Number Properties and Arithmetic Alıştırma Soruları — GMAT — Sayfa 2

If $n$ is any integer, which of the following expressions must be an even integer?

3n^2 + 5n + 2

n^2 + n + 1

2n^2 + 3n

n^3 + 2n

n^2 + 3

Cevabı ve açıklamayı göster

Cevap: $3n^2 + 5n + 2$

Cevap

The expression

3n^2 + 5n + 2

must be an even integer for any integer

n

.

The expression

3n^2 + 5n + 2

can be factored as

3n(n+1) + 2

. Since

n

and

n+1

are consecutive integers, one of them must be even, making

n(n+1)

an even integer. Multiplying an even integer by 3 produces an even integer, and adding 2 maintains even parity. Therefore, this expression is guaranteed to be even for all integer values of

n

.

Adım Adım Çözüm

1

Analyze the parity property of consecutive integers.

For any integer

n

, one of the terms in the pair

\{n, n+1\}

is even. Hence, the product

n(n+1)

is always an even integer.

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

2

Rewrite the target expression

3n^2 + 5n + 2

to isolate the consecutive integer product.

3n^2 + 5n + 2 = 3n^2 + 3n + 2 = 3n(n+1) + 2

.

Algebraic manipulation isolates known parity components.

3

Determine the parity of

3n(n+1) + 2

.

Since

n(n+1)

is even,

3 \times \text{even} = \text{even}

. Then

\text{even} + 2 = \text{even}

.

Multiplying an even integer by an odd integer yields an even integer, and adding an even integer preserves even parity.

Anahtar Kavram

Product of consecutive integers

n(n+1)

is always even; zero is an even integer; basic parity rules under addition and multiplication.

Tahmini Süre:1m 30s

Soru 22Soru →

If $n$ is an integer such that $1 \le n \le 50$ , how many values of $n$ satisfy the condition that $n^2 + 3n$ is an odd integer?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

0

For any integer

n

, the expression

n^2 + 3n

can be rewritten as

n(n + 3)

. Since

n

and

n + 3

differ by 3 (an odd number), one factor must be even and the other must be odd. The product of an even integer and an odd integer is always even. Therefore,

n^2 + 3n

is even for all integer values of

n

, meaning there are exactly 0 values of

n

in the given range for which the expression is odd.

Adım Adım Çözüm

1

Analyze the algebraic expression

n^2 + 3n

by testing parity cases for any integer

n

.

The parity of the expression depends on whether

n

is even or odd.

Every integer is either even or odd.

2

Evaluate

n^2 + 3n

assuming

n

is even.

n^2

is even and

3n

is even; their sum is even.

Even

\times

Even = Even, and Even

+

Even = Even.

3

Evaluate

n^2 + 3n

assuming

n

is odd.

n^2

is odd and

3n

is odd; their sum is even.

Odd

\times

Odd = Odd, and Odd

+

Odd = Even.

4

Count the number of integers

n

in

1 \le n \le 50

for which

n^2 + 3n

is odd.

Because

n^2 + 3n

is even for all integer values of

n

, zero values of

n

make the expression odd.

No integer

n

satisfies the condition.

Anahtar Kavram

Odd and Even Integers (Parity)

Tahmini Süre:1m 0s

Soru 23Soru →

For how many positive integers $n$ less than or equal to $120$ is the expression $n^3 - n + 3^n$ an even integer?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

0

Factoring

n^3 - n

into

(n-1)n(n+1)

shows that this component is the product of three consecutive integers and is therefore always even for any integer

n

. Since 3 is odd,

3^n

is always odd for any positive integer exponent

n

. The sum of an even integer and an odd integer is always odd. Consequently,

n^3 - n + 3^n

is odd for all positive integers

n

, meaning 0 values of

n \le 120

yield an even integer.

Adım Adım Çözüm

1

Analyze the algebraic expression by factoring

n^3 - n

.

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

, which represents the product of three consecutive integers.

The product of consecutive integers contains at least one even number, making

n^3 - n

even for all integers

n

.

2

Analyze the exponential term

3^n

.

Since 3 is odd,

3^n

is odd for every positive integer

n

.

An odd integer raised to any positive integer exponent yields an odd integer.

