Number Properties and Integer Constraints in Data Sufficiency Practice Questions — GMAT

Question 1Question →

Is $x$ an integer?

(1) $3x$ is an integer.
(2) $5x$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Evaluating Statement (1) alone shows

x

could be

\frac{1}{3}

(not an integer) or

1

(an integer), so Statement (1) is insufficient. Evaluating Statement (2) alone shows

x

could be

\frac{1}{5}

(not an integer) or

1

(an integer), so Statement (2) is insufficient. Taking both statements together,

3x = a

and

5x = b

for integers

a

and

b

. Subtracting

5x

from

2(3x)

yields

6x - 5x = x = 2a - b

. Because integers are closed under multiplication and subtraction,

2a - b

must be an integer, confirming that

x

is definitively an integer.

Step-by-Step Solution

1

Evaluate Statement (1) independently

If

3x = a

where

a

is an integer, then

x = \frac{a}{3}

. If

a = 1

,

x = \frac{1}{3}

(not an integer). If

a = 3

,

x = 1

(an integer).

Since

x

can be either an integer or a non-integer, Statement (1) alone is not sufficient.

2

Evaluate Statement (2) independently

If

5x = b

where

b

is an integer, then

x = \frac{b}{5}

. If

b = 1

,

x = \frac{1}{5}

(not an integer). If

b = 5

,

x = 1

(an integer).

Since

x

can be either an integer or a non-integer, Statement (2) alone is not sufficient.

3

Evaluate Statements (1) and (2) together

Since

3x

and

5x

are both integers, their linear combination

2(3x) - 5x = 6x - 5x = x

must also be an integer.

The difference between two integers is always an integer, proving conclusively that

x

is an integer.

Key Concept

Linear combinations of real numbers and integer constraints in Data Sufficiency

Estimated Time:1m 0s

Question 2Question →

If $x$ is a positive real number, is $x$ an integer?

(1) $x^3$ is an integer.
(2) $x^5$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Evaluating each statement independently leaves open the possibility that

x

is an irrational root like

\sqrt[3]{2}

or

\sqrt[5]{2}

, making neither statement alone sufficient. Combining both statements allows us to express

x

as

\frac{(x^3)^2}{x^5}

. Because both

x^3

and

x^5

are integers,

x

must be a rational number. For any rational number whose cube is an integer, its denominator must equal 1, which proves conclusively that

x

is an integer. Thus, both statements together are sufficient.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) is insufficient.

If

x = 2

, then

x^3 = 8

(an integer), so

x

is an integer (Yes). However, if

x = \sqrt[3]{2}

, then

x^3 = 2

(an integer), but

x

is not an integer (No). Since both Yes and No answers are possible, Statement (1) alone is not sufficient.

2

Evaluate Statement (2) independently.

Statement (2) is insufficient.

If

x = 2

, then

x^5 = 32

(an integer), so

x

is an integer (Yes). However, if

x = \sqrt[5]{2}

, then

x^5 = 2

(an integer), but

x

is not an integer (No). Since both Yes and No answers are possible, Statement (2) alone is not sufficient.

3

Combine Statement (1) and Statement (2).

The combined statements are sufficient.

Notice that

\frac{(x^3)^2}{x^5} = \frac{x^6}{x^5} = x

. Since

x^3

is an integer,

(x^3)^2

is also an integer. Since

x^5

is an integer,

x = \frac{(x^3)^2}{x^5}

is the ratio of two integers, meaning

x

must be a rational number. If

x = \frac{a}{b}

in simplest form where

a

and

b

are positive integers with

\gcd(a,b)=1

, then

x^3 = \frac{a^3}{b^3}

being an integer implies

b^3 = 1

, so

b = 1

. Thus,

x = a

, which means

x

must be an integer. This yields a definitive 'Yes' answer.

Key Concept

Integer constraints versus real numbers in Data Sufficiency and quotient relationships of exponent powers

Question 3Question →

If $x$ and $y$ are positive integers, is $x^3 - y^3$ divisible by $3$ ?

(1) $x + 2y$ is divisible by $3$ .
(2) $x^2 - y^2$ is divisible by $3$ .

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

The question asks whether

x^3 - y^3

is divisible by 3. By Fermat's Little Theorem or factoring

a^3 - a = a(a-1)(a+1)

(the product of three consecutive integers, always divisible by 3), any integer cubed has the same remainder when divided by 3 as the integer itself (

a^3 \equiv a \pmod 3

). Therefore,

x^3 - y^3

is a multiple of 3 if and only if

x - y

is a multiple of 3. Statement (1) specifies that

x + 2y

is divisible by 3. We can rewrite

x + 2y

as

(x - y) + 3y

. Because

3y

is automatically a multiple of 3,

(x - y)

must also be divisible by 3. This yields a definitive 'Yes', so Statement (1) alone is sufficient. Statement (2) states that

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

is divisible by 3. Because 3 is prime, this requires 3 to divide either

(x - y)

or

(x + y)

. If

x = 4, y = 1

, 3 divides

(x - y) = 3

, giving a 'Yes'. If

x = 2, y = 1

, 3 divides

(x + y) = 3

while

(x - y) = 1

, giving

x^3 - y^3 = 7

, which is a 'No'. Hence, Statement (2) alone is not sufficient.

Step-by-Step Solution

1

Rephrase the question stem using algebraic modular properties.

Since

a^3 \equiv a \pmod 3

for any integer

a

,

x^3 - y^3 \equiv x - y \pmod 3

. Thus,

x^3 - y^3

is divisible by 3 if and only if

x - y

is divisible by 3.

Simplifying the target expression reduces the problem to determining whether 3 divides

(x - y)

.

2

Evaluate Statement (1):

x + 2y

is divisible by 3.

x + 2y = (x - y) + 3y

. Since

3y

is always a multiple of 3,

(x + 2y)

is divisible by 3 if and only if

(x - y)

is divisible by 3.

