Arithmetic Practice Questions — GRE General Test — Page 5

If $x$ is any integer, which of the following expressions must be divisible by $2$ ?

x^2 + 1

x^2 + x + 4

x^3 + 2x + 1

2x^2 + x + 3

x^2 + 3x + 1

Show answer & explanation

Answer: $x^2 + x + 4$

Answer

The expression

x^2 + x + 4

must be divisible by

2

for all integer values of

x

.

The expression

x^2 + x + 4

can be rewritten as

x(x + 1) + 4

. For any integer

x

, either

x

or

x + 1

is even, so their product

x(x + 1)

is always even (a multiple of

2

). Adding

4

(which is also even) to an even number always yields an even number. Therefore,

x^2 + x + 4

is divisible by

2

for all integer values of

x

.

Step-by-Step Solution

1

Rewrite the expression by factoring out

x

from the first two terms.

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

Grouping terms isolates the product of two consecutive integers.

2

Analyze the parity of the product

x(x + 1)

.

Because

x

and

x + 1

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

x(x + 1)

an even integer.

The product of any integer and its immediate successor is always even.

3

Add

4

to the even product

x(x + 1)

.

The sum of an even integer and an even integer (

4

) is always an even integer.

Adding an even integer to an even integer preserves divisibility by

2

.

Key Concept

Parity of consecutive integer products and integer divisibility rules

Question 82Question →

Let $k$ be a positive integer. The integer $k$ has exactly $6$ positive divisors, $3k$ has exactly $8$ positive divisors, and $5k$ has exactly $12$ positive divisors. Which of the following could be the value of $k$ ? Indicate all such values.

Select all that apply

18

45

63

81

99

Show answer & explanation

Answer: $18$ ; $63$ ; $99$

Answer

The correct values of

k

are

18

,

63

, and

99

.

The integers

18

,

63

, and

99

are all of the form

k = 3^2 \times q^1 = 9q

, where

q

is a prime number other than

3

or

5

(specifically

q = 2, 7, 11

). Each has

(2+1)(1+1) = 6

divisors,

3k = 3^3 \times q^1

has

(3+1)(1+1) = 8

divisors, and

5k = 3^2 \times 5^1 \times q^1

has

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

divisors.

Step-by-Step Solution

1

Analyze the prime factorization form of

k

based on its divisor count

d(k) = 6

.

Since

6 = 6 \times 1

or

3 \times 2

,

k

must be of the form

p^5

or

p^2 q^1

, where

p

and

q

are distinct prime numbers.

The formula for the total number of positive divisors of an integer

n = p_1^{a_1} p_2^{a_2} \dots p_m^{a_m}

is

(a_1 + 1)(a_2 + 1) \dots (a_m + 1)

.

2

Test the prime factorization form

k = p^5

.

k = p^5

is impossible.

If

p \neq 3

, then

3k = 3^1 p^5

, which would have

(1+1)(5+1) = 12

divisors (contradicting

d(3k) = 8

). If

p = 3

,

3k = 3^6

has

7

divisors. If

p = 5

,

5k = 5^6

has

7

divisors.

3

Test the prime factorization form

k = p^2 q^1

under different prime identities.

k

must be of the form

3^2 q^1

, where

q

is a prime number distinct from

3

and

5

.

If

p = 3

and

q \neq 3, 5

, then

3k = 3^3 q^1

has

(3+1)(1+1) = 8

divisors, and

5k = 3^2 \times 5^1 \times q^1

has

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

divisors. Any other choice for

p

or

q

fails to yield

d(3k) = 8

or

d(5k) = 12

.

4

Evaluate the given options against the required form

k = 9q

where

q

is a prime distinct from

3

and

5

.

18 = 9 \times 2

(

q=2

, valid prime),

63 = 9 \times 7

(

q=7

, valid prime), and

99 = 9 \times 11

(

q=11

, valid prime) are correct.

45 = 9 \times 5

(

q=5

, invalid because

q \neq 5

) and

81 = 3^4

(invalid form) are incorrect.

Only primes

q \notin \{3, 5\}

preserve the required divisor counts for

3k

and

5k

.

Key Concept

Divisors formula and impact of prime multiplication on prime factorization exponents

Estimated Time:2m 30s

Question 83Question →

When the positive integer $n$ is divided by $12$ , the remainder is $7$ . What is the remainder when $n^2 + 3n + 5$ is divided by $12$ ?

1

3

4

7

15

Show answer & explanation

Answer: $3$

Answer

The remainder when

n^2 + 3n + 5

is divided by

12

is

3

.

The correct answer is obtained by replacing

n

with its remainder

7

modulo

12

in the expression

n^2 + 3n + 5

. Evaluating gives

7^2 + 3(7) + 5 = 49 + 21 + 5 = 75

. Dividing

75

by

12

yields

6

with a remainder of

3

.

Step-by-Step Solution

1

Express

n

in terms of its quotient and remainder modulo

12

.

n = 12q + 7

for some non-negative integer

q

, meaning

n \equiv 7 \pmod{12}

.

By the division algorithm, any integer

n

leaving a remainder of

7

when divided by

12

can be represented as

12q + 7

.

2

Substitute

n \equiv 7 \pmod{12}

into the polynomial expression

n^2 + 3n + 5

.

n^2 + 3n + 5 \equiv 7^2 + 3(7) + 5 \pmod{12}

.

Properties of modular arithmetic allow direct substitution of equivalent remainders into polynomial expressions.

3

Evaluate the arithmetic expression.