3

Determine the overall parity of the expression

n^3 - n + 3^n

.

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

. The expression is odd for all positive integers

n

.

Adding an odd integer to an even integer results in an odd integer.

4

Count the number of positive integers

n \le 120

for which the expression is even.

There are 0 such integers.

Because the expression is odd for every positive integer

n

, it is never even.

Anahtar Kavram

Parity rules of consecutive integer products, odd exponents, and addition of even/odd integers.

Soru 24Soru →

If $m$ , $n$ , and $p$ are integers such that $m^2 n + n^2 p + p^2 m + m n p$ is an odd integer, which of the following expressions MUST be an even integer?

m n + n p + p m

m + n + p + 1

m n p + 1

m^2 + n^2 + p^2 + 1

m + n + p

Cevabı ve açıklamayı göster

Cevap: $m + n + p$

Cevap

The sum of the three variables,

m + n + p

, MUST be an even integer.

Analyzing the expression

m^2 n + n^2 p + p^2 m + m n p

modulo 2 reveals that

x^2

has the same parity as

x

. Substituting modulo 2 yields

m n + n p + p m + m n p

. Testing all combinations of parity for

m, n, p

shows this expression is odd if and only if exactly two of the three variables are odd and one is even. The sum of two odd integers and one even integer (

m + n + p

) is always even, so

m + n + p

must be an even integer.

Adım Adım Çözüm

1

Analyze the parity of the given expression modulo 2.

Since

x^2 \equiv x \pmod 2

for any integer

x

, the given expression

m^2 n + n^2 p + p^2 m + m n p \equiv m n + n p + p m + m n p \pmod 2

.

Squaring an integer does not change its parity.

2

Determine which parity combinations of

m

,

n

, and

p

result in an odd value.

Evaluating

m n + n p + p m + m n p \pmod 2

across all possible parity combinations shows that the expression is odd if and only if exactly two of the variables are odd and exactly one variable is even.

If all three are even or one is odd and two are even, the expression equals 0 (even). If all three are odd,

1 + 1 + 1 + 1 = 4 \equiv 0

(even). Only when exactly two variables are odd (e.g., 1, 1, 0) does

1\cdot 1 + 1\cdot 0 + 0\cdot 1 + 1\cdot 1\cdot 0 = 1

(odd).

3

Test the parity of

m + n + p

under the condition that exactly two variables are odd and one is even.

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

.

Adding two odd integers produces an even integer, and adding an even integer keeps the sum even.

Anahtar Kavram

Odd and Even Integers (Parity)

Tahmini Süre:2m 0s

Soru 25Soru →

If $p$ and $q$ are integers such that $3p + 2q$ is an odd integer, which of the following statements must be true?

p

must be odd, but

q

can be any integer

p

must be odd, and

q

must be even

Both

p

and

q

must be odd

p

must be even, and

q

must be odd

Both

p

and

q

must be even

Cevabı ve açıklamayı göster

Cevap: $p$ must be odd, but $q$ can be any integer

Cevap

p

must be odd, but

q

can be any integer

Multiplying any integer

q

by 2 guarantees that

2q

is even. For the sum

3p + 2q

to be odd,

3p

must be odd because the sum of an odd integer and an even integer is odd. Since 3 is odd,

3p

is odd if and only if

p

is odd. The variable

q

can be any integer because

2q

remains even whether

q

is even or odd.

Adım Adım Çözüm

1

Analyze the parity of the term

2q

For any integer

q

,

2q

is a multiple of 2, so

2q

is always an even integer.

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

2

Determine the required parity of the term

3p

Since

3p + 2q

is odd and

2q

is even,

3p

must be odd.

An odd integer plus an even integer yields an odd sum (

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

).

3

Determine the parity of

p

Since

3p

is odd and 3 is odd,

p

must be an odd integer.

The product of two integers is odd if and only if both integers are odd (

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

).

4

Evaluate the restrictions on

q

q

can be any integer (even or odd).

Multiplying

q

by 2 forces

2q

to be even regardless of whether

q

is odd or even.

Anahtar Kavram

Parity rules under addition and multiplication of integers

Soru 26Soru →

If $x$ , $y$ , and $z$ are integers, is the expression $x(y + z)$ an odd integer?