This guarantees a definitive 'Yes' to the target question. Statement (1) alone is SUFFICIENT.

3

Evaluate Statement (2):

x^2 - y^2

is divisible by 3.

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

. Since 3 is prime, 3 must divide

(x - y)

or

(x + y)

. If

x = 4, y = 1

, then

x^2 - y^2 = 15

(divisible by 3) and

x^3 - y^3 = 63

(divisible by 3) -> YES. If

x = 2, y = 1

, then

x^2 - y^2 = 3

(divisible by 3), but

x^3 - y^3 = 7

(NOT divisible by 3) -> NO.

Statement (2) yields both 'Yes' and 'No' cases. Statement (2) alone is NOT SUFFICIENT.

Key Concept

Divisibility and Modular Congruences in Integer Properties

Estimated Time:2m 0s

Question 4Question →

If $x$ is a non-zero real number, is $x$ an integer?

(1) $x + \frac{6}{x}$ is an integer.

(2) $x^2 + \frac{36}{x^2}$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statements (1) and (2) TOGETHER are NOT sufficient.

The correct response establishes that neither statement individually nor both statements combined are sufficient to determine whether the real number is an integer. Testing specific real numbers demonstrates that integer values such as 2 and irrational values such as 3 plus the square root of 3 satisfy both algebraic conditions, producing affirmative and negative answers to the question.

Step-by-Step Solution

1

Analyze Statement (1):

x + \frac{6}{x} = k

for some integer

k

.

If

x = 2

, then

2 + \frac{6}{2} = 5

(an integer), and

x

IS an integer. If

x = 3 + \sqrt{3}

, then

x + \frac{6}{x} = 3 + \sqrt{3} + \frac{6(3 - \sqrt{3})}{6} = 6

(an integer), but

x

is NOT an integer.

Since Statement (1) yields both a 'Yes' and a 'No' answer, it is not sufficient.

2

Analyze Statement (2):

x^2 + \frac{36}{x^2} = m

for some integer

m

.

If

x = 2

, then

2^2 + \frac{36}{2^2} = 4 + 9 = 13

(an integer), so

x

IS an integer. If

x = \sqrt{2}

, then

(\sqrt{2})^2 + \frac{36}{(\sqrt{2})^2} = 2 + 18 = 20

(an integer), but

x

is NOT an integer.

Since Statement (2) yields both a 'Yes' and a 'No' answer, it is not sufficient.

3

Analyze Statements (1) and (2) combined.

Note that

\left(x + \frac{6}{x}\right)^2 = x^2 + 12 + \frac{36}{x^2}

. Thus

x^2 + \frac{36}{x^2} = \left(x + \frac{6}{x}\right)^2 - 12

. Whenever

x + \frac{6}{x}

is an integer

k

,

x^2 + \frac{36}{x^2} = k^2 - 12

is automatically an integer. Therefore, Statement (2) provides no extra constraints.

Both

x = 2

(integer) and

x = 3 + \sqrt{3}

(non-integer) satisfy both statements simultaneously, so the combined statements remain insufficient.

Key Concept

Evaluating algebraic expressions for non-integer real number solutions in Data Sufficiency

Question 5Question →

If $k$ is a non-zero real number, is $k$ an integer?

(1) $\frac{12}{k}$ is an integer.

(2) $k^2 - 3k$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to answer the question, but neither statement alone is sufficient.

The correct response demonstrates that neither statement individually restricts

k

exclusively to integer values, but combining them enforces that

k

is rational (from the reciprocal condition) and has a denominator of 1 (from the quadratic condition), proving conclusively that

k

must be an integer.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) states that

\frac{12}{k} = m

for some non-zero integer

m

, so

k = \frac{12}{m}

. If

m = 1

,

k = 12

(an integer). If

m = 24

,

k = 0.5

(not an integer). Since

k

can be either an integer or a non-integer, Statement (1) is NOT sufficient.

Testing specific values reveals that

k

can be fractional while fulfilling the condition.

2

Evaluate Statement (2) independently.

Statement (2) states that

k^2 - 3k = n

for some integer

n

. If

k = 4

,

k^2 - 3k = 4

(an integer). However, if

k = \frac{3 + \sqrt{13}}{2}

,

k^2 - 3k = 1

(an integer), but

k

is irrational. Since

k

can be an integer or irrational, Statement (2) is NOT sufficient.

Quadratic expressions with integer outputs can have irrational inputs.

3

Evaluate Statements (1) and (2) together.

From Statement (1),

k

must be rational. Let

k = \frac{p}{q}

in lowest terms, where

p

and

q

are coprimes (

\gcd(p, q) = 1

) and

q \ge 1

. Substitute

k = \frac{p}{q}

into Statement (2):

k^2 - 3k = \frac{p^2 - 3pq}{q^2}

. For this expression to be an integer,

q^2

must divide

p(p - 3q)

. Since

\gcd(p, q) = 1

,

q

shares no common prime factors with

p

or

p - 3q

. Thus

\gcd(p^2 - 3pq, q^2) = 1

, which means

q^2

can divide

p^2 - 3pq

if and only if

q^2 = 1

, so

q = 1

. Since

q = 1

,

k = p

, which guarantees

k

is an integer. Thus, the combined statements yield a definitive 'Yes'.

Combining rationality from the first condition with the algebraic divisibility constraint of the second condition eliminates all non-integer possibilities.

Key Concept

Data Sufficiency logic with integer constraints and rational number representations

Estimated Time:2m 0s

Question 6Question →

If $x$ and $y$ are positive integers, is the product $xy$ divisible by 4?

