Arithmetic Practice Questions — GRE General Test — Page 4

Question 61Question →

An artisan bakery produces a signature flour blend by combining wheat, rye, and oat flour. Initially, the ratio of wheat flour to rye flour by weight is $5:3$ , and the ratio of rye flour to oat flour by weight is $4:1$ . The baker prepares an initial batch of this mixture weighing exactly $140$ pounds. To adjust the recipe for a special order, pure oat flour is added to the batch until oat flour accounts for exactly $20\%$ of the total weight of the new mixture. How many pounds of pure oat flour must the baker add?

12

16

20

28

32

Show answer & explanation

Answer: 20

Answer

20 pounds

To solve this problem, first express the ratios of wheat, rye, and oat flour with a common term for rye. Since Wheat : Rye = 5 : 3 (or 20 : 12) and Rye : Oat = 4 : 1 (or 12 : 3), the combined ratio Wheat : Rye : Oat is 20 : 12 : 3, giving 35 total parts. In a 140-pound batch, each part equals 4 pounds, meaning the batch contains 12 pounds of oat flour and 128 pounds of non-oat flour. When additional oat flour is added, the non-oat weight stays fixed at 128 pounds. For oat flour to be 20% of the new mixture, non-oat flour must be 80%. Dividing 128 by 0.80 gives a new total weight of 160 pounds. The new oat weight is 20% of 160, which is 32 pounds. Subtracting the initial 12 pounds yields 20 pounds of added oat flour.

Step-by-Step Solution

1

Combine the two given ratios into a single three-part ratio.

Wheat : Rye =

5 : 3 = 20 : 12

, and Rye : Oat =

4 : 1 = 12 : 3

. Thus, Wheat : Rye : Oat =

20 : 12 : 3

.

Scaling the ratios so that the common element (Rye) has equal parts allows unification into a single ratio.

2

Calculate the initial weight of each component in the 140-pound batch.

Total ratio parts =

20 + 12 + 3 = 35

parts. Each part is

140 / 35 = 4

pounds. Initial Oat flour =

3 \times 4 = 12

pounds. Non-oat flour (Wheat + Rye) =

(20 + 12) \times 4 = 128

pounds.

Determining the weight of the constant non-oat portion is key to solving mixture adjustment problems.

3

Set up an equation for the new mixture where oat flour makes up 20% of the total weight.

Since non-oat flour remains unchanged at

128

pounds, it represents

80\%

of the new total weight. New Total Weight =

128 / 0.80 = 160

pounds.

If oat flour is 20% of the new total, the non-oat components must constitute the remaining 80%.

4

Calculate the weight of oat flour added.

New Oat Weight =

20\%

of

160 = 32

pounds. Added Oat Flour =

32 - 12 = 20

pounds.

Subtracting the initial oat weight from the final required oat weight yields the amount added.

Key Concept

Combining multi-part ratios and solving ratio adjustment problems by holding unchanged components constant.

Question 62Question →

A financial forecasting model predicts monthly sales revenue using a sequence where the initial revenue in month 1 is $R_1 = \$80{}000$ , and the revenue for each subsequent month $n$ (for $n \ge 2$ ) is calculated using the recursive formula $R_n = 0.75 R_{n-1} + 5{}000$ . If each monthly revenue figure $R_n$ is rounded to the nearest hundred dollars before summing, what is the estimated total revenue, in dollars, for the first 4 months combined?

Show answer & explanation

Answer: 244100

Answer

The estimated total revenue for the first 4 months combined is 244,100 dollars.

By evaluating the recursive formula

R_n = 0.75 R_{n-1} + 5{}000

for months 1 through 4, we obtain

R_1 = 80{}000

,

R_2 = 65{}000

,

R_3 = 53{}750

, and

R_4 = 45{}312.50

. Rounding each to the nearest hundred yields

80{}000

,

65{}000

,

53{}800

, and

45{}300

, respectively. Summing these four rounded figures gives

244{}100

.

Step-by-Step Solution

1

Find the first term

R_1

and its rounded value.

R_1 = 80{}000

, which rounds to

80{}000

.

Given initial term.

2

Calculate the second term

R_2

using the recursive formula and round it.

R_2 = 0.75(80{}000) + 5{}000 = 65{}000

, which rounds to

65{}000

.

Apply

R_2 = 0.75 R_1 + 5{}000

.

3

Calculate the third term

R_3

using unrounded

R_2

and round to the nearest hundred.

R_3 = 0.75(65{}000) + 5{}000 = 53{}750

, which rounds to

53{}800

.

Apply

R_3 = 0.75 R_2 + 5{}000

and round to the nearest 100.

4

Calculate the fourth term

R_4

using unrounded

R_3

and round to the nearest hundred.

R_4 = 0.75(53{}750) + 5{}000 = 45{}312.50

, which rounds to

45{}300

.