(1) $x^2 + y^2 + z^2$ is an odd integer.
(2) $xy + yz + zx$ 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.

Cevabı ve açıklamayı göster

Cevap: EACH statement ALONE is sufficient.

Cevap

EACH statement ALONE is sufficient.

Each statement independently guarantees that the expression

x(y + z)

must be an even integer. Because a definitive 'NO' answer is obtained from each statement independently, each statement alone is sufficient.

Adım Adım Çözüm

1

Analyze the target expression

x(y+z)

.

The expression

x(y+z)

is odd if and only if both

x

is odd AND

(y+z)

is odd. For

(y+z)

to be odd, exactly one of

y

or

z

must be odd and the other even.

Establishing the precise condition for the expression to be odd determines what parity combinations are needed.

2

Evaluate Statement (1):

x^2 + y^2 + z^2

is an odd integer.

Since

k^2

has the same parity as

k

,

x+y+z

must be odd. This occurs in two parity distributions:
- Case 1: All three of

x, y, z

are odd. Here,

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

, so

x(y+z) = \text{odd} \times \text{even} = \text{even}

.
- Case 2: One variable is odd and two are even.
- Subcase 2a:

x

is odd, while

y

and

z

are even. Then

y+z = \text{even} + \text{even} = \text{even}

, so

x(y+z) = \text{odd} \times \text{even} = \text{even}

.
- Subcase 2b:

x

is even, while one of

y,z

is odd and the other is even. Then

x(y+z) = \text{even} \times \text{odd} = \text{even}

.
In all possible cases,

x(y+z)

is even. Thus, the answer to 'Is

x(y+z)

odd?' is a definitive NO. Statement (1) is sufficient.

Testing all valid parity distributions under Statement (1) shows

x(y+z)

can never be odd.

3

Evaluate Statement (2):

xy + yz + zx

is an even integer.

Note that

x(y+z) = xy + zx

. Therefore,

xy + yz + zx = x(y+z) + yz = \text{even}

.
Suppose for contradiction that

x(y+z)

were odd. Then

yz

would also have to be odd (since

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

). For

yz

to be odd, both

y

and

z

must be odd. But if

y

and

z

are both odd, then

y+z

must be even, which forces

x(y+z) = x \times \text{even} = \text{even}

, contradicting our assumption that

x(y+z)

is odd.
Thus,

x(y+z)

cannot be odd under Statement (2); it must be even. The answer is a definitive NO. Statement (2) is sufficient.

Proof by contradiction demonstrates that

x(y+z)

cannot be odd under Statement (2).

Anahtar Kavram

Parity rules of sums and products, including proof by case analysis and contradiction.

Soru 27Soru →

What is the total number of distinct positive prime factors of the integer $60$ ?

2

3

4

6

12

Cevabı ve açıklamayı göster

Cevap: $3$

Cevap

The total number of distinct positive prime factors of

60

is

3

.

To find the number of distinct positive prime factors of

60

, perform prime factorization:

60 = 2^2 \times 3^1 \times 5^1

. The distinct prime numbers in this expression are

2

,

3

, and

5

. Therefore, there are

3

distinct positive prime factors.

Adım Adım Çözüm

1

Find the prime factorization of

60

.

60 = 2^2 \times 3^1 \times 5^1

Expressing the integer as a product of prime numbers reveals all of its prime factors.

2

Identify the distinct prime bases in the factorization.

The distinct prime factors are

2

,

3

, and

5

.

Prime factors are the prime numbers that divide the given integer evenly.

3

Count the number of distinct prime factors.

There are

3

distinct prime factors.

Counting the elements in the set

\{2, 3, 5\}

gives

3

.

Anahtar Kavram

Prime Factorization and Prime Factors

Tahmini Süre:45s

Soru 28Soru →

For how many integers $n$ satisfying $0 \le n \le 100$ is the expression $n^4 + 2^n + n$ an odd integer?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

There is exactly 1 value of

n

(specifically

n = 0

) in the range

0 \le n \le 100

for which the expression

n^4 + 2^n + n

is an odd integer.

