Number Properties and Integer Constraints in Data Sufficiency Alıştırma Soruları — GMAT — Sayfa 2

Soru 21Soru →

If $a$ and $b$ are positive real numbers such that $ab$ is an integer, is $a$ an integer?

(1) $a + b$ is an integer.
(2) $a - b$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

Combining both statements shows that

2a

and

2b

are integers. This restricts

a

to being either a whole integer or a half-integer. Evaluating the product of two half-integers reveals that their product has a denominator of 4 and can never be an integer. Since the question stem specifies that

ab

is an integer,

a

must be an integer.

Adım Adım Çözüm

1

Analyze the question stem and given constraints

a

and

b

are positive real numbers and

ab = N

, where

N

is an integer. We must determine whether

a

is guaranteed to be an integer.

The variables are defined as real numbers, so we cannot assume they are integers unless forced by the mathematical statements.

2

Evaluate Statement (1):

a + b

is an integer

Statement (1) is NOT sufficient.

Let

a = 2 + \sqrt{3}

and

b = 2 - \sqrt{3}

. Both are positive real numbers. Here

a + b = 4

(an integer) and

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

(an integer). However,

a = 2 + \sqrt{3}

is not an integer. Conversely, if

a = 3

and

b = 1

, then

a + b = 4

and

ab = 3

, and

a

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

a

is an integer.

3

Evaluate Statement (2):

a - b

is an integer

Statement (2) is NOT sufficient.

Let

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

and

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

. Both are positive real numbers. Here

a - b = 1

(an integer) and

ab = \frac{(\sqrt{5}+1)(\sqrt{5}-1)}{4} = 1

(an integer). However,

a

is not an integer. Conversely, if

a = 4

and

b = 1

, then

a - b = 3

and

ab = 4

, and

a

is an integer. Thus, Statement (2) alone is not sufficient.

4

Evaluate Statement (1) and Statement (2) combined

Both statements together are SUFFICIENT.

Let

a + b = k

and

a - b = m

, where

k

and

m

are integers. Adding the two equations gives

2a = k + m \implies a = \frac{k+m}{2}

. Subtracting gives

2b = k - m \implies b = \frac{k-m}{2}

. Since

k

and

m

are integers,

k+m

and

k-m

are integers with the same parity. If

k+m

is even, both

a

and

b

are integers. If

k+m

is odd,

a = \frac{p}{2}

and

b = \frac{q}{2}

for odd integers

p

and

q

. But then

ab = \frac{pq}{4}

, which cannot be an integer because the product of two odd integers is odd and thus not divisible by 4. This contradicts the stem constraint that

ab

is an integer. Therefore,

k+m

must be even, forcing

a

to be an integer.

Anahtar Kavram

Real number vs. integer constraints and parity properties in quadratic system equations

Soru 22Soru →

If $m$ and $n$ are real numbers, is $m + n$ an integer?

(1) $m - n$ is an integer.
(2) $m^2 - n^2$ 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.

Cevabı ve açıklamayı göster

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

Cevap

Statements (1) and (2) together are not sufficient.

Evaluating Statement (1) alone: If

m = \sqrt{2}

and

n = \sqrt{2}

, then

m - n = 0

(an integer), but

m + n = 2\sqrt{2}

is not an integer. If

m = 2

and

n = 1

, then

m - n = 1

(an integer) and

m + n = 3

is an integer. Thus, Statement (1) alone is not sufficient.

Evaluating Statement (2) alone: If

m = \sqrt{2}

and

n = 1

, then

m^2 - n^2 = 1

(an integer), but

m + n = \sqrt{2} + 1

is not an integer. If

m = 2

and

n = 1

, then

m^2 - n^2 = 3

(an integer) and

m + n = 3

is an integer. Thus, Statement (2) alone is not sufficient.

Evaluating Statements (1) and (2) together: Factoring gives

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

. If

m - n = k \neq 0

and

m^2 - n^2 = p

for integers

k

and

p

, then

m + n = \frac{p}{k}

. This establishes that

m + n

must be rational, but a rational number is not necessarily an integer. For example, if

m = \frac{5}{4}

and

n = -\frac{3}{4}

, then

m - n = 2

(an integer) and

m^2 - n^2 = 1

(an integer), but

m + n = \frac{1}{2}

is not an integer. If

m = 2

and

n = 0

, both statements hold and

m + n = 2

is an integer. Since

m + n

can be an integer or a non-integer, both statements together are not sufficient.

Adım Adım Çözüm

1

Analyze the question stem and identify variable constraints.

The variables

m

and

n

are defined as real numbers, NOT necessarily integers.

Failing to account for non-integer real values is a primary trap in GMAT Data Sufficiency number property questions.

2

Evaluate Statement (1) independently.

Statement (1) states

m - n = k

for some integer

k

.
If

m = 2

and

n = 1

,

m - n = 1

(integer) and

m + n = 3

(integer).
If

m = \sqrt{2}

and

n = \sqrt{2}

,

m - n = 0

(integer) and

m + n = 2\sqrt{2}

(not an integer).
Since

m + n

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

Testing both integer and irrational values reveals that a difference being an integer does not constrain the sum to be an integer.

3

Evaluate Statement (2) independently.

Statement (2) states

m^2 - n^2 = p

for some integer

p

.
If

m = 2

and

n = 1

,

m^2 - n^2 = 3

(integer) and

m + n = 3

(integer).
If

m = \sqrt{2}

and

n = 1

,

m^2 - n^2 = 1

(integer) and

m + n = \sqrt{2} + 1

(not an integer).
Since

m + n

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

Testing non-integer values shows that the difference of squares being an integer does not guarantee that the sum is an integer.

4

Evaluate Statements (1) and (2) together.

We have

m - n = k

and

(m - n)(m + n) = p

for integers

k

and

p

.
If

k = 0

,

m = n

, so

m - n = 0

and

m^2 - n^2 = 0

. In this case

m + n = 2m

, which can be non-integer if

m = \sqrt{2}

.
If

k \neq 0

, then

m + n = \frac{p}{k}

. This proves that

m + n

is a rational number, but not necessarily an integer.
For instance, let

m = \frac{5}{4}

and

n = -\frac{3}{4}

:
-

m - n = \frac{5}{4} - \left(-\frac{3}{4}\right) = 2

(integer)
-

m^2 - n^2 = \frac{25}{16} - \frac{9}{16} = 1

(integer)
-

m + n = \frac{5}{4} + \left(-\frac{3}{4}\right) = \frac{1}{2}

(not an integer).
Conversely, if

m = 2

and

n = 0

,

m - n = 2

(integer),

m^2 - n^2 = 4

(integer), and

m + n = 2

(integer).
Because

m + n

can still be an integer or a non-integer, both statements together are NOT sufficient.