Apply

R_4 = 0.75 R_3 + 5{}000

and round to the nearest 100.

5

Add the four rounded terms together.

80{}000 + 65{}000 + 53{}800 + 45{}300 = 244{}100

.

Compute total estimated revenue as specified.

Key Concept

Estimation, Rounding, and Sequences

Question 63Question →

In an industrial mechanical system, Gear $A$ has $16$ teeth, Gear $B$ has $24$ teeth, and Gear $C$ has $40$ teeth. Gear $A$ is meshed directly with Gear $B$ , and Gear $B$ is meshed directly with Gear $C$ . If Gear $A$ rotates at a constant speed of $150$ revolutions per minute ( $\text{rpm}$ ), what is the rotational speed, in revolutions per minute ( $\text{rpm}$ ), of Gear $C$ ?

Show answer & explanation

Answer: 60

Answer

The rotational speed of Gear C is 60 rpm.

For meshed gears, the linear speed of teeth at the point of contact must be identical. Thus, the product of the number of teeth and rotational speed remains constant (

N_A S_A = N_C S_C

). Substituting the known values gives

16 \times 150 = 40 \times S_C

, yielding

2,400 = 40 S_C

, so

S_C = 60

rpm.

Step-by-Step Solution

1

Determine the inverse proportional relationship between number of gear teeth and rotational speed.

The product of teeth count and rotational speed is constant across directly meshed gears:

N_A \times S_A = N_B \times S_B = N_C \times S_C

.

Directly meshed gears engage tooth for tooth, meaning they pass the same total number of teeth per unit time.

2

Calculate the total tooth displacement rate per minute from Gear A.

16 \times 150 = 2,400

teeth per minute.

Gear A has 16 teeth and completes 150 revolutions per minute.

3

Calculate the rotational speed of Gear C.

\text{Speed}_C = \frac{2,400}{40} = 60

rpm.

Gear C has 40 teeth, so dividing the total tooth displacement rate by 40 yields its revolutions per minute.

Key Concept

Inverse Proportionality in Gear Rates

Estimated Time:1m 15s

Question 64Question →

How many positive integer factors of $36$ are prime numbers?

2

3

4

9

1

Show answer & explanation

Answer: $2$

Answer

There are

2

positive integer factors of

36

that are prime numbers:

2

and

3

.

Listing all positive integer factors of

36

yields

1, 2, 3, 4, 6, 9, 12, 18,

and

36

. Evaluating each factor against the definition of a prime number (an integer greater than

1

divisible only by

1

and itself) shows that only

2

and

3

are prime. Therefore, there are exactly

2

prime factors.

Step-by-Step Solution

1

Find all positive integer factors of

36

.

The positive integer factors of

36

are

1, 2, 3, 4, 6, 9, 12, 18,

and

36

.

Dividing

36

by integers from

1

to

36

yields integer quotients for these nine numbers.

2

Identify which of these factors meet the definition of a prime number.

The prime factors are

2

and

3

.

A prime number is an integer greater than

1

with exactly two distinct positive divisors:

1

and itself. The integer

1

is not prime, while

4, 6, 9, 12, 18,

and

36

are composite.

3

Count the identified prime factors.

The total count is

2

.

There are exactly two distinct prime numbers (

2

and

3

) in the list of factors.

Key Concept

Prime Factors and Definition of Prime Numbers

Estimated Time:45s

Question 65Question →

When the positive integer $n$ is divided by $7$ , the remainder is $4$ . What is the remainder when $n + 15$ is divided by $7$ ?

Show answer & explanation

Answer: 5

Answer

5

Adding 15 to

n

increases the remainder by 15. The original remainder is 4, so the new sum of remainders is

4 + 15 = 19

. Dividing 19 by 7 gives a quotient of 2 and a remainder of 5.

Step-by-Step Solution

1

Express

n

algebraically based on the given remainder rule.

n = 7k + 4

for some non-negative integer

k

By the division algorithm, any integer divided by 7 with remainder 4 can be written as a multiple of 7 plus 4.

2

Substitute

n

into the expression

n + 15

.

n + 15 = 7k + 4 + 15 = 7k + 19

We need to find the remainder of this new quantity when divided by 7.

3

Extract the largest multiple of 7 from 19.

7k + 19 = 7k + 14 + 5 = 7(k + 2) + 5

Grouping multiples of 7 isolates the remaining constant term.

4

Identify the final remainder.

The remainder is 5.

7(k + 2)

is completely divisible by 7, leaving 5 as the remainder.

Key Concept

Remainder arithmetic and divisibility properties

Question 66Question →

Which of the following integers is divisible by both $3$ and $4$ ?