For any integer

n \ge 1

,

n^4

and

n

share the exact same parity (both even or both odd), making

n^4 + n

always even. Additionally, for

n \ge 1

,

2^n

is an even integer. The sum of two even integers is always even, so

n^4 + 2^n + n

is even for all

1 \le n \le 100

. When

n = 0

, however,

2^0 = 1

is odd and

0^4 + 0 = 0

is even. The sum

0 + 1 = 1

is odd. Thus,

n = 0

is the single value that satisfies the condition.

Adım Adım Çözüm

1

Analyze parity of

n^4 + n

n^4 + n

is even for all integers

n

An integer

n

and its power

n^4

always have the same parity. The sum of two integers of the same parity is always even.

2

Evaluate

2^n

parity for

n \ge 1

2^n

is even for all

n \ge 1

Any positive integer power of 2 contains 2 as a prime factor and is therefore even.

3

Evaluate the expression for

n = 0

0^4 + 2^0 + 0 = 1

, which is odd

Zero is a non-negative integer within

0 \le n \le 100

. By definition of exponents,

2^0 = 1

, which is odd.

4

Combine results and count valid integers

Total count = 1

For

n \ge 1

, even + even = even. For

n = 0

, even + odd = odd. Hence, only

n = 0

produces an odd integer.

Anahtar Kavram

Parity Rules for Exponent Base 2 Edge Cases and Algebraic Expressions

Soru 29Soru →

If $m$ and $n$ are integers such that $5m + 3n$ is an even integer and $m - 2n$ is an odd integer, which of the following expressions must be an even integer?

mn + 2

m + n + 1

m^2 + n^2

mn + n + 1

2m + n

Cevabı ve açıklamayı göster

Cevap: $m^2 + n^2$

Cevap

The expression

m^2 + n^2

must be an even integer.

Deduce the parities of

m

and

n

:

5m + 3n = 2(2m + n) + (m + n)

, so

m + n

must be even, implying

m

and

n

have the same parity. Next,

m - 2n

is odd, and since

2n

is even,

m

must be odd. Therefore,

n

is also odd. Evaluating

m^2 + n^2

with

m

and

n

both odd yields

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

.

Adım Adım Çözüm

1

Analyze the parity of the expression

5m + 3n

.

m + n

is an even integer.

Rewrite

5m + 3n

as

2(2m + n) + (m + n)

. Since

2(2m + n)

is always even,

5m + 3n

has the same parity as

m + n

. Because

5m + 3n

is even,

m + n

must be even, meaning

m

and

n

share the same parity (both even or both odd).

2

Analyze the parity of the expression

m - 2n

.

m

is an odd integer.

Since

2n

is always even,

m - 2n

has the same parity as

m

. Given that

m - 2n

is odd,

m

must be odd.

3

Determine the parity of

n

and evaluate the options.

Both

m

and

n

are odd integers, so

m^2 + n^2 = \text{odd} + \text{odd} = \text{even}

.

Since

m

is odd and

m + n

is even,

n

must also be odd. The square of an odd integer is odd, so

m^2

and

n^2

are both odd, making their sum

m^2 + n^2

an even integer.

Anahtar Kavram

Parity rules for integer addition, subtraction, and multiplication

Tahmini Süre:1m 30s

Soru 30Soru →

For any integer $k$ , the expression $k(k + 1)(k + 5)$ must be an even integer.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True. The expression

k(k + 1)(k + 5)

is guaranteed to be an even integer for all integer values of

k

.

Because

k

and

k + 1

are consecutive integers, one of them must be even. The product of an even integer and any other integer is always even, making

k(k + 1)(k + 5)

even for every integer

k

.

Adım Adım Çözüm

1

Analyze the parity of the consecutive terms

k

and

k + 1

.

One of the two integers

k

or

k + 1

is always even regardless of whether

k

is even or odd.

Consecutive integers always alternate between even and odd.

2

Determine the parity of the product of an even integer and any other integer.

The product

k(k + 1)

is always an even integer.

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

3

Evaluate the full expression

k(k + 1)(k + 5)

.

Since

k(k + 1)

is even, multiplying by

(k + 5)

produces an even integer.

An even integer multiplied by any integer results in an even integer.