(1) $x$ is divisible by 4.
(2) $x + y$ is an odd integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

The correct answer is that Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (1) establishes that

x

is a multiple of 4, so any integer product

xy

must also be a multiple of 4. Statement (2) only guarantees that one of

x

or

y

is even, which is not enough to determine if the product has two factors of 2.

Step-by-Step Solution

1

Analyze the question stem target

We need to determine whether

xy

is divisible by 4, given that

x

and

y

are positive integers.

Establishing integer constraints ensures we evaluate divisibility conditions properly.

2

Evaluate Statement (1) independently

If

x

is divisible by 4, then

x = 4k

for some positive integer

k

. Then

xy = (4k)y = 4(ky)

, which is divisible by 4. This gives a definitive YES.

Multiplying any multiple of 4 by an integer results in a multiple of 4. Therefore, Statement (1) alone is sufficient.

3

Evaluate Statement (2) independently

If

x + y

is odd, one integer is even and the other is odd. Case 1: If

x = 4

and

y = 1

,

x+y=5

(odd) and

xy = 4

(divisible by 4 -> YES). Case 2: If

x = 2

and

y = 3

,

x+y=5

(odd) and

xy = 6

(not divisible by 4 -> NO).

Since Statement (2) yields both YES and NO outcomes, it alone is not sufficient.

Key Concept

Divisibility and Parity Rules in Data Sufficiency

Question 7Question →

If $m$ is a real number, is $m$ an even integer?

(1) $2m$ is an even integer.
(2) $m + 1$ is an odd integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Answer

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Statement (1) allows

m

to be any integer because multiplying any integer by

2

results in an even integer. Therefore,

m

could be

3

(odd) or

4

(even), making Statement (1) insufficient. Statement (2) specifies that

m + 1

is an odd integer. Subtracting

1

from an odd integer always yields an even integer, so

m

must be an even integer. This provides a single, definitive 'Yes' answer, making Statement (2) alone sufficient.

Step-by-Step Solution

1

Analyze the target question stem.

We need to determine whether

m

is an even integer, given that

m

is a real number.

Rephrasing the question clarifies that a definitive 'Yes' or definitive 'No' answer is required.

2

Evaluate Statement (1):

2m

is an even integer.

If

m = 4

, then

2m = 8

(even integer), and

m

is an even integer (Yes). If

m = 3

, then

2m = 6

(even integer), but

m

is an odd integer (No).

Since Statement (1) yields both 'Yes' and 'No' answers, Statement (1) is NOT sufficient.

3

Evaluate Statement (2):

m + 1

is an odd integer.

If

m + 1

is an odd integer, then

m = (m + 1) - 1

. An odd integer minus

1

is always an even integer.

This guarantees that

m

is an even integer, providing a definitive 'Yes'. Thus, Statement (2) ALONE is sufficient.

Key Concept

Parity and Integer Properties in Data Sufficiency

Question 8Question →

If $n$ is a real number, is $n$ an integer?

(1) $n^2 - 3n$ is an integer.
(2) $n^2 + 5n$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to answer definitively that

n

is an integer, but neither statement alone is sufficient.

Neither statement alone is sufficient because quadratic equations of the form

n^2 - 3n = k

and

n^2 + 5n = m

have non-integer real roots for certain integer values of

k

and

m

. When both statements are combined, subtracting the two equations reveals that

8n

is equal to an integer

p

, so

n = \frac{p}{8}

. Substituting

n = \frac{p}{8}

into

n^2 - 3n

requires

\frac{p(p-24)}{64}

to be an integer. Examining the power of 2 in

p(p-24)

proves that

p

must be divisible by 8, which guarantees that

n

is an integer. Thus, the two statements together provide a definitive 'Yes' answer.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) is NOT sufficient.

If

n^2 - 3n = -1

(an integer), then

n^2 - 3n + 1 = 0

, giving

n = \frac{3 + \sqrt{5}}{2}

, which is not an integer. However, if

n = 3

,

n^2 - 3n = 0

, which is an integer. Thus,

n

can be an integer or a non-integer.

2

Evaluate Statement (2) independently.

Statement (2) is NOT sufficient.

If

n^2 + 5n = 1

(an integer), then

n^2 + 5n - 1 = 0

, giving

n = \frac{-5 + \sqrt{29}}{2}

, which is not an integer. However, if

n = 1

,

n^2 + 5n = 6

, which is an integer. Thus,

n

can be an integer or a non-integer.

3

Combine Statement (1) and Statement (2).

Deduce that

8n

must be an integer.

Let

n^2 - 3n = k

and

n^2 + 5n = m

, where

k

and

m

are integers. Subtracting the first equation from the second yields

(n^2 + 5n) - (n^2 - 3n) = m - k \implies 8n = m - k

. Since

m - k

is an integer, let

8n = p

for some integer

p

. Hence,

n = \frac{p}{8}

.

4

Analyze whether

n = \frac{p}{8}

can be a non-integer fraction.

n

MUST be an integer, making BOTH statements TOGETHER sufficient.

Substitute

n = \frac{p}{8}

back into Statement (1):

n^2 - 3n = \frac{p^2}{64} - \frac{3p}{8} = \frac{p^2 - 24p}{64}

. For this value to be an integer,

p^2 - 24p = p(p - 24)

must be divisible by

64 = 2^6

. If

p

is odd,

p - 24

is odd, so

p(p-24)

is odd and not divisible by 64. If

p

is a multiple of 2 but not 4 (

p = 2b

with

b

odd),

p(p-24) = 4b(b-12)

, containing only

2^2

factors of 2. If

p

is a multiple of 4 but not 8 (

p = 4b

with

b

odd),

p(p-24) = 16b(b-6)

, containing only

2^4

factors of 2. Thus, for

p(p-24)

to contain

2^6

,

p

must be a multiple of 8. If

p

is a multiple of 8, then

n = \frac{p}{8}

must be an integer.

