Number Properties and Arithmetic Practice Questions — GMAT — Page 7

Question 121Question →

If $a$ , $b$ , and $c$ are non-zero real numbers such that $\frac{a}{b} < 0$ , $b c > 0$ , and $a - c > 0$ , which of the following expressions MUST be negative?

a b c

\frac{a + b}{c}

a b^2 c

\frac{b - c}{a}

a^2 b c

Show answer & explanation

Answer: $a b^2 c$

Answer

The expression

a b^2 c

must be negative.

From

\frac{a}{b} < 0

,

a

and

b

have opposite signs. From

b c > 0

,

b

and

c

have the same sign. Thus,

a

and

c

must have opposite signs. The inequality

a - c > 0

implies

a > c

, so

a

must be positive (

a > 0

) and

c

must be negative (

c < 0

). Since

b

shares the sign of

c

,

b

is also negative (

b < 0

). Evaluating

a b^2 c

:

a

is positive,

b^2

is strictly positive for any non-zero real number

b

, and

c

is negative. Therefore,

a b^2 c

is the product of two positive terms and one negative term, which MUST be negative.

Step-by-Step Solution

1

Determine the relative signs of

a

,

b

, and

c

from the given inequalities.

\frac{a}{b} < 0

means

a

and

b

have opposite signs.

b c > 0

means

b

and

c

have the same sign. Therefore,

a

and

c

must have opposite signs.

Quotients of numbers with opposite signs are negative, and products of numbers with the same sign are positive.

2

Use

a - c > 0

to determine the explicit sign of each variable.

a - c > 0 \implies a > c

. Since

a

and

c

have opposite signs and

a > c

,

a

must be positive (

a > 0

) and

c

must be negative (

c < 0

). Since

b

has the same sign as

c

,

b

must also be negative (

b < 0

).

A positive number is always greater than a negative number.

3

Evaluate the sign of

a b^2 c

.

Since

a > 0

,

b^2 > 0

(as

b \neq 0

), and

c < 0

,

a b^2 c = (+) \times (+) \times (-) < 0

.

The product of two positive real numbers and one negative real number is strictly negative.

Key Concept

Deducing signs of variables using properties of products, quotients, and inequalities.

Estimated Time:2m 0s

Question 122Question →

For real numbers $p$ , $q$ , and $r$ , it is given that $p < 0 < q < r$ . If $p^3 q r = -288$ and $\frac{q r}{p^2} = 9$ , what is the value of $p + q r$ ?

Show answer & explanation

Answer: 34

Answer

The value of

p + q r

is 34.

From

\frac{qr}{p^2} = 9

, we get

qr = 9p^2

. Substituting this into

p^3 qr = -288

yields

9p^5 = -288

, so

p^5 = -32

. Taking the fifth root gives

p = -2

, which satisfies

p < 0

. Then

qr = 9(-2)^2 = 36

. Finally,

p + qr = -2 + 36 = 34

.

Step-by-Step Solution

1

Relate

qr

to

p

using the given quotient equality

qr = 9p^2

Multiplying both sides of

\frac{qr}{p^2} = 9

by

p^2

isolates

qr

.

2

Substitute

qr = 9p^2

into the product equation

p^3 (9p^2) = -288 \implies 9p^5 = -288 \implies p^5 = -32

Replacing

qr

with

9p^2

produces a single-variable polynomial in

p

.

3

Solve for

p

enforcing the sign constraint

p < 0

p = -2

Taking the fifth root of

-32

yields

-2

, which satisfies

p < 0

.

4

Calculate the value of

qr

qr = 9(-2)^2 = 36

Squaring a negative number yields a positive value:

(-2)^2 = 4

, so

9 \times 4 = 36

.

5

Evaluate the expression

p + qr

p + qr = -2 + 36 = 34

Adding

p = -2

and

qr = 36

results in

34

.