Anahtar Kavram

Parity of products of consecutive integers

Tahmini Süre:1m 0s

Soru 31Soru →

If $x$ , $y$ , and $z$ are integers such that $(x + 1)(y + 2)$ is an odd integer and $y(z + 3)$ is an even integer, which of the following expressions MUST be an even integer?

x + y + z

xy + z

(y + z)^2 + 1

xz + y

(x + 1)yz

Cevabı ve açıklamayı göster

Cevap: $x + y + z$

Cevap

The expression

x + y + z

MUST be an even integer.

The condition

(x + 1)(y + 2) = \text{odd}

implies both

(x + 1)

and

(y + 2)

are odd, so

x

is even and

y

is odd. Then

y(z + 3) = \text{even}

with

y

being odd requires

(z + 3)

to be even, so

z

is odd. Summing an even integer

x

and two odd integers

y

and

z

yields

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

. Thus, the expression representing the sum of

x

,

y

, and

z

must always be even.

Adım Adım Çözüm

1

Determine the parity of

x

and

y

from the first given condition.

x

is even and

y

is odd.

The product

(x + 1)(y + 2)

is odd, which means both factors

(x + 1)

and

(y + 2)

must be odd. If

x + 1

is odd, then

x

is even. If

y + 2

is odd, then

y

is odd.

2

Determine the parity of

z

using the parity of

y

and the second given condition.

z

is odd.

The product

y(z + 3)

is even. Since

y

is odd,

(z + 3)

must be even for the product to be even. If

z + 3

is even, then

z

must be odd.

3

Evaluate the parity of

x + y + z

.

x + y + z

is even.

Substituting the parities gives

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

.

Anahtar Kavram

Parity rules for addition and multiplication of integers

Tahmini Süre:2m 0s

Soru 32Soru →

Let $f(n) = n^5 + 4n^3 + 3n + 1$ for any positive integer $n$ . For how many integer values of $n$ in the range $1 \le n \le 150$ is the expression $E(n) = (f(n))^n + n^{f(n)}$ an even integer?

Cevabı ve açıklamayı göster

Cevap: 75

Cevap

The total number of integer values of

n

in the given range for which the expression is even is 75.

Analyzing

f(n) \pmod 2

reveals that

n^5 \equiv n \pmod 2

and

3n \equiv n \pmod 2

, making

f(n) \equiv n + 0 + n + 1 \equiv 2n + 1 \equiv 1 \pmod 2

. Since

f(n)

is unconditionally odd,

(f(n))^n

is an odd number raised to a positive integer power, which remains odd. The second term

n^{f(n)}

has positive exponent

f(n) \ge 1

, so its parity matches the base

n

. Therefore,

E(n) = \text{Odd} + n

is even precisely when

n

is odd. In the range

1 \le n \le 150

, exactly half of the 150 integers are odd, yielding 75.

Adım Adım Çözüm

1

Evaluate the parity of

f(n)

f(n) = n^5 + 4n^3 + 3n + 1 \equiv n + 0 + n + 1 \equiv 2n + 1 \equiv 1 \pmod 2

, so

f(n)

is always odd.

Powers of an integer retain the base parity (

n^5 \equiv n \pmod 2

), and coefficients modulo 2 simplify terms.

2

Evaluate the parity of

(f(n))^n

and

n^{f(n)}

(f(n))^n

is always odd because an odd base raised to any positive integer exponent is odd.

n^{f(n)}

has the same parity as

n

because

f(n) \ge 1

.

Exponent parity rules state that

\text{odd}^k = \text{odd}

and

\text{even}^k = \text{even}

for positive integer exponents

k

.

3

Formulate the condition for

E(n)

to be even

E(n) = \text{Odd} + n

. Thus,

E(n)

is even if and only if

n

is odd.

The sum of two integers is even if and only if both integers have the same parity (Odd + Odd = Even).

4

Count odd integers in the specified interval

There are 75 odd integers in

\{1, 2, 3, \dots, 150\}

.

Half of the integers in any consecutive sequence of even length starting at 1 are odd.

Anahtar Kavram

Odd and Even Integers (Parity)

Tahmini Süre:2m 0s

Soru 33Soru →

If $m$ , $n$ , and $p$ are integers such that $m^3 n - n^2 p$ is an odd integer and $m(n + p)$ is an even integer, which of the following expressions MUST be an even integer?

m^2 + n + p

m + np

mn + p

mp + n

n + p + 1

Cevabı ve açıklamayı göster

Cevap: $m^2 + n + p$

Cevap

m^2 + n + p

must be an even integer.