Key Concept

Evaluating integer constraints vs. real numbers in Data Sufficiency through algebraic elimination and prime factorization of rational expressions.

Question 9Question →

If $x$ is a real number, is $x$ an integer?

(1) $x^2 + x$ is an integer.
(2) $x^3 + x^2$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to determine that

x

is an integer, but neither statement alone is sufficient.

Evaluating each statement independently reveals that non-integer real numbers can produce integer values for

x^2 + x

or

x^3 + x^2

. However, combining both statements allows us to factor

x^3 + x^2

as

x(x^2 + x)

. Setting

x^2 + x = m

and

x^3 + x^2 = k

for integers

m

and

k

, we find

x \cdot m = k

. If

m = 0

,

x

is 0 or -1 (integers). If

m \neq 0

,

x = \frac{k}{m}

is a rational number. Writing

x = \frac{a}{b}

in lowest terms and substituting it back into

x^2 + x = m

shows that

b

must divide

a^2

, forcing

b = 1

. Thus,

x

must be an integer, providing a definitive affirmative answer.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) is INSUFFICIENT.

If

x = 2

, then

x^2 + x = 6

, which is an integer (YES). However, if

x = \frac{-1 + \sqrt{5}}{2}

, then

x^2 + x - 1 = 0 \implies x^2 + x = 1

, which is an integer, but

x

is not an integer (NO). Thus, Statement (1) alone is insufficient.

2

Evaluate Statement (2) independently.

Statement (2) is INSUFFICIENT.

If

x = 2

, then

x^3 + x^2 = 12

, an integer (YES). If

x

is the real root of

x^3 + x^2 = 1

(where

x \approx 0.755

),

x^3 + x^2

is an integer, but

x

is not an integer (NO). Thus, Statement (2) alone is insufficient.

3

Evaluate Statements (1) and (2) together.

Statements (1) and (2) together are SUFFICIENT.

Let

x^2 + x = m

where

m \in \mathbb{Z}

, and

x^3 + x^2 = k

where

k \in \mathbb{Z}

. Notice that

x^3 + x^2 = x(x^2 + x) = x \cdot m = k

. If

m = 0

, then

x^2 + x = 0 \implies x = 0

or

x = -1

, both of which are integers. If

m \neq 0

, then

x = \frac{k}{m}

, meaning

x

must be a rational number. Express

x = \frac{a}{b}

in lowest terms, where

a, b \in \mathbb{Z}

,

b > 0

, and

\gcd(a, b) = 1

. Substituting

x = \frac{a}{b}

into

x^2 + x = m

yields

\frac{a^2 + ab}{b^2} = m \implies a^2 + ab = m b^2 \implies a^2 = b(mb - a)

. This implies that

b

must divide

a^2

. Since

\gcd(a, b) = 1

,

b

can only divide

a^2

if

b = 1

. Therefore,

x = \frac{a}{1} = a

, which means

x

MUST be an integer. The answer to the question is a definitive YES.

Key Concept

Testing real vs. integer constraints and applying rational root divisibility properties in Data Sufficiency.

Estimated Time:2m 0s

Question 10Question →

If $x$ and $y$ are positive real numbers, is $x$ an integer?

(1) $x + y$ is an integer, and $x^2 + xy$ is an integer.
(2) $x - y$ is an integer, and $x^2 - y^2$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statements (1) and (2) together are not sufficient to determine whether

x

is an integer.

The correct option identifies that statements (1) and (2) together are not sufficient. Testing half-integers such as

x = 2.5

and

y = 1.5

shows that all given conditions (

x+y=4

,

x^2+xy=10

,

x-y=1

,

x^2-y^2=4

) are integers while

x

is not an integer. Conversely,

x = 3

and

y = 1

satisfies all conditions with an integer

x

. Thus, a definitive Yes/No answer cannot be determined.

Step-by-Step Solution

1

Analyze Statement (1) independently.

Statement (1) states that

x + y = k

(where

k

is an integer) and

x^2 + xy = x(x+y) = m

(where

m

is an integer). Thus,

x \cdot k = m \implies x = \frac{m}{k}

. Since

m

and

k

are integers,

x

is rational, but not necessarily an integer. For example, if

x = 1.5

and

y = 0.5

, then

x + y = 2

(integer) and

x^2 + xy = 1.5(2) = 3

(integer), but

x = 1.5

is not an integer. If

x = 2

and

y = 1

, then

x + y = 3

and

x^2 + xy = 6

, and

x = 2

is an integer. Thus, Statement (1) alone is INSUFFICIENT.

Variable

x

is only constrained to be a rational number

\frac{m}{k}

, which may or may not be an integer.

2

Analyze Statement (2) independently.

Statement (2) states that

x - y = d

(where

d

is an integer) and

x^2 - y^2 = (x-y)(x+y) = p

(where

p

is an integer). Thus

d(x+y) = p \implies x+y = \frac{p}{d}

. Then

2x = (x+y) + (x-y) = \frac{p}{d} + d = \frac{p + d^2}{d} \implies x = \frac{p + d^2}{2d}

. If

x = 2.5

and

y = 1.5

, then

x - y = 1

(integer) and

x^2 - y^2 = 6.25 - 2.25 = 4

(integer), but

x = 2.5

is not an integer. If

x = 2

and

y = 1

,

x - y = 1

and

x^2 - y^2 = 3

, and

x = 2

is an integer. Thus, Statement (2) alone is INSUFFICIENT.

Variable

x

is constrained to a rational form with denominator

2d

, which allows non-integer values.

3

Analyze Statements (1) and (2) combined.