7^2 + 3(7) + 5 = 49 + 21 + 5 = 75

.

Compute the total numeric value before finding the final remainder modulo

12

.

4

Divide the total sum by

12

to find the final remainder.

75 = 12 \times 6 + 3

, so the remainder is

3

.

A remainder modulo

12

must be an integer in the range from

0

to

11

.

Key Concept

Properties of Remainders and Modular Substitution

Estimated Time:1m 15s

Question 84Question →

If $a$ and $b$ are positive rational numbers such that $a < b < 1$ , which of the following expressions must be strictly greater than $1$ ? Select all that apply.

Select all that apply

\frac{b}{a}

\frac{a+1}{b}

\frac{a}{b^2}

\frac{b+1}{a+1}

\frac{a+b}{2b}

Show answer & explanation

Answer: $\frac{b}{a}$ ; $\frac{a+1}{b}$ ; $\frac{b+1}{a+1}$

Answer

The expressions that must be strictly greater than 1 are the ratio of b to a, the fraction (a + 1) over b, and the fraction (b + 1) over (a + 1).

The correct options are those involving ratios where the numerator is demonstrably greater than the denominator. The ratio of b to a compares two positive numbers where b is larger, giving a quotient greater than 1. The fraction with numerator (a + 1) has a value over 1 while denominator b is under 1, ensuring the quotient exceeds 1. The fraction comparing (b + 1) to (a + 1) has a larger numerator because adding 1 preserves the inequality b > a.

Step-by-Step Solution

1

Analyze the ratio of b to a

Since

b > a > 0

,

\frac{b}{a} > 1

is always true.

Dividing any positive number by a smaller positive number results in a value greater than 1.

2

Analyze the fraction (a + 1) over b

Since

a > 0

,

a + 1 > 1

. Since

b < 1

,

a + 1 > b

, which implies

\frac{a+1}{b} > 1

.

The numerator is greater than 1 and the denominator is less than 1, so the fraction exceeds 1.

3

Test counterexamples for the fraction a over b squared

For

a = \frac{1}{8}

and

b = \frac{1}{2}

,

\frac{a}{b^2} = \frac{1/8}{1/4} = \frac{1}{2} \le 1

.

The expression is not guaranteed to be greater than 1 for all valid values of a and b.

4

Analyze the fraction (b + 1) over (a + 1)

Since

b > a

, adding 1 to both quantities yields

b + 1 > a + 1 > 0

, so

\frac{b+1}{a+1} > 1

.

Adding the same positive quantity to two numbers maintains their relative order.

5

Analyze the average fraction (a + b) over 2b

Since

a < b

,

a + b < 2b

, meaning

\frac{a+b}{2b} < 1

.

The numerator is the sum of a and b, which is strictly less than twice b.

Key Concept

Properties of Rational Number Inequalities and Fraction Comparison

Estimated Time:1m 30s

Question 85Question →

Two automated timers are set to chime at regular intervals. The first timer chimes every $12$ minutes, and the second timer chimes every $18$ minutes. If both timers chime simultaneously at 12:00 PM, how many minutes will pass before they next chime at the exact same time?

Show answer & explanation

Answer: 36

Answer

36 minutes

The time interval between simultaneous chimes is given by the least common multiple (LCM) of the two individual intervals,

12

minutes and

18

minutes. Using prime factorization,

12 = 2^2 \times 3

and

18 = 2 \times 3^2

. The LCM takes the highest power of each prime present:

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

. Thus,

36

minutes will elapse before the timers chime together again.

Step-by-Step Solution

1

Find the prime factorizations of 12 and 18

12 = 2^2 \times 3

and

18 = 2 \times 3^2

Expressing numbers as products of prime factors allows us to easily compute the least common multiple.

2

Calculate the Least Common Multiple (LCM)

\text{LCM}(12, 18) = 2^{\max(2,1)} \times 3^{\max(1,2)} = 2^2 \times 3^2 = 36

The next simultaneous occurrence of periodic events is given by the LCM of their periods.

Key Concept

Least Common Multiple (LCM) of periodic events

Question 86Question →

Let $n$ be a positive integer such that $n$ is divisible by $12$ and $n^2$ is divisible by $270$ . Which of the following statements MUST be true? Select all such statements.

Select all that apply

n

is divisible by

180

.

n^2

is divisible by

10,800

.

n

is divisible by

24

.

n^3

is divisible by

250,000

.

n

has at least

18

positive divisors.

Show answer & explanation

Answer: $n$ is divisible by $180$ .; $n^2$ is divisible by $10,800$ .; $n$ has at least $18$ positive divisors.

Answer

The statements asserting that

n

is divisible by

180

,

n^2

is divisible by

10,800

, and

n

has at least

18

positive divisors MUST be true.

Because

12 = 2^2 \cdot 3^1

divides

n

,

n

contains at least two factors of

2

and one factor of

3

. Because

270 = 2^1 \cdot 3^3 \cdot 5^1

divides

n^2

,

n^2

must contain at least three factors of

3

and one factor of

5

, meaning

n

itself must contain at least two factors of

3

and one factor of

5

. Therefore,

n

must be a multiple of

2^2 \cdot 3^2 \cdot 5^1 = 180

. Consequently: (1)

n

is divisible by

180

; (2)

n^2

is a multiple of

180^2 = 32,400

, which is divisible by

10,800

; (3)

n

has at least as many divisors as

180

, which has

(2+1)(2+1)(1+1) = 18

positive divisors.

Step-by-Step Solution

1

Determine the prime factorization constraints on

n

from the given divisibility conditions.