42

54

64

72

86

Show answer & explanation

Answer: $72$

Answer

72

For an integer to be divisible by both

3

and

4

, it must satisfy the rules for both divisors. For

72

, the sum of the digits is

7 + 2 = 9

, which is divisible by

3

. Additionally,

72 \div 4 = 18

, which is an integer. Thus,

72

is divisible by both

3

and

4

.

Step-by-Step Solution

1

Recall the divisibility rule for 3

An integer is divisible by 3 if the sum of its digits is divisible by 3.

This allows quick evaluation of divisibility by 3.

2

Recall the divisibility rule for 4

An integer is divisible by 4 if its last two digits form a number divisible by 4.

This allows quick evaluation of divisibility by 4.

3

Test

72

against both divisibility rules

Sum of digits of

72

is

7 + 2 = 9

, which is divisible by

3

. Also,

72 \div 4 = 18

, which is an integer.

72

meets both required criteria.

Key Concept

Divisibility Rules for 3 and 4

Estimated Time:45s

Question 67Question →

If $x$ and $y$ are positive integers such that $3x + 5y = 101$ and $x$ is a prime number, what is the maximum possible value of the product $xy$ ?

38

108

112

170

180

Show answer & explanation

Answer: 170

Answer

170

The equation

3x + 5y = 101

requires

101 - 3x

to be a multiple of 5. For positive integers

x

and

y

,

x

must be less than 34 and leave a remainder of 2 when divided by 5 (meaning

x

ends in 2 or 7). Testing the primes meeting this condition yields three valid pairs:

(2, 19)

with product 38,

(7, 16)

with product 112, and

(17, 10)

with product 170. Thus, 170 is the maximum product.

Step-by-Step Solution

1

Isolate

5y

and analyze modular divisibility by 5.

5y = 101 - 3x

. For

y

to be an integer,

101 - 3x

must be divisible by 5. Since

101 \equiv 1 \pmod 5

, we require

3x \equiv 1 \pmod 5

, which simplifies to

x \equiv 2 \pmod 5

.

Properties of modular arithmetic determine which values of

x

yield integer values for

y

.

2

Determine the boundary constraints and prime condition for

x

.

Since

y

is a positive integer (

y \ge 1

),

3x < 101 \implies x \le 33

. The prime numbers

x \le 33

that satisfy

x \equiv 2 \pmod 5

(or end in 2 or 7) are

x = 2

,

x = 7

, and

x = 17

.

Only primes satisfying both the inequality bound and divisibility criteria need to be tested.

3

Calculate the corresponding values of

y

and products

xy

for each prime candidate.

For

x = 2

:

5y = 101 - 6 = 95 \implies y = 19

, product

xy = 2 \times 19 = 38

.
For

x = 7

:

5y = 101 - 21 = 80 \implies y = 16

, product

xy = 7 \times 16 = 112

.
For

x = 17

:

5y = 101 - 51 = 50 \implies y = 10

, product

xy = 17 \times 10 = 170

.

Evaluating all valid candidate pairs allows identifying the absolute maximum product.

4

Identify the maximum product.

The maximum possible value of

xy

is 170.

Comparing 38, 112, and 170 confirms that 170 is the greatest value.

Key Concept

Linear Diophantine equations with prime integer constraints

Question 68Question →

If $k$ is a positive integer less than $100$ such that $k$ is divisible by $6$ but not divisible by $8$ , what is the greatest possible value of $k$ ?

Show answer & explanation

Answer: 90

Answer

90

The positive multiples of 6 less than 100 in descending order are 96, 90, 84, etc. The largest candidate, 96, is divisible by 8 (

96 = 8 \times 12

), so it is excluded by the conditions. The next largest candidate, 90, is divisible by 6 (

90 = 6 \times 15

) but leaves a remainder of 2 when divided by 8 (

90 = 8 \times 11 + 2

). Thus, 90 is the greatest integer satisfying all conditions.

Step-by-Step Solution

1

Identify the largest multiples of 6 that are less than 100.

The multiples of 6 less than 100 in descending order are 96, 90, 84, 78, ...

Finding the greatest possible value requires testing candidates starting from the largest possible multiple of 6 below 100.

2

Test 96 against the constraint of not being divisible by 8.

96 / 8 = 12, so 96 is divisible by 8.

The question specifies that k must NOT be divisible by 8, ruling out 96.

3

Test the next candidate, 90, against the constraint.

90 / 8 = 11 R 2, so 90 is not divisible by 8.

90 meets all specified criteria: it is a positive integer less than 100, divisible by 6, and not divisible by 8.

Key Concept

Properties of Multiples and Divisibility Constraints

Question 69Question →

Let $m$ and $n$ be positive integers such that $m^2 n$ is divisible by $72$ and $m n^2$ is divisible by $108$ . What is the minimum possible value of the product $mn$ ?