Algebraic division demonstrates that the sum is guaranteed to be rational, but rational numbers include non-integer fractions.

Anahtar Kavram

Real Number Constraints vs. Integer Constraints in Data Sufficiency

Tahmini Süre:2m 0s

Soru 23Soru →

If $m$ and $n$ are positive integers, is $m^2 - n^2$ a prime number?

(1) $m - n = 1$
(2) $m + n = 7$

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.

Cevabı ve açıklamayı göster

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

Cevap

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

Rephrasing the question stem shows

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

. Statement (1) alone allows

2n + 1

to be prime (for

n=1

) or composite (for

n=4

), so it is insufficient. Statement (2) alone allows pairs like

(4,3)

which gives

7

(prime) and

(5,2)

which gives

21

(composite), so it is insufficient. Combining both statements yields

m = 4

and

n = 3

, giving

m^2 - n^2 = 7

, which definitively answers 'Yes'.

Adım Adım Çözüm

1

Rephrase the question stem using algebraic factorization.

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

. For

m^2 - n^2

to be a prime number, since

m

and

n

are positive integers, the smaller factor

(m - n)

must equal 1 and the larger factor

(m + n)

must be a prime number.

Factoring highlights the conditions required for a product of integers to yield a prime number.

2

Evaluate Statement (1):

m - n = 1

.

Substitute

m = n + 1

into

m^2 - n^2

:

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

. If

n = 1

, then

2(1) + 1 = 3

(prime, Yes). If

n = 4

, then

2(4) + 1 = 9

(composite, No).

Since Statement (1) produces both Yes and No outcomes, it is NOT sufficient.

3

Evaluate Statement (2):

m + n = 7

.

Since

m

and

n

are positive integers, test pairs with

m > n

: If

(m, n) = (4, 3)

, then

m^2 - n^2 = 16 - 9 = 7

(prime, Yes). If

(m, n) = (5, 2)

, then

m^2 - n^2 = 25 - 4 = 21

(composite, No).

Since Statement (2) produces both Yes and No outcomes, it is NOT sufficient.

4

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

System of equations:

m - n = 1

and

m + n = 7

. Adding the equations gives

2m = 8 \implies m = 4

, which leads to

n = 3

. Then

m^2 - n^2 = 4^2 - 3^2 = 7

, which is prime (Definitive Yes).

Combining the statements yields a single unique pair of values and a definitive Yes answer.

Anahtar Kavram

Difference of squares factorization and prime number properties in Data Sufficiency

Tahmini Süre:2m 0s

Soru 24Soru →

If $p$ and $q$ are positive integers, is $p^2 + 3q$ divisible by 6?

(1) $p + q$ is an odd integer.
(2) $p^2 - q^2$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

The question requires determining whether

p^2 + 3q

is divisible by 6, which demands divisibility by both 2 and 3. By parity rules,

p^2 \equiv p \pmod 2

and

3q \equiv q \pmod 2

, meaning

p^2 + 3q

has the same parity as

p + q

. Statement (1) specifies that

p + q

is odd, so

p^2 + 3q

is odd and cannot be divisible by 6. This gives a definitive 'No' answer, making Statement (1) alone sufficient. Statement (2) permits cases where

p^2 + 3q

is divisible by 6 (e.g.,

p=3, q=3

) and cases where it is not (e.g.,

p=4, q=1

), making Statement (2) alone insufficient.

Adım Adım Çözüm

1

Rephrase the question target using number properties and parity rules.

For

p^2 + 3q

to be divisible by 6, it must be divisible by both 2 (even) and 3. For parity,

p^2 \equiv p \pmod 2

and

3q \equiv q \pmod 2

, so

p^2 + 3q \equiv p + q \pmod 2

. Thus,

p^2 + 3q

is even if and only if

p + q

is even.

Simplifying the parity constraint simplifies evaluating statements regarding divisibility by 2.

2

Evaluate Statement (1):

p + q

is an odd integer.

Since

p + q

is odd,

p^2 + 3q

must be odd. An odd integer cannot be divisible by 2, so

p^2 + 3q

CANNOT be divisible by 6. This yields a definitive 'No'.

In Data Sufficiency Yes/No questions, a definitive 'No' answer means the statement is sufficient.

3

Evaluate Statement (2):

p^2 - q^2

is divisible by 3.

Test cases:
- If

p = 3, q = 3

:

p^2 - q^2 = 0

(divisible by 3), and

p^2 + 3q = 18

(divisible by 6 -> YES).
- If

p = 4, q = 1

:

p^2 - q^2 = 15

(divisible by 3), and

p^2 + 3q = 19

(not divisible by 6 -> NO).

Since Statement (2) yields both 'Yes' and 'No' outcomes, it is insufficient.

Anahtar Kavram

Parity rephrasing and Yes/No Data Sufficiency decision logic

Soru 25Soru →

If $n$ is a positive integer, is $n$ a multiple of 18?

(1) $n^2$ is divisible by 108.
(2) $n^3$ is divisible by 243.

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.

Cevabı ve açıklamayı göster

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

Cevap

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

Statement (1) alone is sufficient because prime factor analysis of

n^2

being divisible by

108 = 2^2 \times 3^3

requires the exponent of 3 in

n^2

to be at least 4 (since

n

is an integer and exponents in

n^2

must be even). This guarantees

n

is divisible by

2^1 \times 3^2 = 18

. Statement (2) alone is not sufficient because

n^3

being divisible by

243 = 3^5

only guarantees

n

is a multiple of 9, allowing

n = 9

(not a multiple of 18) or

n = 18

(a multiple of 18).

Adım Adım Çözüm

1

Rephrase the target question using prime factorizations.

The target asks if

n

is a multiple of

18 = 2^1 \times 3^2

. For

n

to be a multiple of 18, its prime factorization must contain at least one factor of 2 and at least two factors of 3.

Decomposing into prime factors simplifies the condition required for sufficiency.

2

Evaluate Statement (1):

n^2

is divisible by

108 = 2^2 \times 3^3

.

Since

n

is a positive integer, any prime factor in the prime factorization of

n^2

must have an even exponent. For

n^2

to be divisible by

2^2 \times 3^3

, the power of 2 in

n^2

must be at least 2, so

n

has at least one factor of 2. The power of 3 in

n^2

must be an even integer greater than or equal to 3, meaning it must be at least 4. Thus,

n

must contain at least two factors of 3 (

3^2 = 9

). Consequently,

n

must be divisible by

2^1 \times 3^2 = 18

. Statement (1) alone yields a definitive 'Yes'.

An integer's square must have even exponents for all its prime factors.

3

Evaluate Statement (2):

n^3

is divisible by

243 = 3^5

.