Key Concept

Positive and Negative Number Properties

Question 123Question →

For three positive integers $a$ , $b$ , and $c$ , the greatest common divisor of any pair among them is $12$ , and the least common multiple of all three integers is $5,040$ . If $a = 60$ , what is the minimum possible value of $b + c$ ?

168

228

300

372

1,020

Show answer & explanation

Answer: $228$

Answer

The minimum possible value of

b + c

is

228

.

The value

228

is correct because prime factor analysis shows

b

and

c

must take exponents

\{4, 2\}

for prime 2,

\{2, 1\}

for prime 3,

\{0, 0\}

for prime 5, and

\{1, 0\}

for prime 7. Combining

2^4 \times 3^2 = 144

and

2^2 \times 3^1 \times 7^1 = 84

satisfies all pairwise GCD and LCM conditions while minimizing the sum to

144 + 84 = 228

.

Step-by-Step Solution

1

Express the given numbers and conditions in prime factorized form.

The pairwise GCD is

12 = 2^2 \times 3^1

. The LCM of

a, b, c

is

5,040 = 2^4 \times 3^2 \times 5^1 \times 7^1

. Given

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

.

Prime factorization allows exact determination of the minimum and maximum required exponent for each prime factor across

a, b,

and

c

.

2

Determine the prime factor exponent constraints for

b

and

c

.

For prime factor 2: min exponent is 2, max is 4. Since

a

has

2^2

, exponents for

(b,c)

for factor 2 must be

\{4, 2\}

to minimize sum.
For prime factor 3: min exponent is 1, max is 2. Since

a

has

3^1

, exponents for

(b,c)

for factor 3 must be

\{2, 1\}

.
For prime factor 5: since

\gcd(a,b) = 12

and

\gcd(a,c) = 12

, 5 cannot divide

b

or

c

. Thus exponents for factor 5 are both 0.
For prime factor 7: max exponent is 1 in LCM, but

a

has

7^0

and pairwise GCD has

7^0

, so exactly one of

b

or

c

has

7^1

and the other has

7^0

.

The pairwise GCD dictates the minimum exponent present in all pairs, while the overall LCM dictates the maximum exponent present across the three numbers.

3

Test allocations of exponents to minimize

b + c

.

We must distribute exponents

\{4, 2\}

for 2,

\{2, 1\}

for 3, and

\{1, 0\}

for 7 between

b

and

c

.
Allocation 1:

b = 2^4 \times 3^2 = 144

and

c = 2^2 \times 3^1 \times 7^1 = 84 \implies b + c = 228

.
Allocation 2:

b = 2^4 \times 3^1 = 48

and

c = 2^2 \times 3^2 \times 7^1 = 252 \implies b + c = 300

.
Allocation 3:

b = 2^4 \times 3^1 \times 7^1 = 336

and

c = 2^2 \times 3^2 = 36 \implies b + c = 372

.
Allocation 4:

b = 2^4 \times 3^2 \times 7^1 = 1008

and

c = 2^2 \times 3^1 = 12 \implies b + c = 1020

.

Comparing all valid exponent combinations reveals the minimum sum.

4

Identify the minimum sum.

The minimum sum is

228

.

The smallest sum among all valid combinations is

144 + 84 = 228

.

Key Concept

Prime factor exponent extraction for pairwise GCD and joint LCM of three numbers

Estimated Time:2m 0s

Question 124Question →

In January, electricity comprised $0.45$ of a facility's total energy consumption. In February, electricity comprised $\frac{3}{5}$ of its total energy consumption. If the facility's total energy consumption was constant at $1,200$ kilowatt-hours in each of the two months, by how many kilowatt-hours did its electricity consumption increase from January to February?

150

kilowatt-hours

180

kilowatt-hours

270

kilowatt-hours

540

kilowatt-hours

720

kilowatt-hours

Show answer & explanation

Answer: $180$ kilowatt-hours

Answer

The electricity consumption increased by

180

kilowatt-hours from January to February.