Combining both statements gives

x + y = k

(integer) and

x - y = d

(integer). Adding these yields

2x = k + d

, so

2x

must be an integer. This means

x

can be an integer or a half-integer (e.g.,

2.5

). Testing

x = 2.5

and

y = 1.5

: both are positive real numbers;

x + y = 4

(integer);

x^2 + xy = 2.5(4) = 10

(integer);

x - y = 1

(integer); and

x^2 - y^2 = 6.25 - 2.25 = 4

(integer). Here

x

is NOT an integer. Testing

x = 3

and

y = 1

: both are positive real numbers;

x + y = 4

(integer);

x^2 + xy = 3(4) = 12

(integer);

x - y = 2

(integer); and

x^2 - y^2 = 9 - 1 = 8

(integer). Here

x

IS an integer. Because both 'Yes' and 'No' answers are possible under all conditions, Statements (1) and (2) together are INSUFFICIENT.

The system of equations allows

x

to be a half-integer (such as

2.5

), satisfying all integer conditions of both statements without requiring

x

itself to be an integer.

Key Concept

Evaluating implicit non-integer constraints and system degree in Data Sufficiency questions.

Question 11Question →

If $k$ is a positive real number, is $k$ an integer?

(1) $\sqrt{k}$ is an integer.
(2) $k^2 - 4k$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (1) alone is sufficient because if the square root of

k

is an integer

m

, then

k = m^2

, which is guaranteed to be an integer. Statement (2) alone is not sufficient because

k^2 - 4k

can equal an integer when

k = 4

(an integer) or when

k = 2 + \sqrt{5}

(a non-integer). Therefore, only Statement (1) is sufficient.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

If

\sqrt{k} = m

where

m

is an integer, then

k = m^2

. Since

m

is an integer,

m^2

must also be an integer. Thus,

k

is definitely an integer (Yes).

The square of any integer is always an integer, so Statement (1) is sufficient.

2

Evaluate Statement (2) independently.

Let

k^2 - 4k = n

, where

n

is an integer. If

n = 0

, then

k(k - 4) = 0 \implies k = 4

(an integer). If

n = 1

, then

k^2 - 4k - 1 = 0 \implies k = 2 + \sqrt{5}

(not an integer).

Since

k

can be either an integer or a non-integer, Statement (2) does not provide a definitive Yes/No answer and is not sufficient.

Key Concept

Algebraic definitions of integers versus real numbers in Data Sufficiency

Estimated Time:2m 0s

Question 12Question →

If $k$ is a positive integer, is $k$ divisible by 6?

(1) $k$ is a multiple of 2.
(2) $k$ is a multiple of 3.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient, but neither statement alone is sufficient.

To determine whether an integer is divisible by 6, it must be divisible by both prime factors of 6, which are 2 and 3. Statement (1) indicates

k

is divisible by 2, and Statement (2) indicates

k

is divisible by 3. Neither statement alone is sufficient because a number can be a multiple of 2 without being a multiple of 3 (e.g., 2, 4) or a multiple of 3 without being a multiple of 2 (e.g., 3, 9). Taking both statements together ensures

k

is a multiple of both 2 and 3, which means

k

is a multiple of 6. Therefore, both statements together are sufficient.

Step-by-Step Solution

1

Rephrase the question target.

For a positive integer

k

to be divisible by 6, it must be divisible by both 2 and 3 because

\text{lcm}(2,3) = 6

.

6 is the product of the prime numbers 2 and 3.

2

Evaluate Statement (1) independently.

If

k = 2

, it is divisible by 2 but not by 6 (No). If

k = 6

, it is divisible by 2 and by 6 (Yes). Insufficient.

Statement (1) alone allows multiple outcomes for divisibility by 6.

3

Evaluate Statement (2) independently.

If

k = 3

, it is divisible by 3 but not by 6 (No). If

k = 6

, it is divisible by 3 and by 6 (Yes). Insufficient.

Statement (2) alone allows multiple outcomes for divisibility by 6.

4

Combine Statement (1) and Statement (2).

Since

k

is divisible by 2 and divisible by 3,

k

must be a multiple of

\text{lcm}(2,3) = 6

. Thus,

k

is definitely divisible by 6 (Definitive Yes).

Both conditions combined uniquely answer the question with a definitive Yes.

Key Concept

Divisibility Rules and Prime Factorization

Question 13Question →

Is the real number $p$ negative?

(1) $p^2 > 0$
(2) $p^3 < 0$

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Answer

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Statement (1) tells us that

p^2 > 0

, which means

p

can be any real number except 0. Since

p

could be positive (e.g.,

p = 5

) or negative (e.g.,

p = -5

), Statement (1) alone cannot answer whether

p

is negative.

Statement (2) tells us that

p^3 < 0

. Because taking an odd power of a real number preserves its sign,

p^3 < 0

implies

p

must be negative. This provides a definitive 'Yes' answer to the question stem, making Statement (2) sufficient alone.

Step-by-Step Solution

1

Evaluate Statement (1):

p^2 > 0

p

can be any non-zero real number (e.g.,

p = 2

or

p = -2

).

Squaring any non-zero real number yields a positive value. Thus,

p

could be positive or negative, so Statement (1) is not sufficient.

2

Evaluate Statement (2):

p^3 < 0

p

must be negative.

An odd exponent preserves the sign of the base. If

p^3 < 0

, then

p

must be negative. This provides a definitive 'Yes' answer to whether

p

is negative, so Statement (2) is sufficient.

Key Concept

Properties of real numbers under even and odd exponents

Estimated Time:1m 0s

Question 14Question →

If $r$ and $s$ are positive integers, is the sum $r + s$ an odd integer?

(1) $r \cdot s$ is an odd integer.
(2) $r$ is an even integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (1) gives enough information to determine that

r + s

must be even, resulting in a definitive 'No' answer to the question. In Data Sufficiency, a definitive 'No' indicates sufficiency. Statement (2) does not fix the parity of

s

, so

r + s

could be either even or odd, making it insufficient.