For

n^3

to be divisible by

3^5

, the exponent of 3 in

n^3

must be a multiple of 3 greater than or equal to 5, which means it must be at least 6. Therefore,

n

must contain at least two factors of 3 (

3^2 = 9

). However, statement (2) places no restriction on whether

n

is even. If

n = 9

,

n^3 = 729 = 243 \times 3

(divisible by 243), but 9 is NOT a multiple of 18. If

n = 18

,

n^3 = 5832 = 243 \times 24

(divisible by 243), and 18 IS a multiple of 18. Since

n

can yield both 'No' and 'Yes', Statement (2) alone is insufficient.

Testing specific values demonstrates that

n

may or may not be even.

Anahtar Kavram

Prime Factor Exponents in Integer Powers and Data Sufficiency Sufficiency Logic

Tahmini Süre:2m 0s

Soru 26Soru →

If $n$ is a positive integer, is $n^2 + 3n + 2$ divisible by 12?

(1) $n$ is a multiple of 3.
(2) $n + 1$ is a prime number.

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.

Cevabı ve açıklamayı göster

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

Cevap

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

Factoring the stem expression gives

n^2 + 3n + 2 = (n + 1)(n + 2)

. Statement (1) tells us

n

is a multiple of 3, which implies neither

n + 1

nor

n + 2

is a multiple of 3. Consequently, their product can never be divisible by 3, and thus can never be divisible by 12. Because Statement (1) conclusively answers 'No' to the question stem, Statement (1) ALONE is sufficient. Statement (2) allows

n = 2

(giving 12, divisible by 12) and

n = 4

(giving 30, not divisible by 12), yielding both 'Yes' and 'No' responses, making Statement (2) alone insufficient.

Adım Adım Çözüm

1

Rephrase the question stem target using factoring.

The expression

n^2 + 3n + 2

factors into

(n + 1)(n + 2)

. The question asks whether

(n + 1)(n + 2)

is divisible by

12 = 2^2 \times 3

.

Factoring quadratic expressions helps reveal divisibility properties of consecutive integers.

2

Evaluate Statement (1):

n

is a multiple of 3.

If

n = 3k

for some positive integer

k

, then

n + 1 = 3k + 1

and

n + 2 = 3k + 2

. Neither factor contains 3 as a prime factor, so their product

(3k + 1)(3k + 2)

leaves a remainder of

1 \times 2 = 2

when divided by 3. Since the product is never divisible by 3, it can NEVER be divisible by 12.

In a Yes/No Data Sufficiency question, a statement that yields a definitive 'No' to the question is SUFFICIENT.

3

Evaluate Statement (2):

n + 1

is a prime number.

If

n = 2

, then

n + 1 = 3

(prime). The expression

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

, which is divisible by 12 (Answer: YES). If

n = 4

, then

n + 1 = 5

(prime). The expression

(4+1)(4+2) = 30

, which is not divisible by 12 (Answer: NO). Since both 'Yes' and 'No' are possible, Statement (2) is NOT sufficient.

Testing specific values shows that Statement (2) does not yield a consistent answer.

Anahtar Kavram

Definitive Yes/No decision logic in Data Sufficiency combined with divisibility and prime factor properties of consecutive integers.

Soru 27Soru →

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

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

(2) $n^2 - n$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

Evaluating each statement alone shows that non-integer real numbers can produce integer values for

n^2 + 5n

or

n^2 - n

. However, combining both statements gives

6n = (n^2 + 5n) - (n^2 - n)

, proving

n = \frac{k}{6}

for some integer

k

. Substituting

n = \frac{k}{6}

back into

n^2 - n

requires

\frac{k^2 - 6k}{36}

to be an integer, which forces

k(k-6)

to be divisible by 36. Examining remainders modulo 6 shows

k

must be a multiple of 6, ensuring

n

is an integer.

Adım Adım Çözüm

1

Evaluate Statement (1) independently.

Statement (1) is NOT sufficient.

If

n^2 + 5n = p

where

p

is an integer,

n

can be an integer (e.g.,

n=1 \implies n^2+5n=6

) or a non-integer real number (e.g.,

n^2+5n-1=0 \implies n = \frac{-5 + \sqrt{29}}{2}

, which gives

n^2+5n=1

).

2

Evaluate Statement (2) independently.

Statement (2) is NOT sufficient.

If

n^2 - n = q

where

q

is an integer,

n

can be an integer (e.g.,

n=2 \implies n^2-n=2

) or a non-integer real number (e.g.,

n^2-n-1=0 \implies n = \frac{1 + \sqrt{5}}{2}

, which gives

n^2-n=1

).

3

Combine Statement (1) and Statement (2).

6n

is equal to an integer

k = p - q

, so

n = \frac{k}{6}

.

Subtracting

(n^2 - n = q)

from

(n^2 + 5n = p)

yields

6n = p - q

. Since

p

and

q

are integers,

k = p - q

must be an integer.

4

Substitute

n = \frac{k}{6}

back into Statement (2) to check integer constraints on

k

.

k

must be a multiple of 6, which implies

n = \frac{k}{6}

is an integer.

Substituting

n = \frac{k}{6}

into

n^2 - n = q

gives

\left(\frac{k}{6}\right)^2 - \frac{k}{6} = \frac{k^2 - 6k}{36} = q

. Thus,

36

must divide

k(k-6)

. Consequently,

6

divides

k(k-6)

. Since

k

and

k-6

have the same remainder

r

modulo 6,

k(k-6) \equiv r^2 \pmod 6

. For

r^2

to be divisible by 6 where

r \in \{0, 1, 2, 3, 4, 5\}

,

r

must be

0

. Hence,

k

is a multiple of 6, making

n

an integer.

Anahtar Kavram

Number properties of non-integer real variables vs. integer constraints when combining polynomial equations.

Soru 28Soru →

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

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

Cevabı ve açıklamayı göster

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

Cevap

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

Evaluating Statement (1) alone allows

x = \sqrt{2}

(not an integer) or

x = 2

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

x = \frac{1}{3}

(not an integer) or

x = 1

(an integer), so Statement (2) is insufficient. Combining both statements establishes that

x = \frac{m}{3}

for an integer

m

, and

x^2 = \frac{m^2}{9} = k

for an integer

k

. This requires

m^2 = 9k

, which implies

m

must be a multiple of 3. Hence

x

must be an integer, giving a definitive 'Yes' answer. Thus, both statements together are sufficient.

Adım Adım Çözüm

1

Analyze the question stem constraint

x

is constrained to be a real number (

x \in \mathbb{R}

). The target question is a Yes/No question: 'Is

x

an integer?'

Without an explicit integer constraint in the stem, non-integer real numbers must be tested as counterexamples.