The correct answer of

180

kilowatt-hours is found by converting both fractions and decimals into comparable parts of the

1,200

kilowatt-hour base. January electricity usage is

0.45 \times 1,200 = 540

kWh. February electricity usage is

\frac{3}{5} \times 1,200 = 720

kWh. The difference is

720 - 540 = 180

kWh. Alternatively, subtracting the proportions first gives

\frac{3}{5} - 0.45 = 0.60 - 0.45 = 0.15

, and

0.15 \times 1,200 = 180

kWh.

Step-by-Step Solution

1

Calculate the January electricity consumption

0.45 \times 1,200 = 540

kilowatt-hours

Electricity represented

0.45

of the

1,200

kilowatt-hour total.

2

Calculate the February electricity consumption

\frac{3}{5} \times 1,200 = 0.60 \times 1,200 = 720

kilowatt-hours

Electricity represented

\frac{3}{5}

(or

0.60

) of the

1,200

kilowatt-hour total.

3

Find the difference between February and January consumption

720 - 540 = 180

kilowatt-hours

Subtracting January usage from February usage yields the net increase.

Key Concept

Fraction and Decimal Conversions and Amount Computations

Question 125Question →

If $k$ is a positive integer such that $k$ is a multiple of $6$ and a factor of $120$ , how many possible values are there for $k$ ?

4

6

8

12

16

Show answer & explanation

Answer: 6

Answer

6

The correct answer is 6. First, express 120 in prime factored form as

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

. Any factor of 120 has the form

2^a \times 3^b \times 5^c

where

0 \le a \le 3

,

0 \le b \le 1

, and

0 \le c \le 1

. To also be a multiple of 6 (

2^1 \times 3^1

),

k

must contain at least one factor of 2 and at least one factor of 3. This restricts

a

to 1, 2, or 3 (3 possibilities),

b

to 1 (1 possibility), and

c

to 0 or 1 (2 possibilities). The total number of valid values for

k

is

3 \times 1 \times 2 = 6

.

Step-by-Step Solution

1

Find the prime factorization of 120 and 6

120 = 2^3 \times 3^1 \times 5^1

and

6 = 2^1 \times 3^1

Prime factorization allows us to express divisibility constraints in terms of exponent bounds.

2

Determine the exponent constraints for

k = 2^a \times 3^b \times 5^c

For

k

to be a factor of 120,

0 \le a \le 3

,

0 \le b \le 1

, and

0 \le c \le 1

. For

k

to be a multiple of 6,

a \ge 1

and

b \ge 1

. Thus:

1 \le a \le 3

,

b = 1

,

0 \le c \le 1

.

Combining factor and multiple requirements yields the exact set of possible exponents.

3

Calculate the total number of combinations for

(a, b, c)

Number of choices for

a

is 3 (1, 2, or 3). Number of choices for

b

is 1 (must be 1). Number of choices for

c

is 2 (0 or 1). Total possible values =

3 \times 1 \times 2 = 6

.

By the fundamental counting principle, multiplying the number of choices for each exponent gives the total number of valid integers

k

.

Key Concept

Determining the number of factors of an integer that meet specific divisibility constraints using prime factorization.

Question 126Question →

In 2025, an architectural design firm divided its working hours among commercial, residential, and municipal projects. Commercial projects accounted for $0.40$ of the total hours. Of the remaining hours, $\frac{5}{12}$ were spent on residential projects, and the rest were spent on municipal projects.

In 2026, the firm's total working hours increased by $25\%$ . Commercial project hours decreased by $15\%$ , while residential project hours increased by $40\%$ . By what percent did the number of hours spent on municipal projects increase from 2025 to 2026?

Show answer & explanation

Answer: 60

Answer

The number of hours spent on municipal projects increased by

60\%

.