Since

12 = 2^2 \cdot 3^1

divides

n

,

v_2(n) \ge 2

and

v_3(n) \ge 1

. Since

270 = 2^1 \cdot 3^3 \cdot 5^1

divides

n^2

,

v_2(n^2) \ge 1

,

v_3(n^2) \ge 3

, and

v_5(n^2) \ge 1

.

Divisibility requires that the exponent of each prime factor in the divisor is less than or equal to the exponent in the dividend.

2

Deduce the minimal exponents of primes in

n

.

v_2(n) \ge 2

,

2 \cdot v_3(n) \ge 3 \implies v_3(n) \ge 2

, and

2 \cdot v_5(n) \ge 1 \implies v_5(n) \ge 1

. Thus,

n

must be a multiple of

2^2 \cdot 3^2 \cdot 5^1 = 180

.

Exponents of prime factors in integer factorizations must be non-negative integers.

3

Evaluate each statement against the minimum required value

n = 180 = 2^2 \cdot 3^2 \cdot 5^1

.

1.

180 \mid n

is true.
2.

n^2

is a multiple of

180^2 = 32,400

, which is divisible by

10,800

(

32,400 = 3 \cdot 10,800

). True.
3.

180

is not divisible by

24

. False.
4.

n^3

has

5^3

, whereas

250,000

requires

5^5

. False.
5. The number of positive divisors of

180

is

(2+1)(2+1)(1+1) = 18

. Any multiple of

180

has at least

18

positive divisors. True.

Testing the baseline integer

n = 180

identifies necessary conditions and counterexamples.

Key Concept

Properties of Integers and Divisibility

Question 87Question →

A community library received a donation of $120$ new books. On Monday, $\frac{3}{8}$ of the donated books were cataloged. On Tuesday, $\frac{1}{3}$ of the remaining uncataloged books were cataloged. How many of the donated books remain to be cataloged?

Show answer & explanation

Answer: 50

Answer

50 books remain to be cataloged.

First, find the number of books cataloged on Monday: 3/8 of 120 is 45 books. Subtracting 45 from 120 leaves 75 books uncataloged. On Tuesday, 1/3 of those remaining 75 books were cataloged, which is 25 books. Subtracting 25 from 75 gives 50 books remaining to be cataloged.

Step-by-Step Solution

1

Calculate the number of books cataloged on Monday.

45 books

Multiply the initial total of 120 books by the fraction cataloged on Monday, which is 3/8.

2

Calculate the uncataloged books remaining after Monday.

75 books

Subtract the 45 cataloged books from the initial 120 books.

3

Calculate the number of books cataloged on Tuesday.

25 books

Multiply the remaining 75 books by the fraction cataloged on Tuesday, which is 1/3.

4

Calculate the final number of uncataloged books.

50 books

Subtract the 25 books cataloged on Tuesday from the 75 uncataloged books remaining after Monday.

Key Concept

Fraction of a Remaining Quantity

Estimated Time:50s

Question 88Question →

Two positive integers $x$ and $y$ have a greatest common divisor (GCD) of $6$ and a least common multiple (LCM) of $108$ . If $x = 18$ , what is the value of $y$ ?

18

36

54

108

648

Show answer & explanation

Answer: $36$

Answer

The value of

y

is

36

.

For any two positive integers

x

and

y

, their product equals the product of their greatest common divisor and least common multiple:

x \cdot y = \text{GCD}(x,y) \cdot \text{LCM}(x,y)

. Substituting

x = 18

,

\text{GCD} = 6

, and

\text{LCM} = 108

gives

18y = 6 \cdot 108 = 648

. Dividing both sides by

18

yields

y = 36

.

Step-by-Step Solution

1

State the fundamental identity relating GCD, LCM, and two positive integers.

x \times y = \text{GCD}(x,y) \times \text{LCM}(x,y)

For any two positive integers, the product of the numbers is equal to the product of their greatest common divisor and least common multiple.

2

Substitute the known values

x = 18

,

\text{GCD} = 6

, and

\text{LCM} = 108

into the equation.

18 \times y = 6 \times 108

Replace the abstract variables with the specific values provided in the question stem.

3

Multiply the numbers on the right side of the equation.

18 \times y = 648

Simplify the product of the GCD and LCM.

4

Divide both sides by

18

to isolate

y

.

y = \frac{648}{18} = 36

Solve for the unknown integer

y

.

Key Concept

Relationship between GCD, LCM, and the product of two positive integers

Estimated Time:1m 0s

Question 89Question →

A university research lab receives a multi-year grant. In the first quarter, the lab spends $\frac{2}{7}$ of the total grant on laboratory equipment. In the second quarter, the lab spends $\frac{3}{5}$ of the remaining funds on research staff salaries. At the beginning of the third quarter, the lab receives a supplementary addition equal to $\frac{1}{4}$ of the funds remaining at the end of the second quarter. If the total amount of money in the grant immediately following the third-quarter addition is $\$45,000$ , what was the initial dollar amount of the research grant?

$84,000

$126,000

$157,500

$210,000

$315,000

Show answer & explanation

Answer: $126,000

Answer

$126,000

The correct answer of

126,000 is found by tracking the fraction of the initial grant

G

remaining after each step. Spending

\frac{2}{7}

leaves

\frac{5}{7}G

. Spending

\frac{3}{5}

of that remainder leaves

\frac{2}{5} \times \frac{5}{7}G = \frac{2}{7}G

. Adding

\frac{1}{4}

of

\frac{2}{7}G

increases the balance by

\frac{1}{14}G

, resulting in a final total of

\frac{2}{7}G + \frac{1}{14}G = \frac{5}{14}G

. Setting

\frac{5}{14}G = 45,000

yields

G = 126,000$.