2

Evaluate Statement (1) alone

If

x = \sqrt{2}

, then

x^2 = 2

(an integer), but

x

is not an integer (Answer: No). If

x = 2

, then

x^2 = 4

(an integer), and

x

is an integer (Answer: Yes). Statement (1) is NOT sufficient.

Statement (1) allows both integer and non-integer values for

x

.

3

Evaluate Statement (2) alone

If

x = \frac{1}{3}

, then

3x = 1

(an integer), but

x

is not an integer (Answer: No). If

x = 1

, then

3x = 3

(an integer), and

x

is an integer (Answer: Yes). Statement (2) is NOT sufficient.

Statement (2) allows fractional values with a denominator of 3 as well as integers.

4

Evaluate Statements (1) and (2) together

From Statement (2),

x = \frac{m}{3}

for some integer

m

. Substituting into Statement (1) yields

x^2 = \left(\frac{m}{3}\right)^2 = \frac{m^2}{9} = k

, where

k

is an integer. Thus,

m^2 = 9k

. Since

9k

is a multiple of 9,

m^2

is divisible by 9, which means

m

must be a multiple of 3. Let

m = 3p

for some integer

p

. Then

x = \frac{3p}{3} = p

, which guarantees that

x

is an integer. The answer is a definitive 'Yes'. Statements (1) and (2) together are SUFFICIENT.

Combining both conditions restricts

x

to rational numbers whose square is an integer, forcing

x

to be an integer.

Anahtar Kavram

Number Properties and Integer Constraints in Data Sufficiency

Tahmini Süre:1m 30s

Soru 29Soru →

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

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

(2) $x^2 + \frac{1}{x}$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

The option stating that both statements together are sufficient is correct. Evaluating Statement (1) alone allows irrational solutions like

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

. Evaluating Statement (2) alone allows irrational roots of

x^3 - 3x + 1 = 0

. However, combining both statements reveals that

x^2 - x

is an integer. Subtracting the resulting linear system forces

x

to be a rational number. For any positive rational

x = \frac{p}{q}

in simplest form,

x + \frac{1}{x}

being an integer requires

p = 1

and

q = 1

, proving that

x = 1

, which is an integer.

Adım Adım Çözüm

1

Evaluate Statement (1) independently.

Statement (1) is NOT sufficient.

If

x = 1

, then

x + \frac{1}{x} = 2

(an integer), and

x

is an integer. However, if

x = \frac{3 + \sqrt{5}}{2} \approx 2.618

(a positive real non-integer), then

x + \frac{1}{x} = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = 3

(an integer). Since

x

can be an integer or a non-integer, Statement (1) alone does not uniquely answer the question.

2

Evaluate Statement (2) independently.

Statement (2) is NOT sufficient.

If

x = 1

, then

x^2 + \frac{1}{x} = 2

(an integer), and

x

is an integer. If

x^2 + \frac{1}{x} = 3

, multiplying by

x

yields

x^3 - 3x + 1 = 0

. Evaluating

f(x) = x^3 - 3x + 1

shows

f(1) = -1 < 0

and

f(2) = 3 > 0

, so by the Intermediate Value Theorem, there exists a real root

x \in (1, 2)

, which is positive but not an integer. Hence, Statement (2) alone is insufficient.

3

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

Both statements together are SUFFICIENT.

Let

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

and

x^2 + \frac{1}{x} = b

, where

a

and

b

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

\left(x^2 + \frac{1}{x}\right) - \left(x + \frac{1}{x}\right) = b - a \implies x^2 - x = k

, where

k = b - a

is an integer. From Statement (1),

x^2 - ax + 1 = 0

, and from the difference,

x^2 - x - k = 0

. Subtracting these two quadratic equations gives

(1 - a)x + (1 + k) = 0

. For any positive real

x

,

a = x + \frac{1}{x} \ge 2

, so

1 - a \neq 0

. Thus

x = \frac{1 + k}{a - 1}

. Because

k

and

a

are integers,

x

must be a rational number. Let

x = \frac{p}{q}

in lowest terms, where

p

and

q

are positive integers with

\gcd(p, q) = 1

. Then

x + \frac{1}{x} = \frac{p}{q} + \frac{q}{p} = \frac{p^2 + q^2}{pq} = a \implies p^2 + q^2 = a p q

. Since

p^2 = q(ap - q)

,

q

must divide

p^2

. But

\gcd(p, q) = 1

, so

q = 1

. Similarly,

p

must divide

q^2

, so

p = 1

. Thus

x = 1

, which is an integer. Both statements together definitively answer YES.

Anahtar Kavram

Deduce integer constraints and rationality by combining non-linear algebraic expressions for real variables in Data Sufficiency.

Soru 30Soru →

If $p$ and $q$ are positive real numbers, is $p$ an integer?

(1) $p^2 + pq$ is an integer.
(2) $q^2 + pq$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

p

is an integer.

The correct answer states that statements (1) and (2) together are not sufficient. Statement (1) alone allows

p = 0.5, q = 1.5

(giving

p^2+pq=1

) as well as

p = 1, q = 1

(giving

p^2+pq=2

). Statement (2) alone similarly allows

p = 0.5, q = 1.5

(giving

q^2+pq=3

) as well as

p = 1, q = 1

(giving

q^2+pq=2

). When both statements are evaluated together, adding the two equations reveals

p+q = \sqrt{(p^2+pq) + (q^2+pq)}

. Choosing

p^2+pq = 1

and

q^2+pq = 3

yields

p+q = 2

and

p = 0.5

, demonstrating that

p

does not have to be an integer even when both statements are satisfied.

Adım Adım Çözüm

1

Analyze the Question Stem and Constraints

Variables

p

and

q

are positive real numbers (not constrained to integers). The goal is to determine if

p

must be an integer.

Recognizing that variables are real numbers prevents premature integer assumptions.

2

Evaluate Statement (1) independently:

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

is an integer

Case 1: If

p = 1

and

q = 1

, then

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

(integer), so

p

IS an integer.
Case 2: If

p = 0.5

and

q = 1.5

, then

p(p+q) = 0.5(2) = 1

(integer), so

p

IS NOT an integer.
Statement (1) is INSUFFICIENT.

Testing non-integer real values tests whether the condition forces

p

to be an integer.

3

Evaluate Statement (2) independently:

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

is an integer

Case 1: If

p = 1

and

q = 1

, then

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

(integer), so

p

IS an integer.
Case 2: If

p = 0.5

and

q = 1.5

, then

q(p+q) = 1.5(2) = 3

(integer), so

p

IS NOT an integer.
Statement (2) is INSUFFICIENT.

Statement (2) primarily constrains

q(p+q)

, leaving

p

unconstrained.

