ and $k - m$ is an odd integer, which of the following expressions must be a positive even integer?

k^2 m^2

(-k)^2 m

(k - m)^2

-k m

k^3 + m^3

Cevabı ve açıklamayı göster

Cevap: $k^2 m^2$

Cevap

The expression

k^2 m^2

must be a positive even integer.

Simplifying

(-k)^5 m < 0

gives

-k^5 m < 0 \implies k^5 m > 0

. This confirms

k \neq 0

and

m \neq 0

, so both

k^2

and

m^2

are positive integers, making

k^2 m^2 > 0

. Additionally,

k - m

being odd requires one variable to be even and the other odd. Squaring an even integer yields an even integer, and multiplying by any integer keeps it even. Hence, the expression

k^2 m^2

is guaranteed to be a positive even integer.

Adım Adım Çözüm

1

Analyze the sign condition

(-k)^5 m < 0

.

Since

(-k)^5 = -k^5

, the inequality becomes

-k^5 m < 0

, which means

k^5 m > 0

. This implies that neither

k

nor

m

is zero, and both

k

and

m

have the same sign (either both positive or both negative).

Raising a negative quantity to an odd power retains the negative sign.

2

Analyze the parity condition

k - m

is odd.

The difference between two integers is odd if and only if one integer is even and the other is odd.

Even minus odd (or odd minus even) produces an odd result.

3

Evaluate the sign and parity of

k^2 m^2

.

Since

k \neq 0

and

m \neq 0

,

k^2 > 0

and

m^2 > 0

, so

k^2 m^2 > 0

(positive). Since one of

k

or

m

is even, its square is also even, so the product

k^2 m^2

must be even. Thus,

k^2 m^2

is guaranteed to be a positive even integer.

The product of non-zero squares is positive, and any integer multiple of an even number is even.

Anahtar Kavram

Parity rules under subtraction/multiplication and sign rules under odd powers.

Soru 18Soru →

Let $a$ , $b$ , and $c$ be non-zero integers such that $\frac{a}{b} < 0$ , $a^3 b c > 0$ , and $a + b$ is an odd integer. Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

c < 0

a b

is an even integer

a c > 0

a^2 + c

is an even integer

b + c < 0

Cevabı ve açıklamayı göster

Cevap: $c < 0$ ; $a b$ is an even integer

Cevap

The statements '

c < 0

' and '

a b

is an even integer' must be true.

The statement '

c < 0

' must be true because

\frac{a}{b} < 0

forces

a b < 0

, and since

a^3 b c = a^2 (a b) c

with

a^2 > 0

,

c

must be negative to yield a positive product. The statement '

a b

is an even integer' must be true because an odd sum

a + b

requires one variable to be even and the other to be odd, making their product even.

Adım Adım Çözüm

1

Determine the sign relationship between

a

and

b

.

a

and

b

have opposite signs, so

a b < 0

.

The quotient

\frac{a}{b} < 0

implies the numerator and denominator have different signs.

2

Determine the sign of

c

using

a^3 b c > 0

.

c < 0

.

Rewrite

a^3 b c

as

a^2 \cdot (a b) \cdot c

. Since

a \neq 0

,

a^2 > 0

. Since

a b < 0

, the product

a^2 (a b) < 0

. For the entire product

a^2 (a b) c

to be positive,

c

must be negative.

3

Analyze the parity of

a

and

b

from

a + b

being odd.

One of

a

or

b

is even and the other is odd, so

a b

must be even.

An odd sum of two integers requires one even and one odd addend. The product of an even integer and any integer is always even.

4

Test the remaining options for counterexamples.

The statements '

a c > 0

', '

a^2 + c

is an even integer', and '

b + c < 0

' can be false under valid assignments.

For example, if

a = 2

,

b = -1

, and

c = -3

, then

\frac{a}{b} = -2 < 0

,

a^3 b c = 8(-1)(-3) = 24 > 0

, and

a + b = 1

(odd). Here,

a c = -6 < 0

,

a^2 + c = 4 - 3 = 1

(odd), and

b + c = -4 < 0

, but setting

b = 5, a = -2, c = -1

gives

b + c = 4 > 0

.

Anahtar Kavram

Deducing sign and parity properties of integers

Tahmini Süre:1m 30s

Soru 19Soru →

Let $m$ and $n$ be integers such that $m < 0$ , $n > 0$ , $(-1)^m = -1$ , and $m + n$ is an even integer. Which of the following expressions must be a positive even integer?

m \cdot n

(-m)^n

n - m

(m + n)^2

m^2 - n

Cevabı ve açıklamayı göster

Cevap: $n - m$

Cevap

The expression

n - m

must be a positive even integer.

The expression

n - m

subtracts a negative odd integer from a positive odd integer, which equals adding two positive odd integers. The sum of two positive odd integers is always a positive even integer.

Adım Adım Çözüm

1

Determine the parity and sign of

m

.

Since

m < 0

and

(-1)^m = -1

,

m

must be a negative odd integer.

An odd exponent on

-1

yields

-1

.

2

Determine the parity and sign of

n

.

Since

n > 0

and

m + n

is even,

n

must be a positive odd integer.

The sum of two integers is even if and only if both integers have the same parity. Since

m

is odd,

n

must also be odd.

3

Evaluate the sign and parity of

n - m

.

n - m = n + (-m)

. Since

n \ge 1

and

-m \ge 1

,

n - m \ge 2

(strictly positive). Also, odd minus odd is always even.

Combining the sign rules (

n > 0

and

-m > 0

) with the even-odd subtraction rule confirms

n - m

is always a positive even integer.

Anahtar Kavram

Even-Odd Properties and Sign Rules

Soru 20Soru →

Three consecutive integers $a$ , $b$ , and $c$ satisfy $a < b < c$ . If $a + b + c = -9$ and $a \cdot b \cdot c < 0$ , what is the value of $(-1)^a + (-1)^b + (-1)^c$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The value of the expression is 1.

The sum of three consecutive integers

a + b + c = 3b = -9

determines

b = -3

, making

a = -4

and

c = -2

. The product

(-4)(-3)(-2) = -24

is negative, confirming the given condition. Applying exponent sign rules, raising

-1

to an even integer power yields

1

, while raising

-1

to an odd integer power yields

-1

. Thus,

(-1)^{-4} = 1

,

(-1)^{-3} = -1

, and

(-1)^{-2} = 1

. Summing these three terms gives

1 + (-1) + 1 = 1

.

Adım Adım Çözüm

1

Find the values of integers

a

,

b

, and

c

.

a = -4

,

b = -3

,

c = -2

Three consecutive integers centered at

b

sum to

3b = -9

, so

b = -3

.

2

Check the sign condition of the product

a \cdot b \cdot c

.

(-4)(-3)(-2) = -24 < 0

The product of three negative numbers is negative.

3

Evaluate

(-1)^n

for each integer power.

(-1)^{-4} = 1

,

(-1)^{-3} = -1

,

(-1)^{-2} = 1

Negative one raised to an even integer power is 1; raised to an odd integer power is -1.

4

Sum the three evaluated terms.

1 + (-1) + 1 = 1

Addition of the resulting values.

Anahtar Kavram

Even-odd exponent rules for negative bases and sign rules for product of signed integers.

Even-Odd Properties and Sign Rules

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