Step-by-Step Solution

1

Analyze the question stem

We are given that

r

and

s

are positive integers and need to answer the Yes/No question: Is

r + s

odd?

Establishing the target question and constraints is essential before evaluating the statements.

2

Evaluate Statement (1):

r \cdot s

is an odd integer

The product of two integers is odd if and only if both integers are odd. Therefore,

r

is odd and

s

is odd. The sum of two odd integers is always an even integer (odd + odd = even). Thus,

r + s

is definitely not odd.

A definitive 'No' answer to a Yes/No Data Sufficiency question means the statement is SUFFICIENT.

3

Evaluate Statement (2):

r

is an even integer

If

r = 2

and

s = 1

, then

r + s = 3

(odd, answer is Yes). If

r = 2

and

s = 2

, then

r + s = 4

(even, answer is No). Because both Yes and No are possible, Statement (2) is INSUFFICIENT.

Since Statement (2) allows multiple outcomes for the target question, it does not provide sufficient information.

Key Concept

Parity rules for addition and multiplication of integers in Data Sufficiency Yes/No decision logic.

Question 15Question →

If $x$ and $y$ are positive real numbers, is $x$ an integer?

(1) $x + y$ is an integer and $xy = 12$ .
(2) $x - y$ is an integer and $x^2 + y^2 = 25$ .

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to determine that

x

must be an integer, but neither statement alone is sufficient.

Evaluating both statements together establishes that

(x+y)^2 = x^2 + y^2 + 2xy = 25 + 2(12) = 49

, meaning

x+y=7

. The system

x+y=7

and

xy=12

yields solutions

x=3, y=4

or

x=4, y=3

. In both outcomes,

x

is guaranteed to be an integer, yielding a definitive 'Yes' answer.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) gives

x + y = k

for some integer

k

, and

xy = 12

. If

k = 7

, the quadratic equation

t^2 - 7t + 12 = 0

gives roots

t = 3

and

t = 4

, which are integers. However, if

k = 9

, the quadratic equation

t^2 - 9t + 12 = 0

gives roots

t = \frac{9 \pm \sqrt{33}}{2}

. Here

x = \frac{9 + \sqrt{33}}{2}

is a positive real number,

y = \frac{9 - \sqrt{33}}{2} > 0

,

x+y = 9

(integer), and

xy = 12

, but

x

is NOT an integer. Thus,

x

can be an integer or a non-integer.

Demonstrate that positive real numbers whose sum and product are integers do not necessarily have to be integers themselves.

2

Evaluate Statement (2) independently.

Statement (2) gives

x - y = m

for some integer

m

, and

x^2 + y^2 = 25

. If

m = 1

, we can have

x = 4

and

y = 3

, so

x

is an integer. If

m = 2

, then

(x-y)^2 = x^2 - 2xy + y^2 \implies 4 = 25 - 2xy \implies 2xy = 21

. Then

(x+y)^2 = x^2 + y^2 + 2xy = 25 + 21 = 46 \implies x+y = \sqrt{46}

. Thus

x = \frac{\sqrt{46} + 2}{2}

, which is positive and real, but NOT an integer. Thus, Statement (2) alone is not sufficient.

Test whether integer difference and fixed sum of squares guarantees integer values.

3

Evaluate Statements (1) and (2) together.

From Statement (1),

xy = 12

and

x + y = k

(where

k

is an integer). From Statement (2),

x^2 + y^2 = 25

. Using the algebraic identity

x^2 + y^2 = (x+y)^2 - 2xy

, we substitute the known values:

25 = k^2 - 2(12) \implies 25 = k^2 - 24 \implies k^2 = 49

. Since

x > 0

and

y > 0

,

x + y = k > 0

, so

k = 7

. Now we have

x + y = 7

and

xy = 12

. The possible values for

x

and

y

are the solutions to

t^2 - 7t + 12 = 0 \implies (t-3)(t-4) = 0 \implies t \in \{3, 4\}

. Therefore,

x

must be either 3 or 4. In both cases,

x

is definitively an integer.

Combine the conditions to solve for

x+y

and prove

x

must be an integer.

Key Concept

Integer constraints vs. real number assumptions in Data Sufficiency algebraic systems

Question 16Question →

For a real number $p$ , is $p$ an integer?

(1) $p^3 - p$ is a positive prime number.
(2) $p^2$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (1) is sufficient because factoring

p^3 - p

gives

(p-1)p(p+1)

, which represents the product of three consecutive integers whenever

p

is an integer. The product of any three consecutive integers must be divisible by 6. Since no prime number is a multiple of 6,

p

cannot be an integer if

p^3 - p

is prime. This provides a definitive 'No' answer to the question 'Is

p

an integer?', establishing sufficiency. Statement (2) is insufficient because

p

could be an integer like 3 (yielding 'Yes') or an irrational number like

\sqrt{3}

(yielding 'No').

Step-by-Step Solution

1

Analyze Statement (1):

p^3 - p

is a positive prime number.

p

cannot be an integer, yielding a definitive 'No' to the question stem.

Factor

p^3 - p

as

(p-1)p(p+1)

. If

p

were an integer, this product would represent three consecutive integers. Any three consecutive integers contain at least one even factor and exactly one multiple of 3, making their product divisible by 6. A prime number is a positive integer greater than 1 with no positive divisors other than 1 and itself, so no prime number can be a multiple of 6. Thus, no integer

p

can make

p^3 - p

a prime number. Since

p^3 - p

is given to be a positive prime,

p

must be a non-integer real number. A definitive 'No' answer establishes that Statement (1) alone is sufficient.

2

Analyze Statement (2):

p^2

is an integer.

p

could be an integer or a non-integer, so the question cannot be answered uniquely.