By setting the total initial hours to a variable

T

(or a convenient constant like

100

), we find that in 2025, commercial projects comprised

0.40T

, residential comprised

\frac{5}{12} \times 0.60T = 0.25T

, and municipal comprised

0.35T

. In 2026, total hours rose to

1.25T

, commercial dropped to

0.34T

, and residential rose to

0.35T

, leaving

0.56T

for municipal. The percent increase in municipal hours is

\frac{0.56T - 0.35T}{0.35T} = \frac{0.21}{0.35} = 60\%

.

Step-by-Step Solution

1

Determine the baseline breakdown of working hours for 2025 in terms of total hours

T

.

Commercial hours =

0.40T

, Residential hours =

0.25T

, Municipal hours =

0.35T

.

Commercial is given as

0.40T

. The remaining fraction

0.60T

is split such that

\frac{5}{12} \times 0.60T = 0.25T

goes to residential, leaving

0.60T - 0.25T = 0.35T

for municipal.

2

Calculate the updated working hours for each category in 2026.

Total hours =

1.25T

, Commercial hours =

0.34T

, Residential hours =

0.35T

.

A

25\%

total increase yields

1.25T

. A

15\%

decrease in commercial hours yields

0.40T \times 0.85 = 0.34T

. A

40\%

increase in residential hours yields

0.25T \times 1.40 = 0.35T

.

3

Find the municipal project hours for 2026 by subtracting commercial and residential hours from total 2026 hours.

Municipal hours (2026) =

0.56T

.

Municipal hours in 2026 equal

1.25T - 0.34T - 0.35T = 0.56T

.

4

Compute the percent change in municipal project hours from 2025 to 2026.

Percent Increase =

60\%

.

Percent increase is

\frac{\text{Municipal}_{2026} - \text{Municipal}_{2025}}{\text{Municipal}_{2025}} \times 100\% = \frac{0.56T - 0.35T}{0.35T} \times 100\% = \frac{0.21}{0.35} \times 100\% = 60\%

.

Key Concept

Multi-step percentage change and fraction-decimal conversions

Estimated Time:2m 0s

Question 127Question →

At the start of a fiscal year, a sovereign wealth fund allocated its capital into three asset classes: Stocks, Bonds, and Commodities. Exactly $0.40$ of the total capital was allocated to Stocks. Of the remaining capital, $\frac{1}{3}$ was allocated to Bonds and the rest was allocated to Commodities. Over the course of the fiscal year, the value of Stocks increased by $15\%$ and the value of Bonds decreased by $10\%$ . If the total value of the fund increased by $5\%$ by the end of the fiscal year, by what percent did the value of Commodities change?

Show answer & explanation

Answer: 2.5

Answer

The value of Commodities increased by

2.5\%

.

By setting the initial total capital to

V

, the initial values are Stocks =

0.40V

, Bonds =

\frac{1}{3}(0.60V) = 0.20V

, and Commodities =

0.60V - 0.20V = 0.40V

. Applying the percentage changes gives final values of Stocks =

0.46V

, Bonds =

0.18V

, and Total Fund =

1.05V

. The final Commodities value is

1.05V - 0.64V = 0.41V

. The percent change in Commodities is

\frac{0.41V - 0.40V}{0.40V} \times 100\% = 2.5\%

.

Step-by-Step Solution

1

Determine initial asset allocations as proportions of total capital

V

Stocks (

S_0

) =

0.40V

, Bonds (

B_0

) =

0.20V

, Commodities (

C_0

) =

0.40V

Stocks take

0.40V

, leaving

0.60V

. Bonds take

\frac{1}{3}

of

0.60V = 0.20V

, leaving

0.40V

for Commodities.

2

Compute final values for Stocks, Bonds, and Total Fund after percentage changes

S_1 = 0.46V

,

B_1 = 0.18V

, and Total Fund

V_1 = 1.05V

Stocks increase by

15\%

(

0.40V \times 1.15 = 0.46V

), Bonds decrease by

10\%

(

0.20V \times 0.90 = 0.18V

), and the total fund grows by

5\%

(

1.05V

).