12

18

24

36

72

Show answer & explanation

Answer: 36

Answer

36

The product

(m^2 n)(m n^2) = (mn)^3

must be divisible by

72 \times 108 = 2^5 \times 3^5

. For

(mn)^3

to be a valid cube of an integer, the exponents of 2 and 3 in its prime factorization must be multiples of 3 greater than or equal to 5. The smallest such multiples of 3 are 6, meaning

(mn)^3 \ge 2^6 \times 3^6 = (36)^3

, so

mn \ge 36

. Setting

m = 6

and

n = 6

confirms that

36

satisfies all requirements.

Step-by-Step Solution

1

Express the given divisibility conditions in terms of prime factorizations.

72 = 2^3 \times 3^2

and

108 = 2^2 \times 3^3

. Therefore,

m^2 n

is a multiple of

2^3 \times 3^2

, and

m n^2

is a multiple of

2^2 \times 3^3

.

Prime factorization allows us to analyze the minimum exponent requirements for 2 and 3.

2

Multiply the two expressions to find a lower bound on

(mn)^3

.

(m^2 n)(m n^2) = m^3 n^3 = (mn)^3

must be divisible by

(2^3 \times 3^2)(2^2 \times 3^3) = 2^5 \times 3^5

.

Combining the expressions yields a perfect cube

(mn)^3

.

3

Determine the minimum prime powers needed for

mn

.

Since

(mn)^3

is a perfect cube divisible by

2^5 \times 3^5

, the exponents of 2 and 3 in

(mn)^3

must be multiples of 3 that are at least 5. The smallest such exponents are 6 for both 2 and 3. Thus,

(mn)^3 \ge 2^6 \times 3^6 = (2^2 \times 3^2)^3 = 36^3

, so

mn \ge 36

.

Exponents in a perfect cube factorization must be multiples of 3.

4

Verify that

mn = 36

is achievable with integer values of

m

and

n

.

Setting

m = 6

and

n = 6

gives

mn = 36

. Then

m^2 n = 6^3 = 216 = 3 \times 72

(divisible by 72) and

m n^2 = 6^3 = 216 = 2 \times 108

(divisible by 108).

Constructing valid integers

m

and

n

confirms that 36 is achievable.

Key Concept

Properties of Integers and Prime Factorization

Question 70Question →

If $k$ is an integer greater than $1$ such that $k^3 - k$ is a multiple of $240$ , what is the least possible value of $k$ ?

Show answer & explanation

Answer: 9

Answer

The least possible value of

k

is 9.

Factoring

k^3 - k

gives

(k-1)k(k+1)

, representing the product of three consecutive integers. The prime factorization of

240

is

2^4 \times 3 \times 5

. To be divisible by

240

, the product

(k-1)k(k+1)

must contain at least four factors of 2, one factor of 3, and one factor of 5. Testing integers

k > 1

: for

k=9

, the product is

(8)(9)(10) = 720

, which is

3 \times 240

. Checking all integers

1 < k < 9

confirms that no smaller integer satisfies the condition.

Step-by-Step Solution

1

Factor the algebraic expression

k^3 - k

.

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

.

This expresses the polynomial as a product of three consecutive integers.

2

Determine the prime factorization of 240.

240 = 2^4 \times 3 \times 5 = 16 \times 3 \times 5

.

For

(k-1)k(k+1)

to be a multiple of 240, the product must be divisible by 16, 3, and 5 simultaneously.

3

Test candidate values of

k > 1

to find the smallest valid integer.

For

k=9

,

(k-1)k(k+1) = (8)(9)(10) = 720

, which equals

3 \times 240

.

Testing smaller values:

k=5

yields

120

,

k=6

yields

210

,

k=7

yields

336

, and

k=8

yields

504

. None of these are divisible by 240.

Key Concept

Divisibility rules and prime factor distribution across consecutive integers

Estimated Time:2m 0s

Question 71Question →

When the positive integer $m$ is divided by $8$ , the remainder is $5$ . What is the remainder when $3m + 7$ is divided by $8$ ?

2

4

5

6

7

Show answer & explanation

Answer: $6$

Answer

The remainder is

6

.

Since

m

leaves a remainder of

5

when divided by

8

, we can evaluate

3m + 7

modulo

8

by replacing

m

with

5

. Calculating

3(5) + 7

gives

22

. Dividing

22

by

8

gives a quotient of

2

with a remainder of

6

because

22 = 8 \times 2 + 6

.

Step-by-Step Solution

1

Express integer

m

using the division algorithm.

m = 8k + 5

for some non-negative integer

k

.

Dividing

m

by

8

yields a remainder of

5

.

2

Substitute the expression for

m

into

3m + 7

.

3m + 7 = 3(8k + 5) + 7 = 24k + 15 + 7 = 24k + 22

.

This expresses the new quantity algebraically in terms of

k

.

3

Determine the remainder when

24k + 22

is divided by

8

.