4

Evaluate Statements (1) and (2) Combined

Let

p(p+q) = k_1

and

q(p+q) = k_2

where

k_1, k_2

are positive integers.
Adding gives

(p+q)^2 = k_1 + k_2 \implies p+q = \sqrt{k_1 + k_2}

.
Thus,

p = \frac{k_1}{\sqrt{k_1 + k_2}}

.
Case 1: Let

k_1 = 1

and

k_2 = 3

. Then

p+q = \sqrt{4} = 2 \implies p = \frac{1}{2} = 0.5

(not an integer).
Case 2: Let

k_1 = 2

and

k_2 = 2

. Then

p+q = \sqrt{4} = 2 \implies p = \frac{2}{2} = 1

(an integer).
Since

p

can still be either an integer or a non-integer, both statements combined are INSUFFICIENT.

Algebraic combination yields

p = \frac{k_1}{\sqrt{k_1 + k_2}}

, which produces non-integers for appropriate choices of integer constants

k_1

and

k_2

.

Anahtar Kavram

Avoiding Implicit Integer Assumptions in Real-Valued Data Sufficiency

Tahmini Süre:2m 0s

Soru 31Soru →

If $n$ is a positive integer, is $n$ divisible by 18?

(1) $n^2$ is divisible by 54.
(2) $n^3$ is divisible by 108.

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.

Cevabı ve açıklamayı göster

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

Cevap

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

Statement (1) establishes that

n^2

is divisible by

54 = 2^1 \times 3^3

. Because the exponent of 3 in

n^2

is

2b

, the condition

2b \geq 3

implies

b \geq 1.5

, which forces

b \geq 2

since

b

must be an integer. Similarly,

2a \geq 1 \implies a \geq 1

. Therefore,

n

must contain at least

2^1 \times 3^2 = 18

as a factor, guaranteeing that

n

is divisible by 18 (Definitive Yes). Statement (2) establishes that

n^3

is divisible by

108 = 2^2 \times 3^3

, which only forces

a \geq 1

and

b \geq 1

, so

n

must only be a multiple of 6. Since

n = 6

yields a 'No' and

n = 18

yields a 'Yes', Statement (2) is insufficient.

Adım Adım Çözüm

1

Rephrase the question target in terms of prime factorization

For a positive integer

n

to be divisible by

18 = 2^1 \times 3^2

, the prime factorization of

n

must contain at least one factor of 2 and at least two factors of 3.

Decomposing the target integer into prime factors establishes clear minimum exponent thresholds for evaluation.

2

Evaluate Statement (1):

n^2

is divisible by

54 = 2^1 \times 3^3

If

n = 2^a \times 3^b \times k

, then

n^2 = 2^{2a} \times 3^{2b} \times k^2

. Divisibility by

2^1 \times 3^3

requires

2a \geq 1 \implies a \geq 1

and

2b \geq 3 \implies b \geq 2

(since

a

and

b

are integers). Thus

n

must contain at least

2^1 \times 3^2 = 18

, giving a definitive YES.

Integer constraints on exponents force fractional lower bounds to round up to the next integer.

3

Evaluate Statement (2):

n^3

is divisible by

108 = 2^2 \times 3^3

If

n^3 = 2^{3a} \times 3^{3b} \times k^3

, then

3a \geq 2 \implies a \geq 1

and

3b \geq 3 \implies b \geq 1

. Thus

n

must be divisible by

2^1 \times 3^1 = 6

. Testing values: if

n = 6

,

6^3 = 216

(divisible by 108), but 6 is NOT divisible by 18 (NO). If

n = 18

,

18^3 = 5832

(divisible by 108), and 18 IS divisible by 18 (YES). Statement (2) is INSUFFICIENT.

Testing specific values demonstrates that Statement (2) allows both 'Yes' and 'No' outcomes.

Anahtar Kavram

Prime Factor Exponent Analysis under Integer Constraints in Data Sufficiency

Tahmini Süre:2m 0s

Soru 32Soru →

If $m$ and $n$ are positive real numbers, is $m + n$ an integer?

(1) $m^2 - n^2$ is a prime number.
(2) $m - n$ is a positive 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.

Cevabı ve açıklamayı göster

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

Cevap

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

The correct answer is that both statements combined are insufficient. A common high-level trap in GMAT Data Sufficiency is assuming variables must be integers when the stem specifies 'real numbers'. Combining

(m - n)(m + n) = p

(prime) with

m - n = k

(positive integer) gives

m + n = \frac{p}{k}

. If

m = 2.25

and

n = 0.25

, we get

m - n = 2

(integer) and

m^2 - n^2 = 5

(prime), but

m + n = 2.5

(not an integer). Conversely,

m = 2

and

n = 1

gives

m - n = 1

and

m^2 - n^2 = 3

(prime), where

m + n = 3

(an integer). Because both an affirmative and a negative answer are possible, the information remains insufficient.

Adım Adım Çözüm

1

Evaluate Statement (1) independently.

Statement (1) is NOT sufficient.

If

m^2 - n^2 = 2

(a prime number), we could have

m = 2

and

n = √2

, giving

m + n = 2 + √2

(not an integer). Alternatively,

m = 2

and

n = 1

gives

m^2 - n^2 = 3

(prime) and

m + n = 3

(an integer). Hence, Statement (1) alone is insufficient.

2

Evaluate Statement (2) independently.

Statement (2) is NOT sufficient.

If

m - n = 2

, we could have

m = 3.5

and

n = 1.5

, giving

m + n = 5

(an integer). Or we could have

m = 3.2

and

n = 1.2

, giving

m + n = 4.4

(not an integer). Hence, Statement (2) alone is insufficient.

3

Evaluate Statements (1) and (2) combined.

Test for consistency with both integer and non-integer outcomes for

m + n

.

We know

m^2 - n^2 = (m - n)(m + n) = p

, where

p

is prime and

m - n = k

is a positive integer. Thus

k(m + n) = p

, so

m + n = \frac{p}{k}

.

4

Construct concrete counterexamples under both combined statements.

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

Case 1 (YES): Let

m = 2

and

n = 1

. Then

m^2 - n^2 = 3

(prime) and

m - n = 1

(positive integer). Here,

m + n = 3

, which IS an integer.
Case 2 (NO): Let

m = 2.25

and

n = 0.25

. Both are positive real numbers.

m - n = 2.25 - 0.25 = 2

(positive integer).

m^2 - n^2 = (2.25 - 0.25)(2.25 + 0.25) = 2 \times 2.5 = 5

(prime number). But

m + n = 2.25 + 0.25 = 2.5

, which is NOT an integer. Since

m + n

can be an integer or a non-integer, both statements together are not sufficient.

Anahtar Kavram

Implicit Integer Constraints vs. Real Number Definitions in Data Sufficiency

Soru 33Soru →

If $r$ and $s$ are positive integers, is $r^2 - s^2$ divisible by 8?