If

p = 3

,

p^2 = 9

(an integer), and

p

is an integer (Yes). If

p = \sqrt{3}

,

p^2 = 3

(an integer), but

p

is not an integer (No). Because both 'Yes' and 'No' outcomes are possible, Statement (2) alone is not sufficient.

3

Determine the overall Data Sufficiency choice.

Statement (1) alone is sufficient, but Statement (2) alone is not sufficient.

Statement (1) provides a conclusive answer on its own, whereas Statement (2) remains ambiguous.

Key Concept

Number Properties and Integer Constraints in Data Sufficiency

Question 17Question →

If $x$ is a real number, is $x$ an integer?

(1) $x^2 + 2x$ is an integer.
(2) $x^3 + 2x^2$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

Both statements together are sufficient to determine that

x

is an integer, but neither statement alone is sufficient.

The correct response is that both statements together are sufficient, but neither statement alone is sufficient. Statement (1) permits irrational values like

x = -1 + \sqrt{2}

, making it insufficient on its own. Statement (2) permits irrational values like

x = \frac{-1 + \sqrt{5}}{2}

, making it insufficient on its own. Combining both statements allows us to express

x

as the quotient of two integers

b/a

when

x^2 + 2x \neq 0

, which mathematically forces

x

to be an integer.

Step-by-Step Solution

1

Evaluate Statement (1) independently:

x^2 + 2x

is an integer.

Statement (1) is NOT sufficient.

Let

x^2 + 2x = 1

. Solving

x^2 + 2x - 1 = 0

gives

x = -1 + \sqrt{2}

, which is a real non-integer. However, if

x = 1

,

x^2 + 2x = 3

is also an integer. Since

x

could be an integer or a non-integer, Statement (1) alone is insufficient.

2

Evaluate Statement (2) independently:

x^3 + 2x^2

is an integer.

Statement (2) is NOT sufficient.

Let

x^3 + 2x^2 = 1

. The equation

x^3 + 2x^2 - 1 = 0

factors as

(x + 1)(x^2 + x - 1) = 0

. Setting

x^2 + x - 1 = 0

yields non-integer root

x = \frac{-1 + \sqrt{5}}{2}

. For this non-integer value,

x^3 + 2x^2 = 1

, which is an integer. Since

x

can also be the integer

-1

, Statement (2) alone is insufficient.

3

Evaluate Statements (1) and (2) together.

Both statements together are SUFFICIENT.

From Statement (1), let

x^2 + 2x = a

, where

a

is an integer. From Statement (2), let

x^3 + 2x^2 = b

, where

b

is an integer. Notice that

x^3 + 2x^2 = x(x^2 + 2x) = x \cdot a = b

. Case 1: If

a = 0

, then

x^2 + 2x = 0 \implies x(x + 2) = 0 \implies x = 0

or

x = -2

, both of which are integers. Case 2: If

a \neq 0

, then

x = \frac{b}{a}

, meaning

x

is a rational number. Let

x = \frac{m}{n}

in lowest terms where

\gcd(m, n) = 1

and

n > 0

. Substituting

x = \frac{m}{n}

into

x^2 + 2x = a

gives

\frac{m^2}{n^2} + \frac{2m}{n} = a \implies m^2 + 2mn = a n^2 \implies m(m + 2n) = a n^2

. If

n > 1

, any prime factor

p

of

n

must divide

m(m + 2n)

, which implies

p

divides

m^2

and thus

p

divides

m

. This contradicts

\gcd(m, n) = 1

. Thus,

n

must equal

1

, proving

x

is an integer.

Key Concept

Using algebraic combination of Data Sufficiency statements and integer polynomial constraints to establish sufficiency without assuming variables are integers.

Question 18Question →

If $n$ is a positive real number, is $n$ an integer?

(1) $n + \frac{6}{n}$ is an integer.
(2) $n^2 + \frac{36}{n^2}$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statements (1) and (2) TOGETHER are NOT sufficient to determine whether

n

is an integer.

The correct answer identifies that neither statement alone is sufficient, and combining them offers no additional constraint because the second statement is algebraically derived by squaring the first. Since non-integer real numbers such as

n = 3 + \sqrt{3}

satisfy both statements, the statements together remain insufficient.

Step-by-Step Solution

1

Evaluate Statement (1) by setting

n + \frac{6}{n} = k

, where

k

is an integer.

Multiplying by

n

gives

n^2 - kn + 6 = 0

, which yields

n = \frac{k \pm \sqrt{k^2 - 24}}{2}

. For

n

to be a real number, we need

k^2 \ge 24

, so

k \ge 5

.

We must test whether all positive real solutions for

n

under this condition are integers.

2

Test specific integer values for

k

in Statement (1).

If

k = 5

, then

n = \frac{5 \pm 1}{2}

, giving

n = 3

or

n = 2

(both are integers, yielding a YES answer). However, if

k = 6

, then

n = \frac{6 \pm \sqrt{12}}{2} = 3 \pm \sqrt{3}

, which is a positive real number but NOT an integer (yielding a NO answer).

Since Statement (1) permits both integer and non-integer values for

n

, Statement (1) ALONE is NOT sufficient.

3

Evaluate Statement (2) independently and analyze its algebraic relationship to Statement (1).

If

n + \frac{6}{n} = k

, then squaring both sides gives

\left(n + \frac{6}{n}\right)^2 = n^2 + 12 + \frac{36}{n^2} = k^2

, which implies

n^2 + \frac{36}{n^2} = k^2 - 12

. If

k

is an integer,

k^2 - 12

is automatically an integer.

This shows Statement (2) is logically and algebraically implied by Statement (1).

4

Evaluate Statements (1) and (2) together.

Because Statement (2) provides zero new information beyond Statement (1), combining them does not eliminate the non-integer solution

n = 3 + \sqrt{3}

. Thus, the question cannot be answered definitively.

Redundant statements cannot resolve insufficiency present in the individual statements.