Factoring

m^3 n - n^2 p

as

n(m^3 - np)

shows that

n

is odd and

m^3 - np

is odd. Because

n

is odd,

m^3 - np

has the same parity as

m - p

, meaning

m

and

p

have opposite parities. Testing the condition that

m(n + p)

is even reveals that

p

must be odd and

m

must be even (if

p

were even,

m

would be odd and

m(n + p)

would be odd, a contradiction). With

m

even,

n

odd, and

p

odd, the expression

m^2 + n + p

calculates as

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

, which must be an even integer.

Adım Adım Çözüm

1

Analyze the parity of the expression

m^3 n - n^2 p

.

n

is odd and

m^3 - np

is odd.

The expression can be factored as

n(m^3 - np)

. For a product of two integers to be odd, both factors must be odd. Hence,

n

must be odd, and

m^3 - np

must also be odd.

2

Determine the relative parities of

m

and

p

.

m

and

p

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

Since

n

is odd,

np

has the same parity as

p

. The expression

m^3 - np

has the same parity as

m - p

. For

m - p

to be odd,

m

and

p

must have opposite parities.

3

Use the second given condition

m(n + p)

is even to determine the exact parity of

m

and

p

.

m

is even and

p

is odd.

If

p

were even, then

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

. Since

m

and

p

have opposite parities,

m

would be odd, making

m(n + p) = \text{odd} \times \text{odd} = \text{odd}

, which contradicts the condition that

m(n + p)

is even. Thus,

p

cannot be even;

p

must be odd, which implies

m

is even.

4

Evaluate the parity of the options using

m = \text{even}

,

n = \text{odd}

, and

p = \text{odd}

.

m^2 + n + p = \text{even} + \text{odd} + \text{odd} = \text{even}

.

m^2

is even since

m

is even. Summing an even number (

m^2

) and two odd numbers (

n

and

p

) yields an even integer.

Anahtar Kavram

Parity rules for integer addition and multiplication: the product of integers is odd if and only if all factors are odd, and the sum of two odd integers is even.

Tahmini Süre:2m 0s

Soru 34Soru →

What is the total number of positive factors of the integer $36$ ?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The total number of positive factors of

36

is 9.

The prime factorization of

36

is

2^2 \times 3^2

. By the total factors formula, adding 1 to each exponent and multiplying gives

(2+1)(2+1) = 9

. The factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Adım Adım Çözüm

1

Find the prime factorization of 36

36 = 4 \times 9 = 2^2 \times 3^2

Decomposing the number into prime factors allows the application of the factor count formula.

2

Apply the exponent formula for counting factors

(2 + 1)(2 + 1) = 3 \times 3 = 9

Each factor of

36

is of the form

2^x \times 3^y

, where

x \in \{0, 1, 2\}

(3 choices) and

y \in \{0, 1, 2\}

(3 choices).

Anahtar Kavram

Calculating the total number of positive factors from prime factorization exponents.

Tahmini Süre:45s

Soru 35Soru →

A positive integer $N$ has no prime factors other than $2$ and $3$ . If $N$ is a multiple of $12$ and has exactly $18$ positive divisors, what is the sum of all possible values of $N$ ?

Cevabı ve açıklamayı göster

Cevap: 2028

Cevap

2028

The correct numerical answer is 2028, obtained by determining all exponent combinations of 2 and 3 that yield 18 total factors while guaranteeing divisibility by 12.

Adım Adım Çözüm

1

Set up the prime factorization of

N

with constraints.

N = 2^a \cdot 3^b

with

a \ge 2

and

b \ge 1

.

Since the only prime factors are

2

and

3

, and

12 = 2^2 \cdot 3^1

divides

N

, the exponents must satisfy

a \ge 2

and

b \ge 1

.

2

Apply the total number of divisors formula.

(a+1)(b+1) = 18

.

The number of positive divisors of

p_1^{e_1} p_2^{e_2}

is

(e_1 + 1)(e_2 + 1)

.