Step-by-Step Solution

1

Determine the remaining fraction after the first quarter.

Fraction remaining =

1 - \frac{2}{7} = \frac{5}{7}

of the initial grant

G

.

The lab spent

\frac{2}{7}

of the total grant.

2

Calculate the remaining fraction after the second quarter.

Fraction remaining =

\left(1 - \frac{3}{5}\right) \times \frac{5}{7}G = \frac{2}{5} \times \frac{5}{7}G = \frac{2}{7}G

.

The lab spent

\frac{3}{5}

of the remaining funds, leaving

\frac{2}{5}

of those funds.

3

Account for the third-quarter supplementary addition.

Final fraction =

\frac{2}{7}G + \frac{1}{4}\left(\frac{2}{7}G\right) = \frac{2}{7}G + \frac{1}{14}G = \frac{4}{14}G + \frac{1}{14}G = \frac{5}{14}G

.

An additional

\frac{1}{4}

of the second-quarter remaining funds was added to the grant.

4

Solve for the initial grant amount

G

.

\frac{5}{14}G = 45,000 \implies G = 45,000 \times \frac{14}{5} = 9,000 \times 14 = 126,000

.

The final remaining balance of

\$45,000

corresponds to

\frac{5}{14}

of the original grant.

Key Concept

Multi-step sequential fraction operations and base tracking in word problems

Question 90Question →

Let $n$ be a positive integer such that $n$ has exactly 15 positive divisors. If $n$ is divisible by 18, but $n$ is NOT divisible by 8, what is the remainder when $n$ is divided by 7?

1

2

3

4

5

Show answer & explanation

Answer: 2

Answer

The remainder when

n

is divided by 7 is 2.

The total number of divisors of a positive integer with prime factorization

p_1^{a} p_2^{b} \dots

is

(a+1)(b+1)\dots

. Given that

n

has 15 divisors, the possible exponent forms are 14 (since

14+1=15

) or 4 and 2 (since

(4+1)(2+1)=15

). Since

n

is divisible by

18 = 2 \cdot 3^2

, its prime factors must be 2 and 3. The form

p^{14}

is eliminated because it contains only one prime factor. Testing the two permutations for

p^4 \cdot q^2

:
1. If

n = 2^4 \cdot 3^2 = 144

,

n

is divisible by 8 (

144 / 8 = 18

), which contradicts the condition that

n

is NOT divisible by 8.
2. If

n = 2^2 \cdot 3^4 = 324

,

n

is divisible by 18 (

324 / 18 = 18

) and is NOT divisible by 8 (

324 / 8 = 40.5

).
Dividing 324 by 7 gives

324 = 7 \times 46 + 2

, so the remainder is 2.

Step-by-Step Solution

1

Analyze the divisor count formula for

n

The prime factorization of

n

must be either

p^{14}

or

p^4 \cdot q^2

for distinct prime numbers

p

and

q

.

The total number of positive divisors of an integer with prime factorization

p_1^{a} p_2^{b} \dots

is given by

(a+1)(b+1)\dots = 15

. Since 15 factors as

15 \times 1

or

5 \times 3

, the exponent structures are 14 or 4 and 2.

2

Apply the divisibility conditions by 18 and 8

n

must equal

2^2 \cdot 3^4 = 324

.

Because

n

is divisible by

18 = 2^1 \cdot 3^2

, its prime factors must include both 2 and 3, ruling out

p^{14}

. Thus

\{p, q\} = \{2, 3\}

. If

n = 2^4 \cdot 3^2 = 144

, then

n

is divisible by 8 (

2^3

), violating the given constraint. Therefore,

n

must be

2^2 \cdot 3^4 = 324

, which is divisible by 18 (

324 / 18 = 18

) and not divisible by 8 (

324 / 8 = 40.5

).

3

Compute the remainder when 324 is divided by 7

324 divided by 7 yields a quotient of 46 and a remainder of 2.

324 = 7 \times 46 + 2

because

7 \times 46 = 322

.

Key Concept

Prime factorization, divisor count formula, and divisibility constraints

Question 91Question →

An industrial chemical reservoir is emptied by three pumps, $P_1$ , $P_2$ , and $P_3$ , operating independently at different constant rates. Working alone, $P_1$ can empty the full reservoir in $6$ hours, $P_2$ can empty it in $8$ hours, and $P_3$ can empty it in $12$ hours. Initially, the reservoir is completely full. First, $P_1$ and $P_2$ work together for $2$ hours. Then $P_1$ is turned off, and $P_3$ is turned on to work alongside $P_2$ . Additionally, while $P_2$ and $P_3$ are working together, a defect causes liquid to leak out of the bottom of the reservoir at a constant rate equal to $\frac{1}{6}$ of the combined emptying rate of $P_2$ and $P_3$ . How many additional hours will it take to completely empty the remaining liquid from the reservoir?

\frac{12}{7}

hours

2

hours

\frac{10}{9}

hours

\frac{12}{5}

hours

\frac{144}{49}

hours

Show answer & explanation

Answer: $\frac{12}{7}$ hours

Answer

\frac{12}{7}

hours

The option stating '

\frac{12}{7}

hours' is correct because during the first 2 hours,

P_1

and

P_2

empty

\frac{7}{12}

of the reservoir, leaving

\frac{5}{12}

remaining. Then,

P_2

and

P_3

have a combined rate of

\frac{5}{24}

, and the defect adds an extra

\frac{5}{144}

per hour, giving a total emptying rate of

\frac{35}{144}

reservoirs per hour. Dividing

\frac{5}{12}

by

\frac{35}{144}

yields

\frac{12}{7}

hours.