24k + 22 = 8(3k + 2) + 6

, so the remainder is

6

.

The term

8(3k + 2)

is a multiple of

8

, leaving

6

as the non-negative remainder less than

8

.

Key Concept

Remainder properties of linear combinations of integers

Estimated Time:45s

Question 72Question →

For all positive integers $n$ , what is the greatest integer that MUST divide $n^5 - n$ ?

6

10

15

30

60

Show answer & explanation

Answer: 30

Answer

The greatest integer that must divide

n^5 - n

for all positive integers

n

is 30.

Factoring the expression yields

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

. The factor

(n-1)n(n+1)

consists of three consecutive integers, ensuring divisibility by both 2 and 3 (and thus by 6). Furthermore, the units digit of

n^5

is always identical to the units digit of

n

for any positive integer

n

, which guarantees that

n^5 - n

is divisible by 5. Since 6 and 5 share no common factors,

n^5 - n

must be divisible by

6 \times 5 = 30

. Testing

n = 2

gives

2^5 - 2 = 30

, proving that 30 is the largest integer that divides the expression for all positive integers

n

.

Step-by-Step Solution

1

Factor the algebraic expression

n^5 - n

.

n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = (n - 1)n(n + 1)(n^2 + 1)

.

Factoring helps isolate product terms with known divisibility rules.

2

Analyze divisibility by 2 and 3.

The sub-expression

(n - 1)n(n + 1)

is the product of three consecutive integers.

Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3. Therefore,

(n - 1)n(n + 1)

is always divisible by

2 \times 3 = 6

.

3

Analyze divisibility by 5.

By Fermat's Little Theorem (or analyzing last-digit repeating cycles of powers), for any integer

n

,

n^5 \equiv n \pmod 5

, which implies

5

divides

n^5 - n

.

The units digit of

n^5

is always equal to the units digit of

n

, so

n^5 - n

always ends in 0 or 5, making it a multiple of 5.

4

Combine the common prime factors and evaluate the maximum lower bound.

Since 2, 3, and 5 are pairwise coprime, any number divisible by 2, 3, and 5 must be divisible by

2 \times 3 \times 5 = 30

. Evaluating at

n = 2

yields

2^5 - 2 = 30

, showing no integer greater than 30 can divide

n^5 - n

for all

n

.

The greatest common divisor across all generated values of

n^5 - n

is 30.

Key Concept

Divisibility properties of consecutive integers and power mod rules

Estimated Time:1m 30s

Question 73Question →

A container originally holds a liquid mixture in which $\frac{2}{5}$ of the total volume is pure juice. After $6$ liters of pure juice are added to the container, pure juice makes up $\frac{1}{2}$ of the new total volume. How many liters of mixture were originally in the container?

14

20

24

30

36

Show answer & explanation

Answer: 30

Answer

30 liters

The original mixture volume

V

contains

\frac{2}{5}V

liters of pure juice. When

6

liters of pure juice are added, the new volume of juice becomes

\frac{2}{5}V + 6

and the new total volume becomes

V + 6

. Setting

\frac{2}{5}V + 6 = \frac{1}{2}(V + 6)

leads to

\frac{1}{10}V = 3

, giving

V = 30

liters.

Step-by-Step Solution

1

Define variables for the original volume and pure juice volume.

Let

V

represent the original volume of the mixture in liters. The original volume of pure juice is

\frac{2}{5}V

.

Expressing the initial quantity of juice in terms of the unknown total volume establishes the baseline algebraic relationship.

2

Set up an equation incorporating the added quantity of pure juice.

New juice volume

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

, and new total volume

= V + 6

. The equation is

\frac{2}{5}V + 6 = \frac{1}{2}(V + 6)

.

Adding pure juice increases both the pure juice component and the total volume of the mixture by

6

liters.

3

Solve the linear equation for

V

.

Expanding the right side gives

\frac{2}{5}V + 6 = \frac{1}{2}V + 3

. Subtracting

\frac{2}{5}V

from both sides gives

6 - 3 = \left(\frac{1}{2} - \frac{2}{5}\right)V \Rightarrow 3 = \frac{1}{10}V \Rightarrow V = 30

.

Finding a common denominator for the fractions

\frac{1}{2} = \frac{5}{10}

and

\frac{2}{5} = \frac{4}{10}

allows solving for

V

directly.

Key Concept

Solving linear equations involving fractions of whole quantities and mixture relationships.

Estimated Time:1m 0s

Question 74Question →

If $a$ and $b$ are positive integers such that $15a$ is divisible by $24$ and $10b$ is divisible by $14$ , which of the following statements MUST be true? Select all such statements.