3

Identify all valid integer solution pairs for

(a+1, b+1)

.

Valid pairs are

(9,2)

,

(6,3)

, and

(3,6)

, corresponding to

(a,b) = (8,1), (5,2), (2,5)

.

We require

a+1 \ge 3

and

b+1 \ge 2

. Pairs

(18,1)

and

(2,9)

violate these lower bounds.

4

Compute the corresponding values of

N

and find their sum.

768 + 288 + 972 = 2028

.

Evaluating

2^8 \cdot 3^1 = 768

,

2^5 \cdot 3^2 = 288

, and

2^2 \cdot 3^5 = 972

yields a total sum of

2028

.

Anahtar Kavram

Divisor Count Formula & Multiplicativity Constraints

Soru 36Soru →

For how many integer values of $k$ in the range $1 \le k \le 60$ is the expression $3k^2 + 5k + 7$ an even integer?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

0

The expression

3k^2 + 5k + 7

can be evaluated for parity by considering the cases for

k

.
If

k

is even,

3k^2

is even,

5k

is even, and

7

is odd. The sum of two even integers and an odd integer (even + even + odd) is always odd.
If

k

is odd,

3k^2

is odd,

5k

is odd, and

7

is odd. The sum of three odd integers (odd + odd + odd) is always odd.
Because the expression yields an odd integer for every integer

k

, there are no integer values of

k

in the specified range for which the expression is even. Therefore, the correct count is 0.

Adım Adım Çözüm

1

Examine parity by testing even and odd cases for k

If k is even: 3(even)^2 + 5(even) + 7 = even + even + odd = odd. If k is odd: 3(odd)^2 + 5(odd) + 7 = odd + odd + odd = odd.

Covering both cases establishes the parity of the expression for all integer inputs.

2

Count the number of values of k in 1 <= k <= 60 that yield an even result

Since the expression is odd for all integer values of k, zero values yield an even integer.

The question specifically asks for the number of integer values of k that make the expression even.

Anahtar Kavram

Parity Rules for Addition, Multiplication, and Algebraic Expressions

Soru 37Soru →

For any integer $n$ , which of the following expressions must be divisible by $2$ ?

n^2 + n

n^2 + 1

n^3 + 1

(n + 1)^2

2n + 1

Cevabı ve açıklamayı göster

Cevap: $n^2 + n$

Cevap

The expression

n^2 + n

is always divisible by

2

for any integer

n

.

The expression

n^2 + n

factors into

n(n + 1)

. Because

n

and

n + 1

are consecutive integers, one of them must be even. Any integer multiplied by an even integer produces an even number, guaranteeing that

n^2 + n

is divisible by

2

for every integer

n

.

Adım Adım Çözüm

1

Factor the given algebraic expression.

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

Factoring out

n

reveals the product of two consecutive integers.

2

Analyze the parity of consecutive integers

n

and

n + 1

.

In any pair of consecutive integers

(n, n + 1)

, exactly one number is even.

Even and odd integers alternate sequentially.

3

Determine divisibility by

2

.

Because one of the factors is even (divisible by

2

), their product

n(n + 1)

is always divisible by

2

.

Any integer multiplied by an even number yields an even result.

Anahtar Kavram

The product of any two consecutive integers is always even and therefore divisible by 2.

Tahmini Süre:45s

Soru 38Soru →

Let $A = 12^6 \cdot 35^4$ and $B = 18^4 \cdot 14^6$ . If $d = \text{gcd}(A, B)$ , how many positive factors of $d^2$ are not factors of $d$ ?

Cevabı ve açıklamayı göster

Cevap: 2072

Cevap

The number of positive factors of

d^2

that are not factors of

d

is 2072.

Decomposing

A

and

B

into prime factors yields

A = 2^{12} \cdot 3^6 \cdot 5^4 \cdot 7^4

and

B = 2^{10} \cdot 3^8 \cdot 7^6

. Taking the minimum power of each common prime gives

d = \text{gcd}(A, B) = 2^{10} \cdot 3^6 \cdot 7^4

, which has

(10+1)(6+1)(4+1) = 385

positive factors. For

d^2 = 2^{20} \cdot 3^{12} \cdot 7^8

, the total number of positive factors is

(20+1)(12+1)(8+1) = 2457

. Subtracting the factors of

d

yields

2457 - 385 = 2072

.