Step-by-Step Solution

1

Calculate the combined emptying rate of

P_1

and

P_2

and determine the fraction emptied in the first

2

hours.

Rate of

P_1 = \frac{1}{6}

reservoir/hr, Rate of

P_2 = \frac{1}{8}

reservoir/hr. Combined rate = \frac{1}{6} + \frac{1}{8} = \frac{7}{24} reservoir/hr. In

2

hours, amount emptied =

2 \times \frac{7}{24} = \frac{7}{12}

of the reservoir.

Determines how much work was completed during the initial stage.

2

Find the remaining fraction of liquid in the reservoir.

Remaining fraction =

1 - \frac{7}{12} = \frac{5}{12}

of the reservoir.

Establishes the remaining volume that must be emptied in the second stage.

3

Calculate the combined rate of

P_2

,

P_3

, and the defect rate.

Combined rate of

P_2

and

P_3 = \frac{1}{8} + \frac{1}{12} = \frac{5}{24}

reservoir/hr. Defect rate = \frac{1}{6} \times \frac{5}{24} = \frac{5}{144} reservoir/hr. Total emptying rate = \frac{5}{24} + \frac{5}{144} = \frac{30 + 5}{144} = \frac{35}{144} reservoir/hr.

Combines all simultaneous emptying processes during the second stage.

4

Divide the remaining fraction by the total emptying rate to find the additional time needed.

Time = \frac{5/12}{35/144} = \frac{5}{12} \times \frac{144}{35} = \frac{5}{35} \times \frac{144}{12} = \frac{1}{7} \times 12 = \frac{12}{7} hours.

Applies the rate formula

\text{Time} = \frac{\text{Work}}{\text{Rate}}

.

Key Concept

Compound Fraction Work Rates and Rational Operations

Question 92Question →

Let $x$ be a positive integer. If the expression $x^2 + 3x$ is divisible by $6$ , which of the following statements MUST be true?

x

is divisible by

3

x

is divisible by

6

x

is an even integer

x

is an odd integer

x + 1

is a prime number

Show answer & explanation

Answer: $x$ is divisible by $3$

Answer

x

is divisible by

3

Factoring

x^2 + 3x

yields

x(x+3)

. For any integer

x

, one of

x

or

x+3

is even and the other is odd, so their product

x(x+3)

is always divisible by

2

. For the product to also be divisible by

3

, either

x

or

x+3

must be divisible by

3

. Because

x

and

x+3

differ by

3

,

x+3

is a multiple of

3

if and only if

x

is a multiple of

3

. Thus,

x

must be divisible by

3

.

Step-by-Step Solution

1

Factor the given algebraic expression

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

Factoring allows us to analyze the parity and prime factors of each component separately.

2

Analyze parity (divisibility by 2)

If

x

is even,

x+3

is odd

\Rightarrow

product is even. If

x

is odd,

x+3

is even

\Rightarrow

product is even.

The expression

x(x+3)

is guaranteed to be divisible by

2

for all integers

x

.

3

Analyze divisibility by 3

For

x(x+3)

to be divisible by

3

, at least one factor must be divisible by

3

. Since

x+3 \equiv x \pmod 3

,

x+3

is divisible by

3

if and only if

x

is divisible by

3

.

Since divisibility by

2

is always satisfied, divisibility by

6

requires

x(x+3)

to be divisible by

3

, which requires

x

to be a multiple of

3

.

Key Concept

Divisibility properties of factored integer expressions and modular arithmetic

Estimated Time:1m 15s

Question 93Question →

Let $n$ be a positive integer of the form $n = 2^a \cdot 3^b$ , where $a$ and $b$ are positive integers. If $n$ is divisible by $6$ and $n^2$ has exactly $35$ positive divisors, how many positive divisors does $6n$ have?

12

18

20

24

70

Show answer & explanation

Answer: 20

Answer

The integer

6n

has 20 positive divisors.

The correct answer is 20. Since

n = 2^a \cdot 3^b

, we have

n^2 = 2^{2a} \cdot 3^{2b}

. The number of divisors of

n^2

is

(2a+1)(2b+1) = 35

. Given

n

is divisible by 6,

a, b \ge 1

, which means

2a+1 \ge 3

and

2b+1 \ge 3

. The factorizations of 35 are

5 \times 7

and

7 \times 5

, giving exponent values of 2 and 3. Then

6n = 2^{a+1} \cdot 3^{b+1}

, which has

(a+2)(b+2) = (2+2)(3+2) = 20

positive divisors.

Step-by-Step Solution

1

Express

n^2

in terms of its prime factorization.

n^2 = (2^a \cdot 3^b)^2 = 2^{2a} \cdot 3^{2b}

Squaring an integer doubles the exponents of its prime factors.

2

Set up the equation for the number of positive divisors of

n^2

.

(2a + 1)(2b + 1) = 35

The number of positive divisors of

p_1^{e_1} p_2^{e_2}

is given by

(e_1 + 1)(e_2 + 1)

.

3

Solve for the exponents

a

and

b

given that

n

is divisible by 6.

Since

n

is divisible by 6, both

a \ge 1

and

b \ge 1

. Thus

2a + 1 > 1

and

2b + 1 > 1

. The only factor pairs of 35 into two integers greater than 1 are

5 \times 7

or

7 \times 5

. Therefore,

\{2a+1, 2b+1\} = \{5, 7\}

, which yields

\{a, b\} = \{2, 3\}

.