3

Calculate the end-of-year value of Commodities

C_1 = 0.41V

Subtract final Stocks and Bonds from total fund:

C_1 = 1.05V - (0.46V + 0.18V) = 0.41V

.

4

Calculate the percent change of Commodities relative to its initial value

2.5\%

increase

Percent change

= \frac{0.41V - 0.40V}{0.40V} \times 100\% = \frac{0.01}{0.40} \times 100\% = 2.5\%

.

Key Concept

Weighted percentage change and fractional portion modeling

Estimated Time:1m 40s

Question 128Question →

A commercial coffee roastery prepares a batch of specialty coffee beans containing Arabica, Robusta, and Liberica beans. Initially, Arabica beans make up $0.50$ of the total weight of the batch, Robusta beans make up $\frac{3}{10}$ of the total weight, and Liberica beans make up the remaining weight. After a specialized roasting process, the weight of the Arabica beans decreases by $10\%$ , the weight of the Robusta beans decreases by $50\%$ , and the weight of the Liberica beans remains unchanged. What percent of the final total weight of the coffee batch is made up of Liberica beans?

Show answer & explanation

Answer: 25

Answer

25%

To find the percentage of Liberica beans in the final mixture, first determine the initial weight breakdown assuming a total weight of 100 units: Arabica is 50 units, Robusta is 30 units, and Liberica is 20 units. After roasting, Arabica weight decreases by 10% to 45 units, Robusta weight decreases by 50% to 15 units, and Liberica weight stays at 20 units. The new total weight is 45 + 15 + 20 = 80 units. The proportion of Liberica in the final batch is 20 out of 80 units, which equals 1/4 or 25%.

Step-by-Step Solution

1

Find the initial fractional and percentage composition of the coffee blend.

Arabica accounts for 50% (0.50), Robusta accounts for 30% (3/10), and Liberica accounts for 20% (1 - 0.50 - 0.30 = 0.20).

The sum of all components in the initial blend must equal 1 (or 100%).

2

Assume an initial reference weight of 100 units to represent the batch.

Initial Arabica weight = 50 units, initial Robusta weight = 30 units, and initial Liberica weight = 20 units.

Using a convenient base value like 100 simplifies multi-step percentage change calculations without loss of generality.

3

Calculate the post-roasting weight for each component.

New Arabica weight = 50 * (1 - 0.10) = 45 units. New Robusta weight = 30 * (1 - 0.50) = 15 units. New Liberica weight = 20 units.

Apply the respective percentage decreases to each individual component weight.

4

Determine the new total weight of the batch.

Total final weight = 45 + 15 + 20 = 80 units.

The final total weight is the sum of the remaining weights of all three bean types.

5

Compute the final percentage of Liberica beans.

(20 / 80) * 100% = 25%.

Divide the final weight of Liberica beans by the final total weight of the batch and convert to a percentage.

Key Concept

Combining initial fractions and percentages to compute multi-step composition changes

Question 129Question →

What is the remainder when $17^{17^{17}}$ is divided by $7$ ?

Show answer & explanation

Answer: 5

Answer

The remainder when

17^{17^{17}}

is divided by

7

is

5

.

First, reduce the base modulo

7

:

17 \equiv 3 \pmod 7

, turning the expression into

3^{17^{17}} \pmod 7

. Next, evaluate the pattern of powers of

3 \pmod 7

:

3^1 \equiv 3

,

3^2 \equiv 2

,

3^3 \equiv 6

,

3^4 \equiv 4

,

3^5 \equiv 5

, and

3^6 \equiv 1

, showing a period of

6

. To find which term of the cycle corresponds to the exponent

17^{17}

, evaluate

17^{17} \pmod 6

. Since

17 \equiv -1 \pmod 6

,

17^{17} \equiv (-1)^{17} = -1 \equiv 5 \pmod 6

. Finally, the

5

th term of the cycle gives

3^5 = 243 \equiv 5 \pmod 7

. Therefore, the remainder is

5

.