(1) $r$ and $s$ are both odd integers.
(2) $r + s$ is divisible by 4.

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.

Cevabı ve açıklamayı göster

Cevap: EACH statement ALONE is sufficient.

Cevap

Each statement alone is sufficient to answer the question with a definitive 'Yes'.

The correct response identifies that each statement alone gives a definitive 'Yes' answer. Statement (1) shows that the square of any odd integer is

1 \pmod 8

, so subtracting two odd squares gives

0 \pmod 8

. Statement (2) shows

r+s

is a multiple of 4, which forces

r

and

s

to share parity, making

r-s

even and their product

(r+s)(r-s)

a multiple of

4 \times 2 = 8

.

Adım Adım Çözüm

1

Rephrase the question target using algebraic factorization.

Note that

r^2 - s^2 = (r + s)(r - s)

. The question asks whether

(r + s)(r - s)

is a multiple of 8.

Factoring difference of squares isolates the sum and difference components to easily test parity and divisibility.

2

Evaluate Statement (1):

r

and

s

are both odd integers.

Any odd integer can be written as

2k + 1

. Squaring gives

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

. Since

k(k + 1)

is the product of two consecutive integers, it is always even, making

4k(k + 1)

a multiple of 8. Thus, any odd integer squared leaves a remainder of 1 when divided by 8. Therefore,

r^2 \equiv 1 \pmod 8

and

s^2 \equiv 1 \pmod 8

, so

r^2 - s^2 \equiv 1 - 1 = 0 \pmod 8

.

Statement (1) alone yields a definitive 'Yes', so Statement (1) is sufficient.

3

Evaluate Statement (2):

r + s

is divisible by 4.

If

r + s

is divisible by 4, then

r + s = 4m

for some integer

m

. Since

r + s

is even,

r

and

s

must have the same parity (either both even or both odd). Consequently, their difference

r - s = (r + s) - 2s = 4m - 2s = 2(2m - s)

is an even integer (a multiple of 2). Multiplying

(r + s)

by

(r - s)

yields

(\text{multiple of } 4) \times (\text{multiple of } 2) = \text{multiple of } 8

.

Statement (2) alone yields a definitive 'Yes', so Statement (2) is sufficient.

Anahtar Kavram

Divisibility Rules of Difference of Squares and Parity of Integers

Tahmini Süre:2m 0s

Soru 34Soru →

If $a$ and $b$ are positive integers, is $a^2b$ divisible by $12$ ?

(1) $ab^2$ is divisible by $18$ .
(2) $a^3b$ is divisible by $72$ .

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.

Cevabı ve açıklamayı göster

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

Cevap

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

Statement (2) alone guarantees that

a^3b

contains at least three factors of

2

and two factors of

3

. Analyzing the exponent bounds for non-negative integers proves that

a^2b

must contain at least two factors of

2

and one factor of

3

, ensuring

a^2b

is divisible by

12

. Statement (1) alone is insufficient because

a=3, b=6

makes

ab^2=108

(divisible by

18

) but

a^2b=54

(not divisible by

12

). Thus, Statement (2) ALONE is sufficient.

Adım Adım Çözüm

1

Rephrase the target question in terms of prime factorization

For

a^2b

to be divisible by

12 = 2^2 \times 3^1

, we need

2 v_2(a) + v_2(b) \ge 2

and

2 v_3(a) + v_3(b) \ge 1

, where

v_p(n)

is the exponent of prime

p

in the prime factorization of

n

.

Decomposing divisibility into prime factor exponent inequalities allows definitive evaluation of sufficiency.

2

Evaluate Statement (1):

ab^2

is divisible by

18 = 2^1 \times 3^2

This gives

v_2(a) + 2 v_2(b) \ge 1

and

v_3(a) + 2 v_3(b) \ge 2

. Test counterexample

a=3, b=6

:

ab^2 = 3 \times 36 = 108

(divisible by 18), but

a^2b = 9 \times 6 = 54

, which is not divisible by 12.

A single valid counterexample proves Statement (1) is NOT sufficient.

3

Evaluate Statement (2):

a^3b

is divisible by

72 = 2^3 \times 3^2

This gives

3 v_2(a) + v_2(b) \ge 3

and

3 v_3(a) + v_3(b) \ge 2

. If

v_2(a) = 0

, then

v_2(b) \ge 3 \implies 2 v_2(a) + v_2(b) \ge 3 \ge 2

. If

v_2(a) \ge 1

, then

2 v_2(a) + v_2(b) \ge 2(1) + 0 = 2

. Similarly for prime 3: if

v_3(a) = 0

,

v_3(b) \ge 2 \implies 2 v_3(a) + v_3(b) \ge 2 \ge 1

; if

v_3(a) \ge 1

,

2 v_3(a) + v_3(b) \ge 2 \ge 1

. Thus

a^2b

is always divisible by 12.

Statement (2) strictly guarantees that the prime factor counts for 2 and 3 in

a^2b

meet or exceed the required thresholds, yielding a definitive 'Yes'.

Anahtar Kavram

Divisibility analysis using prime factor exponent inequalities in Data Sufficiency.

Soru 35Soru →

If $x$ and $y$ are non-zero real numbers, is $\frac{x}{y}$ an integer?

(1) $x^2 + y^2 = 5xy$
(2) $x$ is a prime number and $xy$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

Statement (1) allows us to set up a quadratic equation for the ratio

\frac{x}{y}

, yielding

\left(\frac{x}{y}\right)^2 - 5\left(\frac{x}{y}\right) + 1 = 0

. The roots of this quadratic equation are

\frac{5 \pm \sqrt{21}}{2}

, which are irrational numbers. Therefore,

\frac{x}{y}

cannot be an integer under any circumstance, providing a definitive 'No' answer to the question. Statement (1) is therefore sufficient. Statement (2) allows

\frac{x}{y}

to be an integer (e.g.,

x=3, y=1

) or a non-integer (e.g.,

x=3, y=2

), so it is not sufficient. Thus, the option stating that Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient is the correct choice.

Adım Adım Çözüm

1

Analyze the target question stem

The target asks if the quotient

\frac{x}{y}

is an integer, where

x

and

y

are real numbers (not restricted to integers).

Establishing that

x

and

y

are real numbers prevents making unjustified integer assumptions.

2

Evaluate Statement (1):

x^2 + y^2 = 5xy

Divide both sides by

y^2

(since

y \neq 0

):

\left(\frac{x}{y}\right)^2 + 1 = 5\left(\frac{x}{y}\right)

, which rearranges to

\left(\frac{x}{y}\right)^2 - 5\left(\frac{x}{y}\right) + 1 = 0

. Setting

k = \frac{x}{y}

, we get

k^2 - 5k + 1 = 0

. Solving for

k

using the quadratic formula gives

k = \frac{5 \pm \sqrt{21}}{2}

.