Divisibility by 6 guarantees both prime factors 2 and 3 are present in

n

.

4

Find the prime factorization of

6n

and calculate its number of positive divisors.

6n = (2^1 \cdot 3^1) \cdot (2^a \cdot 3^b) = 2^{a+1} \cdot 3^{b+1}

. Using

\{a, b\} = \{2, 3\}

, the exponents are

a+1

and

b+1

, which equal 3 and 4 (in either order). The number of divisors is

(3+1)(4+1) = 4 \times 5 = 20

.

Adding 1 to each exponent in

n

gives the exponents for

6n

, leading to

(a+2)(b+2) = (2+2)(3+2) = 20

.

Key Concept

Divisor Count Formula and Prime Factorization Constraints

Estimated Time:2m 0s

Question 94Question →

Which of the following statements about the positive integer $180$ are true? Select all that apply.

Select all that apply

The number of distinct prime factors of

180

is

3

.

The greatest common divisor (GCD) of

180

and

75

is

15

.

The number

1

is the smallest prime factor of

180

.

The least common multiple (LCM) of

180

and

30

is

5400

.

The prime factorization of

180^2

can be written as

(2 + 3 + 5)^4

.

Show answer & explanation

Answer: The number of distinct prime factors of $180$ is $3$ .; The greatest common divisor (GCD) of $180$ and $75$ is $15$ .

Answer

The true statements are that

180

has

3

distinct prime factors and that the greatest common divisor of

180

and

75

is

15

.

The prime factorization of

180

is

2^2 \times 3^2 \times 5^1

, showing exactly

3

distinct prime factors (

2, 3, 5

). Additionally, comparing this with

75 = 3^1 \times 5^2

confirms that their greatest common divisor is

3^1 \times 5^1 = 15

.

Step-by-Step Solution

1

Find the prime factorization of

180

.

180 = 2^2 \times 3^2 \times 5^1

.

Decomposing

180

into prime factors provides the foundation to evaluate statements regarding prime factors, GCD, and LCM.

2

Count the distinct prime factors of

180

.

The distinct prime bases are

2

,

3

, and

5

, giving

3

distinct prime factors.

Exponents represent multiplicity, so counting distinct bases yields the number of distinct prime factors.

3

Calculate the GCD of

180

and

75

.

Since

75 = 3^1 \times 5^2

,

\text{GCD}(180, 75) = 3^{\min(2,1)} \times 5^{\min(1,2)} = 3^1 \times 5^1 = 15

.

The GCD is found by taking the minimum exponent for each shared prime factor.

4

Verify definition of prime numbers and divisibility rules for remaining choices.

1

is not prime (smallest prime factor is

2

), and

\text{LCM}(180, 30) = 180

because

30

is a factor of

180

.

Eliminate false statements based on arithmetic definitions and divisibility properties.

Key Concept

Prime Factorization, Greatest Common Divisor (GCD), and Least Common Multiple (LCM)

Question 95Question →

Let $a$ and $b$ be rational numbers such that $-1 < a < 0 < b < 1$ . Which of the following expressions MUST be strictly greater than $1$ ? Select all such expressions.

Select all that apply

\frac{b - a}{b}

\frac{1 + b}{1 + a}

\frac{a + b}{b}

\frac{b - a}{1 - a}

\frac{1}{b - a}

Show answer & explanation

Answer: $\frac{b - a}{b}$ ; $\frac{1 + b}{1 + a}$

Answer

The expressions

\frac{b - a}{b}

and

\frac{1 + b}{1 + a}

MUST be strictly greater than 1.

The expression representing the difference

(b - a)

divided by

b

simplifies to

1 + \frac{-a}{b}

. Because

-a

and

b

are both positive, this term is strictly greater than 1. Similarly, the expression

\frac{1 + b}{1 + a}

divides a numerator greater than 1 by a positive denominator less than 1, which always yields a value strictly greater than 1.

Step-by-Step Solution

1

Analyze the expression

\frac{b - a}{b}

by splitting the fraction.

\frac{b - a}{b} = \frac{b}{b} - \frac{a}{b} = 1 - \frac{a}{b}

.

Since

a < 0

and

b > 0

, the quotient

\frac{a}{b}

is strictly negative. Subtracting a negative number from 1 is equivalent to adding a positive number, so

1 - \frac{a}{b} > 1

.

2

Analyze the expression

\frac{1 + b}{1 + a}

by establishing bounds for the numerator and denominator.

Since

b > 0

,

1 + b > 1

. Since

-1 < a < 0

, adding 1 yields

0 < 1 + a < 1

.

Dividing any real number greater than 1 by a positive real number less than 1 produces a value strictly greater than 1.

3

Evaluate the remaining options with counterexamples or algebraic bounds to verify they are not guaranteed to be greater than 1.

The expression

\frac{a + b}{b} = 1 + \frac{a}{b} < 1

; the expression

\frac{b - a}{1 - a} < 1

because

b < 1 \implies b - a < 1 - a

; and

\frac{1}{b - a}

can be less than 1 when

b - a > 1

(e.g.,

b = 0.8, a = -0.5 \implies b - a = 1.3

).

Demonstrating that an expression is less than 1 or can be less than 1 eliminates it from being strictly greater than 1 for all valid rational numbers

a

and

b

.