Select all that apply

a

is divisible by

8

b

is divisible by

14

The product

ab

is divisible by

56

The sum

a + b

must be an even integer

a

is divisible by

24

Show answer & explanation

Answer: $a$ is divisible by $8$ ; The product $ab$ is divisible by $56$

Answer

The statements asserting that

a

is divisible by

8

and that the product

ab

is divisible by

56

must be true.

Because

24 = 2^3 \times 3

and

15 = 3 \times 5

,

a

must provide three factors of

2

, meaning

a

is divisible by

8

. Similarly,

14 = 2 \times 7

and

10 = 2 \times 5

, so

b

must provide a factor of

7

, meaning

b

is divisible by

7

. Multiplying the minimum properties (

a

is a multiple of

8

and

b

is a multiple of

7

) guarantees that

ab

is a multiple of

56

.

Step-by-Step Solution

1

Analyze the condition

15a

is divisible by

24

Prime factorizations:

15 = 3 \times 5

and

24 = 2^3 \times 3

. For

24

to divide

15a

,

a

must supply

2^3 = 8

, so

a

is a multiple of

8

.

Determines the prime factors required for

a

.

2

Analyze the condition

10b

is divisible by

14

Prime factorizations:

10 = 2 \times 5

and

14 = 2 \times 7

. For

14

to divide

10b

,

b

must supply

7

, so

b

is a multiple of

7

.

Determines the prime factors required for

b

.

3

Evaluate each statement against the deduced minimum properties of

a

and

b

Since

a

is a multiple of

8

, the statement regarding

a

being divisible by

8

is always true. Since

a

is a multiple of

8

and

b

is a multiple of

7

,

ab

is a multiple of

8 \times 7 = 56

. Counterexamples disprove the remaining choices (

b=7

disproves divisibility of

b

by

14

and evenness of

a+b

;

a=8

disproves divisibility of

a

by

24

).

Identifies which statements MUST be true for all valid values of

a

and

b

.

Key Concept

Divisibility and Prime Factorization Constraints

Question 75Question →

When the positive integer $n$ is divided by $7$ , the remainder is $4$ . What is the remainder when $n + 18$ is divided by $7$ ?

Show answer & explanation

Answer: 1

Answer

The remainder when

n + 18

is divided by

7

is

1

.

Because

n

leaves a remainder of

4

when divided by

7

,

n

can be represented as

7k + 4

. Adding

18

gives

n + 18 = 7k + 22

. Since

7k

is a multiple of

7

, the remainder of

(7k + 22) \div 7

depends entirely on

22 \div 7

. Dividing

22

by

7

gives a quotient of

3

with a remainder of

1

.

Step-by-Step Solution

1

Represent the positive integer

n

algebraically based on the given remainder.

n = 7k + 4

for some non-negative integer

k

.

By the division algorithm, any integer

n

divided by

7

with remainder

4

can be written as

7k + 4

.

2

Add

18

to

n

.

n + 18 = (7k + 4) + 18 = 7k + 22

.

Substitute

7k + 4

for

n

in the expression

n + 18

.

3

Determine the remainder when

7k + 22

is divided by

7

.

Since

7k

is divisible by

7

, the remainder is

22 \pmod 7 = 1

.

Dividing

22

by

7

yields a quotient of

3

and a remainder of

1

(

22 = 7 \times 3 + 1

).

Key Concept

Remainder Properties under Addition

Question 76Question →

If $n$ is a positive integer such that $n$ is divisible by $12$ and $n^2$ is divisible by $180$ , what is the least possible value of $n$ ?

30

60

90

180

360

Show answer & explanation

Answer: 60

Answer

60

The correct answer is 60. The prime factorization of 12 is

2^2 \times 3^1

, requiring

n

to have at least two factors of 2 and one factor of 3. The prime factorization of 180 is

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

. For

n^2

to be divisible by 180,

n

must contribute a factor of 5 (which becomes

5^2

in

n^2

). Multiplying the minimal prime factors yields

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

.

Step-by-Step Solution

1

Analyze the prime factorization required for

n

to be divisible by 12

Since

12 = 2^2 \times 3^1

, any positive integer

n

divisible by 12 must contain at least

2^2

and

3^1

in its prime factorization.

An integer must contain all prime factors of its divisor with at least equal exponents.

2

Analyze the prime factorization required for

n^2

to be divisible by 180

Since

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

,

n^2

must contain at least

2^2

,

3^2

, and

5^1

. Because

n^2

doubles all prime exponents of

n

,

n

must contribute at least

5^1

, which makes the factor of 5 in

n^2

equal to

5^2

.

The exponent of any prime factor in a perfect square

n^2

must be even.

3

Combine the minimal prime factor requirements for

n

The minimum prime factorization for

n

is

2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60

.

Taking the minimum required power of each prime factor yields the smallest positive integer fulfilling both conditions.