Since

\sqrt{21}

is irrational,

k = \frac{x}{y}

is an irrational number and can NEVER be an integer.

3

Determine sufficiency for Statement (1)

Statement (1) yields a definitive 'No' to the question 'Is

\frac{x}{y}

an integer?'. Thus, Statement (1) alone is SUFFICIENT.

In Data Sufficiency Yes/No questions, a definitive 'No' answer is a sufficient result.

4

Evaluate Statement (2):

x

is a prime number and

xy

is an integer

Case A: Let

x = 3

and

y = 1

. Then

xy = 3

(an integer), and

\frac{x}{y} = 3

(an integer) -> YES.
Case B: Let

x = 3

and

y = 2

. Then

xy = 6

(an integer), and

\frac{x}{y} = \frac{3}{2}

(not an integer) -> NO.

Because

\frac{x}{y}

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

Anahtar Kavram

Data Sufficiency Yes/No decision logic combined with irrational root analysis and real number constraints.

Tahmini Süre:2m 0s

Soru 36Soru →

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

(1) $x^2 + y^2$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

Statement (1) alone is sufficient because the square of any integer leaves a remainder of 0 or 1 when divided by 3. The only way for

x^2 + y^2

to be a multiple of 3 is if both

x^2

and

y^2

leave remainders of 0, meaning both

x

and

y

are multiples of 3. Therefore,

x + y

must be a multiple of 3. Statement (2) alone is not sufficient because

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

being divisible by 3 allows

x + y

to be divisible by 3 (e.g.,

x = 2, y = 1

) or not divisible by 3 (e.g.,

x = 4, y = 1

).

Adım Adım Çözüm

1

Analyze remainder properties of perfect squares modulo 3.

For any integer

z

,

z^2 \equiv 0 \pmod 3

if

z

is a multiple of 3, and

z^2 \equiv 1 \pmod 3

if

z

is not a multiple of 3.

Establishing the possible values of perfect squares modulo 3 simplifies the evaluation of both statements.

2

Evaluate Statement (1):

x^2 + y^2

is divisible by 3.

Since

x^2 \pmod 3 \in \{0, 1\}

and

y^2 \pmod 3 \in \{0, 1\}

, the sum

x^2 + y^2 \equiv 0 \pmod 3

can only occur if

x^2 \equiv 0 \pmod 3

and

y^2 \equiv 0 \pmod 3

. This implies both

x

and

y

are divisible by 3, so

x + y

must also be divisible by 3.

Statement (1) yields a definitive 'Yes' answer, making it sufficient.

3

Evaluate Statement (2):

x^2 - y^2

is divisible by 3.

Factor as

(x - y)(x + y)

. If

x = 2

and

y = 1

,

x^2 - y^2 = 3

(divisible by 3) and

x + y = 3

(divisible by 3, answer Yes). If

x = 4

and

y = 1

,

x^2 - y^2 = 15

(divisible by 3) and

x + y = 5

(not divisible by 3, answer No).

Since statement (2) yields both 'Yes' and 'No' outcomes, it is not sufficient.

Anahtar Kavram

Quadratic Remainder Constraints in Data Sufficiency

Soru 37Soru →

If $m$ and $n$ are positive real numbers, is $m$ an integer?

(1) $m^2 n$ is an integer and $m n^2$ is an integer.
(2) $m^3 n$ is an integer and $n$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

Statements (1) and (2) together are not sufficient because setting

m = \frac{1}{2}

and

n = 8

satisfies all given conditions (

m^2 n = 2

,

m n^2 = 32

,

m^3 n = 1

,

n = 8

) while yielding a 'No' to whether

m

is an integer. Meanwhile,

m = 1

and

n = 8

also satisfies all conditions and yields a 'Yes'. Therefore, the two statements combined do not uniquely determine if

m

is an integer.

Adım Adım Çözüm

1

Evaluate Statement (1) independently.

Statement (1) is insufficient.

If

m = 1

and

n = 2

, then

m^2 n = 2

and

m n^2 = 4

(both integers), and

m = 1

is an integer. However, if

m = \sqrt[3]{2}

and

n = \sqrt[3]{4}

, then

m^2 n = 2

and

m n^2 = 2

(both integers), but

m = \sqrt[3]{2}

is not an integer.

2

Evaluate Statement (2) independently.

Statement (2) is insufficient.

If

m = 1

and

n = 1

, then

m^3 n = 1

and

n = 1

(both integers), and

m = 1

is an integer. However, if

m = \frac{1}{2}

and

n = 8

, then

n = 8

is an integer and

m^3 n = (\frac{1}{8})(8) = 1

is an integer, but

m = \frac{1}{2}

is not an integer.

3

Evaluate Statements (1) and (2) together.

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

Test the candidate case

m = \frac{1}{2}

and

n = 8

against both statements simultaneously:
- Under Statement (1):

m^2 n = (\frac{1}{4})(8) = 2

(integer) and

m n^2 = (\frac{1}{2})(64) = 32

(integer).
- Under Statement (2):

n = 8

(integer) and

m^3 n = (\frac{1}{8})(8) = 1

(integer).
Here, all conditions from both statements are satisfied, but

m = \frac{1}{2}

is NOT an integer. Conversely,

m = 1

and

n = 8

also satisfies both statements and

m = 1

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

Anahtar Kavram

Implicit real number constraints versus integer constraints in Data Sufficiency polynomial expressions.

Soru 38Soru →

If $x$ is a positive integer, is $x^2 + 5x$ divisible by 6?

(1) $x$ is a prime number.
(2) $x + 1$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

Factoring the stem expression gives

x(x+5)

. For any positive integer

x

, one of

x

or

x+5

is even, making

x(x+5)

divisible by 2. Thus, the question reduces to whether

x(x+5)

is divisible by 3. Statement (1) allows

x=2

(

x^2+5x=14

, not divisible by 6) and

x=3

(

x^2+5x=24

, divisible by 6), yielding both 'No' and 'Yes' responses, so it is insufficient. Statement (2) states that

x+1

is a multiple of 3, meaning

x \equiv 2 \pmod 3

. Evaluating

x(x+5) \equiv 2(2+5) = 14 \equiv 2 \pmod 3

shows

x^2+5x

is never divisible by 3, and therefore never divisible by 6. This consistent 'No' answer makes Statement (2) alone sufficient.

Adım Adım Çözüm

1

Analyze and rephrase the question stem target expression.

Factor

x^2 + 5x = x(x + 5)

. Since

x

and

x + 5

have opposite parity, their product is always even (divisible by 2). Thus,

x^2 + 5x

is divisible by 6 if and only if

x(x + 5)

is divisible by 3.