Key Concept

Number Properties and Integer Constraints in Data Sufficiency

Estimated Time:2m 30s

Question 19Question →

If $k$ is a real number, is $k$ an integer?

(1) $k^2 - k$ is an integer.
(2) $k^3 - k$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Neither statement alone is sufficient because non-integer real numbers can produce integer outputs for

k^2 - k

or

k^3 - k

. However, when both statements are combined, expressing

k^3 - k

as

m(k + 1)

proves that

k

must be a rational number. For any rational number

\frac{a}{b}

in lowest terms,

\left(\frac{a}{b}\right)^2 - \frac{a}{b}

can only be an integer if the denominator

b = 1

, which proves definitively that

k

must be an integer.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) is NOT sufficient.

We are given that

k^2 - k = m

for some integer

m

. If

m = 1

, then

k^2 - k - 1 = 0

, which gives

k = \frac{1 + \sqrt{5}}{2}

. Here,

k

is not an integer, but

k^2 - k = 1

is an integer. Thus, Statement (1) alone does not determine whether

k

is an integer.

2

Evaluate Statement (2) independently.

Statement (2) is NOT sufficient.

We are given that

k^3 - k = n

for some integer

n

. If

n = 1

, then

k^3 - k - 1 = 0

. By the Intermediate Value Theorem, there exists a real number

k \in (1, 2)

satisfying this equation. Since

1 < k < 2

,

k

cannot be an integer, yet

k^3 - k = 1

is an integer. Thus, Statement (2) alone does not determine whether

k

is an integer.

3

Evaluate Statements (1) and (2) combined.

The combined statements are SUFFICIENT.

Let

k^2 - k = m \in \mathbb{Z}

and

k^3 - k = n \in \mathbb{Z}

. Note that

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

. Thus,

n = m(k + 1)

. If

m = 0

, then

k^2 - k = 0 \implies k = 0

or

k = 1

, both of which are integers. If

m \neq 0

, then

mk = n - m \implies k = \frac{n - m}{m}

. Since

m

and

n

are integers,

k

must be a rational number. Let

k = \frac{a}{b}

in reduced form with

\gcd(a, b) = 1

and

b \ge 1

. Substituting into

k^2 - k = m

yields

\frac{a^2 - ab}{b^2} = m \implies a(a - b) = m b^2

. This means

b^2

must divide

a(a - b)

. Since

\gcd(a, b^2) = 1

,

b^2

must divide

a - b

, which implies

b = 1

. Therefore,

k = a

, so

k

MUST be an integer.

Key Concept

Testing implicit real number constraints vs integer constraints using polynomial reduction and rational root properties.

Question 20Question →

If $r$ is a positive real number, is $r$ an integer?

(1) $r^2 + 2r$ is an integer.
(2) $r^2 - 2r$ is an integer.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

Show answer & explanation

Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Neither statement alone is sufficient because irrational numbers like

\sqrt{2}-1

and

\sqrt{2}+1

can satisfy statements (1) and (2) respectively. However, combining both statements proves that

4r

is an integer

k

, and substituting

r = k/4

into the quadratic equation shows that

k

must be a multiple of 4, forcing

r

to be a positive integer.

Step-by-Step Solution

1

Evaluate Statement (1) independently.

Statement (1) is INSUFFICIENT.

If

r = 1

, then

r^2 + 2r = 3

(an integer), and

r

is an integer (YES). If

r = \sqrt{2} - 1

, then

r^2 + 2r = (3 - 2\sqrt{2}) + (2\sqrt{2} - 2) = 1

(an integer), but

r

is NOT an integer (NO). Since

r

can be an integer or a non-integer, Statement (1) alone is not sufficient.

2

Evaluate Statement (2) independently.

Statement (2) is INSUFFICIENT.

If

r = 3

, then

r^2 - 2r = 3

(an integer), and

r

is an integer (YES). If

r = \sqrt{2} + 1

, then

r^2 - 2r = (3 + 2\sqrt{2}) - (2\sqrt{2} + 2) = 1

(an integer), but

r

is NOT an integer (NO). Since

r

can be an integer or a non-integer, Statement (2) alone is not sufficient.

3

Evaluate Statement (1) and Statement (2) together.

Statements (1) and (2) together are SUFFICIENT.

Let

r^2 + 2r = a

and

r^2 - 2r = b

, where

a

and

b

are integers. Subtracting Statement (2) from Statement (1) gives

(r^2 + 2r) - (r^2 - 2r) = a - b \implies 4r = a - b

. Since

a

and

b

are integers,

a - b = k

for some integer

k

. Thus,

r = \frac{k}{4}

. Now substitute

r = \frac{k}{4}

back into Statement (1):

\left(\frac{k}{4}\right)^2 + 2\left(\frac{k}{4}\right) = a \implies \frac{k^2}{16} + \frac{k}{2} = a \implies \frac{k^2 + 8k}{16} = a

. This means

16

must divide

k^2 + 8k = k(k+8)

. If

k

were odd,

k(k+8)

would be odd and thus not divisible by 16. So

k

must be even. Let

k = 2m

. Then

k(k+8) = 2m(2m+8) = 4m(m+4)

. For

16

to divide

4m(m+4)

,

4

must divide

m(m+4)

. If

m

were odd,

m(m+4)

would be odd, which is not divisible by 4. Thus,

m

must be even. Let

m = 2p

. Then

k = 2m = 4p

, which means

k

is a multiple of 4. Therefore,

r = \frac{k}{4} = \frac{4p}{4} = p

, an integer. Thus,

r

MUST be an integer.

Key Concept

Testing integer constraints in Data Sufficiency requires evaluating whether non-integer real numbers satisfy given equations, and combining equations to determine if rational and integer bounds strictly hold.

Number Properties and Integer Constraints in Data Sufficiency

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