Key Concept

Prime Factorization and Divisibility of Powers

Question 77Question →

When the positive integer $n$ is divided by 7, the remainder is 3, and when $n$ is divided by 11, the remainder is 5. If $n$ is a three-digit integer less than 200 that is divisible by 6, what is the value of $n$ ?

Show answer & explanation

Answer: 192

Answer

The value of

n

is 192.

To find

n

, combine the remainder requirements

n \equiv 3 \pmod 7

and

n \equiv 5 \pmod{11}

. The smallest positive integer solution is

38

. Because

\text{lcm}(7, 11) = 77

, all valid integers take the form

n = 77m + 38

. Restricting

n

to three-digit numbers less than 200 gives candidates

n = 115

(

m=1

) and

n = 192

(

m=2

). Among these, only 192 is divisible by 6 (since

192 = 6 \times 32

).

Step-by-Step Solution

1

Formulate remainder congruences for the given conditions

n \equiv 3 \pmod{7}

and

n \equiv 5 \pmod{11}

Dividing

n

by 7 leaves remainder 3, and dividing by 11 leaves remainder 5.

2

Find the smallest positive integer satisfying both congruences

n = 38

Checking values

11k + 5

:

5, 16, 27, 38

.

38 \div 7 = 5

remainder 3, so 38 satisfies both conditions.

3

Determine the general solution for

n

using the Chinese Remainder Theorem logic

n = 77m + 38

for integer

m \ge 0

Since 7 and 11 are coprime, the solutions repeat every

\text{lcm}(7, 11) = 77

.

4

Evaluate candidate values for

n

such that

100 \le n < 200

For

m=1

,

n = 115

; for

m=2

,

n = 192

These are the only three-digit integers less than 200 of the form

77m + 38

.

5

Apply the final constraint that

n

must be divisible by 6

192

is divisible by 6 (

192 = 6 \times 32

)

115

is odd, so it is not divisible by 6.

192

is even and the sum of its digits (

1+9+2=12

) is a multiple of 3, so it is divisible by 6.

Key Concept

Simultaneous congruences and combined divisibility rules (Chinese Remainder Theorem)

Estimated Time:2m 0s

Question 78Question →

Let $N = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers such that $N$ is a multiple of $8$ . If $N^2$ has $105$ distinct positive divisors and $\frac{N}{6}$ has $12$ distinct positive divisors, how many distinct positive divisors does $10N$ have?

30

36

45

48

60

Show answer & explanation

Answer: 45

Answer

The number of distinct positive divisors of

10N

is 45.

The correct answer is 45. Given

N = 2^a \cdot 3^b \cdot 5^c

, the number of positive divisors of

N^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}

is

(2a+1)(2b+1)(2c+1) = 105

. The unique factor decomposition of 105 into three factors greater than 1 is

3 \times 5 \times 7

. Because

N

is a multiple of 8,

a \ge 3

, so

2a+1 \ge 7

, forcing

2a+1 = 7

, which means

a = 3

. The remaining factor set

\{2b+1, 2c+1\} = \{3, 5\}

gives two cases for

(b, c)

:

(2, 1)

or

(1, 2)

. For

\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c

, the divisor count is

a \cdot b \cdot (c+1) = 3 \cdot b \cdot (c+1) = 12

, so

b(c+1) = 4

. Substituting

(b, c) = (2, 1)

yields

2 \times 2 = 4

, which satisfies this relation. Therefore,

(a, b, c) = (3, 2, 1)

. Now

10N = 2^{a+1} \cdot 3^b \cdot 5^{c+1} = 2^4 \cdot 3^2 \cdot 5^2

. Its total number of positive divisors is

(4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45

.

Step-by-Step Solution

1

Express the divisor count formula for

N^2

d(N^2) = (2a+1)(2b+1)(2c+1) = 105

For an integer with prime factorization

p_1^{e_1} p_2^{e_2} \cdots

, the number of divisors is

(e_1+1)(e_2+1)\cdots

. Here

N^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}

.

2

Determine the value of

a

using the divisibility condition

a = 3

Since

N

is a multiple of

8 = 2^3

, we must have

a \ge 3

, which implies

2a+1 \ge 7

. The prime factorization of

105

into three odd factors greater than

1

is

3 \times 5 \times 7

. Thus,

2a+1 = 7

, giving

a = 3

.

3

Solve for exponents

b

and

c

using the divisor count of

\frac{N}{6}

b = 2

and

c = 1

Since

\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c

, its divisor count is

a \cdot b \cdot (c+1) = 3b(c+1) = 12

, so

b(c+1) = 4

. From step 2,

\{2b+1, 2c+1\} = \{3, 5\}

, so either

(b, c) = (2, 1)

or

(b, c) = (1, 2)

. Testing

(b, c) = (2, 1)

gives

2(1+1) = 4

, which satisfies the equation.