Determining divisibility by 6 requires checking divisibility by both 2 and 3.

2

Evaluate Statement (1):

x

is a prime number.

If

x = 2

,

x^2 + 5x = 14

, which is not divisible by 6 (Answer: NO). If

x = 3

,

x^2 + 5x = 24

, which is divisible by 6 (Answer: YES). Since both 'Yes' and 'No' are possible, Statement (1) alone is INSUFFICIENT.

A statement must yield a single consistent answer ('Yes' or 'No') to be sufficient.

3

Evaluate Statement (2):

x + 1

is divisible by 3.

If

x + 1

is divisible by 3, then

x = 3k - 1

for some integer

k \ge 1

. Substituting

x = 3k - 1

into

x(x + 5)

yields

(3k - 1)(3k + 4) = 9k^2 + 9k - 4

. Dividing this expression by 3 gives a remainder of 2. Because

x(x + 5)

is never divisible by 3,

x^2 + 5x

is NEVER divisible by 6.

A definitive 'No' answer for all permitted values of

x

establishes sufficiency.

Anahtar Kavram

Number Properties and Integer Constraints in Data Sufficiency

Alternatif Yöntem

Test explicit values for Statement (2):

x \in \{2, 5, 8, 11\}

. Evaluating

x^2 + 5x

yields

14, 50, 104, 176

, none of which are divisible by 6. Recognizing this uniform 'No' outcome confirms Statement (2) is sufficient.

Tahmini Süre:2m 0s

Soru 39Soru →

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

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

Cevabı ve açıklamayı göster

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

Cevap

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

The correct option states that both statements together are sufficient, but neither alone is sufficient. Statement (1) allows irrational square roots like

\sqrt{2}

, making it insufficient. Statement (2) allows irrational cube roots like

\sqrt[3]{2}

, making it insufficient. Combined, taking the ratio

\frac{k^3}{k^2} = k

shows that

k

is the quotient of two integers and therefore rational. A rational number whose square is an integer must be an integer itself, giving a definitive 'Yes'.

Adım Adım Çözüm

1

Evaluate Statement (1):

k^2

is an integer.

If

k = 2

,

k^2 = 4

(an integer) and

k

is an integer (Yes). If

k = \sqrt{2}

,

k^2 = 2

(an integer), but

k

is not an integer (No). Statement (1) is NOT sufficient.

The question specifies

k

is a real number, so non-integer square roots must be tested.

2

Evaluate Statement (2):

k^3

is an integer.

If

k = 2

,

k^3 = 8

(an integer) and

k

is an integer (Yes). If

k = \sqrt[3]{2}

,

k^3 = 2

(an integer), but

k

is not an integer (No). Statement (2) is NOT sufficient.

Cube roots of non-cube integers yield irrational real numbers.

3

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

If

k = 0

,

k

is an integer (Yes). If

k \neq 0

, then

k = \frac{k^3}{k^2}

. Since

k^3

and

k^2

are both integers,

k

must be a rational number. Let

k = \frac{p}{q}

in simplest form where

p, q \in \mathbb{Z}

and

\gcd(p, q) = 1

. Since

k^2 = \frac{p^2}{q^2}

is an integer and

\gcd(p^2, q^2) = 1

,

q^2

must equal

1

, so

q = \pm 1

. Thus,

k

is an integer. Combined, the statements are SUFFICIENT.

Combining algebraic operations reveals that

k

is rational, and a rational number with an integer square must itself be an integer.

Anahtar Kavram

Data Sufficiency evaluation of number properties with real number constraints and rational number properties

Soru 40Soru →

If $x$ and $y$ are positive real numbers, is the product $xy$ an odd integer?

(1) $x^2y$ is an odd integer.
(2) $xy^2$ 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.

Cevabı ve açıklamayı göster

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

Cevap

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

The correct option is the choice stating that Statements (1) and (2) together are NOT sufficient. Since

x

and

y

are positive real numbers, setting

x = y = \sqrt[3]{3}

satisfies both

x^2y = 3

and

xy^2 = 3

(both odd integers), but gives

xy = \sqrt[3]{9}

, which is not an integer. Conversely, setting

x = 1/\sqrt[3]{3}

and

y = \sqrt[3]{9}

yields

x^2y = 3

and

xy^2 = 9

(both odd integers), giving

xy = 3

, which is an odd integer. Because both a 'No' and a 'Yes' answer are possible, the information remains insufficient even when both statements are combined.

Adım Adım Çözüm

1

Analyze the stem constraints and target question.

The target asks whether

xy

is an odd integer, given that

x

and

y

are positive real numbers (not necessarily integers).

Recognizing that

x

and

y

are not restricted to integers is essential to avoid invalid integer assumption traps.

2

Evaluate Statement (1) independently.

Statement (1) gives

x^2y = k

for some odd integer

k

. If

x = \sqrt{3}

and

y = 1

, then

x^2y = 3

(an odd integer), but

xy = \sqrt{3}

(not an integer). If

x = 3

and

y = 1

, then

x^2y = 9

(an odd integer) and

xy = 3

(an odd integer). Thus, Statement (1) alone is NOT sufficient.

Testing non-integer values demonstrates that

xy

may or may not be an integer.

3

Evaluate Statement (2) independently.

Statement (2) gives

xy^2 = m

for some odd integer

m

. If

x = 1

and

y = \sqrt{3}

, then

xy^2 = 3

(an odd integer), but

xy = \sqrt{3}

(not an integer). If

x = 1

and

y = 3

, then

xy^2 = 9

(an odd integer) and

xy = 3

(an odd integer). Thus, Statement (2) alone is NOT sufficient.

Symmetry with Statement (1) confirms that Statement (2) alone produces variable outcomes.

4

Evaluate Statements (1) and (2) combined.

Multiplying the two expressions gives

(x^2y)(xy^2) = x^3y^3 = (xy)^3 = km

. Since

k

and

m

are odd integers, their product

km

is also an odd integer, say

P

. Thus,

(xy)^3 = P \implies xy = \sqrt[3]{P}

. If

k = 3

and

m = 3

, then

P = 9

, so

xy = \sqrt[3]{9}

(not an integer). If

k = 3

and

m = 9

, then

P = 27

, so

xy = \sqrt[3]{27} = 3

(an odd integer). Because

xy

can be either an odd integer or a non-integer, the combined statements are NOT sufficient.

Combining the statements yields

(xy)^3

equal to an odd integer, which does not guarantee that

xy

itself is an integer.

Anahtar Kavram

Implicit Real vs. Integer Constraints in Data Sufficiency

Tahmini Süre:2m 0s

Number Properties and Integer Constraints in Data Sufficiency

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Alternatif Yöntem

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