Step-by-Step Solution

1

Reduce the base modulo 7

17 \equiv 3 \pmod 7

, so

17^{17^{17}} \equiv 3^{17^{17}} \pmod 7

.

Modular arithmetic permits replacing the base of an exponential expression with its remainder upon division by the modulus.

2

Determine the cyclicity period of powers of 3 modulo 7

The remainders of

3^n \pmod 7

repeat in a cycle of length 6:

(3, 2, 6, 4, 5, 1)

.

By Fermat's Little Theorem,

3^6 \equiv 1 \pmod 7

, meaning the sequence of remainders repeats every 6 integer powers.

3

Evaluate the exponent

17^{17}

modulo the period length 6

17 \equiv -1 \pmod 6

, so

17^{17} \equiv (-1)^{17} = -1 \equiv 5 \pmod 6

.

The position within the 6-term cyclicity pattern depends on the exponent modulo 6. Using negative remainders simplifies calculating odd powers of

-1

.

4

Calculate the final remainder using the 5th position in the cyclicity pattern

3^5 = 243 = 7 \times 34 + 5 \equiv 5 \pmod 7

.

Since the exponent leaves a remainder of 5 when divided by 6, the overall remainder corresponds to

3^5 \pmod 7

.

Key Concept

Modular cyclicity and tower exponent remainder reduction

Question 130Question →

Set $S$ consists of $n$ consecutive integers. The product of the smallest and largest integers in $S$ is $-144$ . If the sum of all elements in $S$ is $91$ , what is the value of $n$ ?

Show answer & explanation

Answer: 26

Answer

The number of consecutive integers

n

in Set

S

is 26.

By combining the consecutive set sum formula

n(a + b) = 182

with the term span relation

b - a = n - 1

and the product constraint

ab = -144

, testing the factors of

182

uniquely determines

n = 26

, corresponding to the set of consecutive integers from

-9

to

16

.

Step-by-Step Solution

1

Relate the number of elements

n

to the smallest element

a

and largest element

b

.

b - a = n - 1

In any set of

n

consecutive integers, the distance between the maximum and minimum elements is

n - 1

.

2

Use the sum formula for an evenly spaced set.

n(a + b) = 182

The sum is

\text{Sum} = n \times \frac{a + b}{2} = 91

, which simplifies to

n(a + b) = 182

.

3

Factor

182

to find integer solutions satisfying

a \cdot b = -144

.

n = 26

,

a = -9

,

b = 16

For

n = 26

,

a + b = 7

and

b - a = 25

, giving

a = -9

and

b = 16

. Their product is

(-9)(16) = -144

.

Key Concept

Properties of consecutive integer sets, average/median sum formula, and term boundary indexing

Question 131Question →

If $x$ is a real number such that $\sqrt[4]{3^x + 3^x + 3^x} = 27$ , what is the value of $x$ ?

Show answer & explanation

Answer: 11

Answer

The value of

x

is 11.

Summing three terms of

3^x

yields

3 \cdot 3^x = 3^{x+1}

. Expressing the fourth root of

3^{x+1}

as a fractional power gives

3^{\frac{x+1}{4}}

. Equating this to

27 = 3^3

results in

\frac{x+1}{4} = 3

, leading directly to

x = 11

.

Step-by-Step Solution

1

Combine the repeated addition terms inside the fourth root.

3^x + 3^x + 3^x = 3 \cdot 3^x = 3^{x+1}

Adding three identical terms

3^x

is equivalent to multiplying

3^x

by 3, which increments the exponent by 1.

2

Rewrite the fourth root expression using a fractional exponent.

\sqrt[4]{3^{x+1}} = 3^{\frac{x+1}{4}}

Applying the property

\sqrt[n]{a^m} = a^{m/n}

.