Adım Adım Çözüm

1

Find the prime factorizations of

A

and

B

A = 2^{12} \cdot 3^6 \cdot 5^4 \cdot 7^4

and

B = 2^{10} \cdot 3^8 \cdot 7^6

Converting composite bases into prime factors allows determination of common divisor properties.

2

Determine the greatest common divisor

d = \text{gcd}(A, B)

d = 2^{10} \cdot 3^6 \cdot 7^4

The GCD takes the minimum exponent for each common prime factor between

A

and

B

.

3

Calculate the total number of positive factors of

d

385 positive factors

Adding 1 to each prime exponent of

d

and multiplying gives

(10+1)(6+1)(4+1) = 385

.

4

Find the prime factorization and number of positive factors of

d^2

d^2 = 2^{20} \cdot 3^{12} \cdot 7^8

, which has 2457 positive factors

Squaring

d

doubles all prime exponents. The number of factors is

(20+1)(12+1)(8+1) = 2457

.

5

Subtract the number of factors of

d

from the number of factors of

d^2

2457 - 385 = 2072

Since every factor of

d

is also a factor of

d^2

, the factors of

d^2

that are not factors of

d

equal the total factors of

d^2

minus the factors of

d

.

Anahtar Kavram

Prime Factorization, Greatest Common Divisor (GCD), and the Divisor Count Formula

Soru 39Soru →

If $m$ and $n$ are integers such that $m^2 n + m$ is an odd integer, which of the following expressions MUST be an even integer?

m^2 + n

m + 2n

m(n + 1)

m^2 + n^2 + 1

3m - n

Cevabı ve açıklamayı göster

Cevap: $m^2 + n^2 + 1$

Cevap

The expression

m^2 + n^2 + 1

must be an even integer.

Factoring the given expression gives

m(mn + 1) = \text{odd}

, which implies

m

is odd and

mn + 1

is odd. Consequently,

mn

is even, which forces

n

to be even (since

m

is odd). Substituting

m = \text{odd}

and

n = \text{even}

into the expression

m^2 + n^2 + 1

yields

\text{odd} + \text{even} + 1 = \text{even}

. Thus, this expression must be an even integer.

Adım Adım Çözüm

1

Factor the given expression to analyze its parity.

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

.

Factoring out

m

isolates the product of two factors.

2

Determine the parity of each factor.

Since

m(mn + 1)

is odd, both

m

and

(mn + 1)

must be odd integers.

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

3

Determine the parity of

n

.

Since

mn + 1

is odd,

mn

must be even. Because

m

is odd,

n

must be even.

For the product

mn

to be even when

m

is odd,

n

must be an even integer (including 0).

4

Evaluate the target expression

m^2 + n^2 + 1

.

m^2

is odd,

n^2

is even, so

m^2 + n^2 + 1 = \text{odd} + \text{even} + \text{odd} = \text{even}

.

Summing two odd integers and one even integer always results in an even integer.

Anahtar Kavram

Parity properties under addition, subtraction, and multiplication

Tahmini Süre:1m 30s

Soru 40Soru →

If $a$ , $b$ , and $c$ are integers such that $a(b + c)$ is an odd integer, which of the following expressions MUST be an even integer?

a b + c

a + b c

a b + b c

a + b + c

a c + b

Cevabı ve açıklamayı göster

Cevap: $a + b + c$

Cevap

The expression

a + b + c

must be an even integer.

For the product

a(b + c)

to be odd, both

a

and

(b + c)

must be odd integers. Regrouping the sum

a + b + c

as

a + (b + c)

shows it is the sum of two odd integers, which is guaranteed to be an even integer.

Adım Adım Çözüm

1

Analyze the condition

a(b + c)

is odd

For a product of two integers to be odd, both factors must be odd. Therefore,

a

is odd and

(b + c)

is odd.

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

2

Evaluate the parity of

a + b + c

a + b + c = a + (b + c) = \text{odd} + \text{odd} = \text{even}

.

The sum of any two odd integers is always an even integer.

Anahtar Kavram

Parity rules for integer addition and multiplication

Number Properties and Arithmetic

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