Key Concept

Properties and Inequalities of Rational Numbers

Question 96Question →

A jacket originally priced at $\$80$ is on sale for $20\%$ off the original price. What is the sale price of the jacket?

\$64

\$16

\$96

\$60

\$72

Show answer & explanation

Answer: $\$64$

Answer

\$64

To find the sale price after a

20\%

discount, calculate

20\%

of the original price (

0.20 \times 80 = 16

) and subtract it from the original price (

80 - 16 = 64

). Alternatively, multiply the original price by

(1 - 0.20) = 0.80

, giving

80 \times 0.80 = 64

.

Step-by-Step Solution

1

Calculate the dollar value of the discount

\text{Discount} = 0.20 \times \$80 = \$16

A

20\%

discount means reducing the original price by

20

hundredths of its total value.

2

Subtract the discount from the original price

\text{Sale Price} = \$80 - \$16 = \$64

The final sale price equals the original price minus the discount amount.

Key Concept

Percentage Discount and Net Price

Estimated Time:45s

Question 97Question →

Given $x = 3.6 \times 10^{-3}$ and $y = 1.2 \times 10^{-4}$ , which of the following statements are true? Select all that apply.

Select all that apply

x

is greater than

y

.

The ratio

\frac{x}{y}

is equal to

30

.

The sum

x + y

is equal to

4.8 \times 10^{-7}

.

The product

x \times y

is equal to

4.32 \times 10^{12}

.

The value of

y

expressed as a standard decimal is

0.0012

.

Show answer & explanation

Answer: $x$ is greater than $y$ .; The ratio $\frac{x}{y}$ is equal to $30$ .

Answer

The statement that

x

is greater than

y

and the statement that the ratio

\frac{x}{y}

is equal to

30

are both correct.

The statement asserting that

x

is greater than

y

is true because

3.6 \times 10^{-3} = 0.0036

is larger than

1.2 \times 10^{-4} = 0.00012

. The statement regarding the ratio

\frac{x}{y}

is true because dividing coefficients

\frac{3.6}{1.2} = 3

and subtracting exponents

-3 - (-4) = 1

yields

3 \times 10^1 = 30

.

Step-by-Step Solution

1

Convert both numbers to standard decimal notation to compare them.

x = 0.0036

and

y = 0.00012

.

Converting to standard form makes place value comparison direct.

2

Compare

x

and

y

.

0.0036 > 0.00012

, so

x > y

.

The thousandths digit of

x

(

3

) is greater than the thousandths digit of

y

(

0

).

3

Calculate the ratio

\frac{x}{y}

using properties of exponents.

\frac{3.6 \times 10^{-3}}{1.2 \times 10^{-4}} = \left(\frac{3.6}{1.2}\right) \times 10^{-3 - (-4)} = 3 \times 10^1 = 30

.

Divide the coefficients and subtract the exponents for scientific notation division.

Key Concept

Decimals and Scientific Notation Operations

Question 98Question →

An investment fund allocates its total portfolio assets among three asset classes: stocks, bonds, and real estate. Initially, $\frac{5}{12}$ of the total portfolio is invested in stocks and $\frac{1}{3}$ of the total portfolio is invested in bonds, with the remaining fraction invested in real estate. If $\frac{2}{5}$ of the stock portfolio and $\frac{1}{4}$ of the bond portfolio are subsequently liquidated, what fraction of the fund's remaining total portfolio is invested in real estate?

\frac{1}{4}

\frac{3}{10}

\frac{1}{3}

\frac{4}{9}

\frac{5}{12}

Show answer & explanation

Answer: $\frac{1}{3}$

Answer

The fraction of the fund's remaining total portfolio invested in real estate is

\frac{1}{3}

.

Let the original total portfolio be

T

. Initially, stocks are

\frac{5}{12}T

, bonds are

\frac{4}{12}T

, and real estate is

1 - \frac{9}{12}T = \frac{3}{12}T

. After liquidating

\frac{2}{5}

of stocks and

\frac{1}{4}

of bonds, the remaining stock value is

\frac{3}{5} \times \frac{5}{12}T = \frac{3}{12}T

and the remaining bond value is

\frac{3}{4} \times \frac{4}{12}T = \frac{3}{12}T

. Real estate remains

\frac{3}{12}T

. The new total portfolio is

\frac{3}{12}T + \frac{3}{12}T + \frac{3}{12}T = \frac{9}{12}T = \frac{3}{4}T

. Therefore, real estate represents

\frac{3/12 T}{9/12 T} = \frac{3}{9} = \frac{1}{3}

of the remaining total portfolio.

Step-by-Step Solution

1

Determine the initial fraction invested in real estate.

Real estate accounts for

1 - \left(\frac{5}{12} + \frac{1}{3}\right) = 1 - \frac{9}{12} = \frac{3}{12} = \frac{1}{4}

of the original total portfolio.

The sum of all three asset fractions must equal 1.

2

Calculate the remaining fractions for stocks and bonds after liquidation.

Remaining stocks =

\left(1 - \frac{2}{5}\right) \times \frac{5}{12} = \frac{3}{5} \times \frac{5}{12} = \frac{3}{12}

. Remaining bonds =

\left(1 - \frac{1}{4}\right) \times \frac{1}{3} = \frac{3}{4} \times \frac{4}{12} = \frac{3}{12}

.

Multiplying the unliquidated fraction by each asset's initial fraction gives its remaining fraction relative to the original total.

3

Calculate the total remaining portfolio fraction.