4

Calculate the prime factorization and divisor count of

10N

d(10N) = (4+1)(2+1)(2+1) = 45

Since

10N = 2 \cdot 5 \cdot (2^3 \cdot 3^2 \cdot 5^1) = 2^4 \cdot 3^2 \cdot 5^2

, applying the divisor formula yields

5 \times 3 \times 3 = 45

.

Key Concept

Divisor Count Formula & Prime Factorization Constraints

Estimated Time:2m 30s

Question 79Question →

A storage tank is initially $\frac{2}{3}$ full of liquid. After $\frac{1}{4}$ of the liquid inside the tank is drained out, $15$ gallons of liquid are added to the tank. If the tank is then $\frac{5}{6}$ full, what is the total capacity of the tank, in gallons?

30

36

45

60

90

Show answer & explanation

Answer: 45

Answer

45 gallons

The correct answer of 45 gallons is found by calculating the liquid remaining after draining

\frac{1}{4}

of the initial

\frac{2}{3} C

, which leaves

\frac{1}{2} C

. Adding 15 gallons brings the total volume to

\frac{5}{6} C

. Solving

\frac{5}{6} C - \frac{1}{2} C = 15

gives

\frac{1}{3} C = 15

, so

C = 45

.

Step-by-Step Solution

1

Determine the fraction of liquid in the tank after draining.

The initial amount of liquid is

\frac{2}{3} C

, where

C

is the total capacity. Draining

\frac{1}{4}

of this liquid leaves

1 - \frac{1}{4} = \frac{3}{4}

of the initial liquid. Thus, the remaining liquid is

\frac{3}{4} \times \frac{2}{3} C = \frac{1}{2} C

.

The fraction drained applies to the existing liquid volume, not the total container capacity.

2

Set up an equation incorporating the added liquid and final tank fraction.

\frac{1}{2} C + 15 = \frac{5}{6} C

Adding

15

gallons to the remaining liquid yields a volume equal to

\frac{5}{6}

of the total capacity.

3

Solve for the total capacity

C

.

Subtracting

\frac{1}{2} C

(which is

\frac{3}{6} C

) from both sides gives

15 = \frac{5}{6} C - \frac{3}{6} C = \frac{2}{6} C = \frac{1}{3} C

. Multiplying both sides by

3

yields

C = 45

.

Isolating

C

determines the overall volume of the tank.

Key Concept

Multi-step operations with fractions and fractional parts of a whole

Estimated Time:1m 30s

Question 80Question →

A laboratory vessel contains a liquid solution consisting of water and ethanol, where water makes up $\frac{3}{8}$ of the total volume. First, $\frac{1}{5}$ of the total volume of the solution is drained and replaced with an equal volume of pure ethanol. Next, $\frac{1}{4}$ of the resulting mixture is evaporated, removing water and ethanol in proportion to their presence. Finally, pure water is added to fill the vessel back to its original total volume. What fraction of the final solution is water? Express your answer as a decimal.

Show answer & explanation

Answer: 0.475

Answer

The fraction of the final solution that is water is 0.475 (or 19/40).

By following the multi-step fractional changes to the liquid volume, the remaining water prior to refilling is

\frac{9}{40}

of the original capacity. Refilling the missing

\frac{1}{4}

(or

\frac{10}{40}

) volume with pure water yields

\frac{19}{40} = 0.475

of the total solution as water.

Step-by-Step Solution

1

Track water content after the initial replacement.

Water fraction becomes

\frac{3}{10}

of the original volume.

Let the original total volume be

V

. Initially, water volume is

\frac{3}{8}V

. Removing

\frac{1}{5}

of the solution leaves

\frac{4}{5}

of the original solution, so the water volume becomes

\frac{4}{5} \times \frac{3}{8}V = \frac{3}{10}V

. Replacing the removed volume with pure ethanol brings total volume back to

V

, with water occupying

\frac{3}{10}V

.

2

Track water content after evaporation.

Water volume becomes

\frac{9}{40}V

and total solution volume becomes

\frac{3}{4}V

.

Evaporating

\frac{1}{4}

of the solution leaves

\frac{3}{4}

of the mixture intact. The remaining water volume is

\frac{3}{4} \times \frac{3}{10}V = \frac{9}{40}V

.

3

Calculate the final water fraction after refilling with pure water.

Final water volume is

\frac{19}{40}V = 0.475V

.

To restore the total volume from

\frac{3}{4}V

back to

V

, an amount equal to

V - \frac{3}{4}V = \frac{1}{4}V

of pure water is added. Adding this to the existing water gives

\frac{9}{40}V + \frac{1}{4}V = \frac{9}{40}V + \frac{10}{40}V = \frac{19}{40}V

. Dividing by total volume

V

yields

\frac{19}{40} = 0.475

.

Key Concept

Sequential fractional reduction and component tracking

Estimated Time:2m 30s

Arithmetic

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