Total remaining portfolio =

\frac{3}{12} \text{ (stocks)} + \frac{3}{12} \text{ (bonds)} + \frac{3}{12} \text{ (real estate)} = \frac{9}{12} = \frac{3}{4}

of the original total portfolio.

Sum the remaining portions of all three asset categories.

4

Compute the ratio of real estate to the remaining total portfolio.

\frac{\text{Real Estate}}{\text{Remaining Total}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.

The question asks for the fraction relative to the new reduced total portfolio.

Key Concept

Sequential Fraction Allocation and Reduced Base Ratios

Estimated Time:1m 30s

Question 99Question →

Let

a

,

b

, and

c

be positive integers, and define three rational numbers

x

,

y

, and

z

as follows:

x = \frac{a}{a+b}, \quad y = \frac{b}{b+c}, \quad \text{and} \quad z = \frac{c}{c+a}

Which of the following statements MUST be true for all such positive integers

a

,

b

, and

c

? Select all that apply.

Select all that apply

The sum

x + y + z

satisfies the inequality

1 < x + y + z < 2

.

The product

(1 - x)(1 - y)(1 - z)

is identically equal to

xyz

.

The product

xyz

cannot exceed

\frac{1}{8}

.

The sum

x + y + z

can be equal to an integer.

If

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

, then it must be that

x = y = z = \frac{1}{2}

.

Show answer & explanation

Answer: The sum $x + y + z$ satisfies the inequality $1 < x + y + z < 2$ .; The product $(1 - x)(1 - y)(1 - z)$ is identically equal to $xyz$ .; The product $xyz$ cannot exceed $\frac{1}{8}$ .

Answer

The statements asserting that

1 < x + y + z < 2

, that

(1 - x)(1 - y)(1 - z) = xyz

, and that

xyz \le \frac{1}{8}

are all true.

The statement bounding the sum between 1 and 2 is correct because adjusting denominators to a common sum

a+b+c

reveals that the sum is strictly greater than 1 and strictly less than 2. The identity statement is correct because subtracting each fraction from 1 shifts the numerators cyclically without changing the overall product. The statement placing an upper bound of

\frac{1}{8}

on the product is correct by applying the AM-GM inequality to each denominator term.

Step-by-Step Solution

1

Establish strict bounds on the sum

x + y + z

.

Since

a, b, c > 0

, we have

\frac{a}{a+b} > \frac{a}{a+b+c}

. Summing all three terms gives

x + y + z > \frac{a+b+c}{a+b+c} = 1

. To find the upper bound, rewrite

\frac{a}{a+b} < \frac{a+c}{a+b+c}

. Summing all three terms gives

x + y + z < \frac{(a+c)+(a+b)+(b+c)}{a+b+c} = 2

. Thus,

1 < x + y + z < 2

.

This proves that the sum of the three fractions must always lie strictly between 1 and 2.

2

Compute the product

(1-x)(1-y)(1-z)

algebraically.

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

,

1 - y = \frac{c}{b+c}

, and

1 - z = \frac{a}{c+a}

. The product is

\frac{b \cdot c \cdot a}{(a+b)(b+c)(c+a)} = \frac{abc}{(a+b)(b+c)(c+a)} = xyz

.

This proves the structural algebraic identity between

(1-x)(1-y)(1-z)

and

xyz

.

3

Apply the AM-GM inequality to find the upper bound of

xyz

.

Since

a+b \ge 2\sqrt{ab}

,

b+c \ge 2\sqrt{bc}

, and

c+a \ge 2\sqrt{ca}

, their product satisfies

(a+b)(b+c)(c+a) \ge 8\sqrt{a^2b^2c^2} = 8abc

. Taking the reciprocal gives

xyz = \frac{abc}{(a+b)(b+c)(c+a)} \le \frac{abc}{8abc} = \frac{1}{8}

.

This confirms that

xyz

has a maximum possible value of

\frac{1}{8}

(achieved when

a=b=c

).

4

Test counterexamples for the remaining statements.

Since

1 < x + y + z < 2

, no integer value is possible, ruling out integer sums. Additionally, substituting

a=1, b=2, c=1

gives

x=\frac{1}{3}, y=\frac{2}{3}, z=\frac{1}{2}

, whose sum is

\frac{3}{2}

despite

x, y, z

not being equal to

\frac{1}{2}

.

This disproves the statements claiming the sum can be an integer or that a sum of

\frac{3}{2}

requires all three fractions to be equal.

Key Concept

Bounding and algebraic properties of cyclic rational fractions

Estimated Time:3m 0s

Question 100Question →

What is the sum of all distinct prime factors of $420$ ?

15

17

18

19

21

Show answer & explanation

Answer: $17$

Answer

The sum of all distinct prime factors of

420

is

17

.

The prime factorization of

420

is

2^2 \times 3 \times 5 \times 7

. The distinct prime numbers in this factorization are

2

,

3

,

5

, and

7

. Adding these values together yields

2 + 3 + 5 + 7 = 17

.

Step-by-Step Solution

1

Find the prime factorization of

420

.

420 = 2^2 \times 3 \times 5 \times 7

Decompose

420

into prime components:

420 = 10 \times 42 = (2 \times 5) \times (2 \times 3 \times 7) = 2^2 \times 3 \times 5 \times 7

.

2

Identify the set of distinct prime factors.

The distinct prime factors are

2, 3, 5,

and

7

.

Exponents are ignored when selecting unique prime factors.

3

Calculate the sum of the distinct prime factors.

2 + 3 + 5 + 7 = 17

Add the unique prime factors together.

Key Concept

Prime Factorization

Estimated Time:45s

Arithmetic

Answer

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