3

Express the constant on the right side of the equation with base 3.

27 = 3^3

Rewriting both sides of the equation with the same base allows equating exponents.

4

Equate the exponents and solve for

x

.

\frac{x+1}{4} = 3 \implies x + 1 = 12 \implies x = 11

Since the bases are equal, the powers must be equal.

Key Concept

Combining like exponential terms and equating fractional exponents

Estimated Time:1m 30s

Question 132Question →

If $x$ , $y$ , and $z$ are non-zero real numbers such that $x y z^2 < 0$ , $x - y > 0$ , and $\frac{x}{z} < \frac{y}{z}$ , which of the following MUST be true?

\frac{x - z}{y} < 0

\frac{x - y}{z} > 0

x^2 - y^2 > 0

\frac{y - z}{x} < 0

\frac{x + z}{y} > 0

Show answer & explanation

Answer: $\frac{x - z}{y} < 0$

Answer

\frac{x - z}{y} < 0

From

z^2 > 0

, the inequality

x y z^2 < 0

implies

x y < 0

, so

x

and

y

have opposite signs. Since

x - y > 0 \implies x > y

,

x

must be positive (

x > 0

) and

y

must be negative (

y < 0

). Next, comparing

x > y

with

\frac{x}{z} < \frac{y}{z}

shows that dividing by

z

reversed the inequality sign, which proves

z

is negative (

z < 0

). Evaluating

\frac{x - z}{y}

: the numerator

x - z

is positive minus negative, which equals positive plus positive (strictly positive), while the denominator

y

is strictly negative. A positive value divided by a negative value is always negative, so

\frac{x - z}{y} < 0

MUST be true.

Step-by-Step Solution

1

Determine the relative signs of

x

and

y

using

x y z^2 < 0

.

x

and

y

must have opposite signs (

x y < 0

).

Since

z \neq 0

,

z^2 > 0

is always positive. Dividing

x y z^2 < 0

by

z^2

yields

x y < 0

.

2

Determine the individual signs of

x

and

y

using

x - y > 0

.

x > 0

(positive) and

y < 0

(negative).

x - y > 0 \implies x > y

. Since

x

and

y

have opposite signs and

x

is strictly greater than

y

,

x

must be positive and

y

must be negative.

3

Determine the sign of

z

using

\frac{x}{z} < \frac{y}{z}

.

z < 0

(negative).

We know

x > y

. When dividing both sides of

x > y

by

z

, the inequality direction reverses to

<

. An inequality sign flips if and only if the divisor

z

is negative.

4

Evaluate the expression

\frac{x - z}{y}

.

\frac{x - z}{y} < 0

Because

x > 0

and

z < 0

,

-z > 0

, so the numerator

x - z = x + (-z)

is the sum of two positive numbers, which is positive. The denominator

y

is negative. Dividing a positive number by a negative number yields a negative result.

Key Concept

Deducing variable signs from product conditions and inequality sign reversal rules

Estimated Time:2m 0s

Question 133Question →

Set $S$ consists of consecutive integers. The sum of all positive integers in set $S$ is $105$ , and the sum of all integers in set $S$ is $-31$ . How many negative integers are contained in set $S$ ?

Show answer & explanation

Answer: 16

Answer

There are 16 negative integers in set

S

.

The correct response is 16. By setting up the sum of consecutive positive integers starting at 1, we determine that the set contains positive integers up to 14, which sum to 105. Subtracting 105 from the total set sum of -31 reveals that the negative integers must sum to -136. The consecutive negative integers -1, -2, ..., -p sum to -136 when p = 16, since 16 × 17 / 2 = 136.

Step-by-Step Solution

1

Find the maximum positive integer

k

in set

S

The largest positive integer in set

S

is

14

Because set

S

consists of consecutive integers, the positive integers are

1, 2, \dots, k

. The sum formula

\frac{k(k+1)}{2} = 105

leads to

k(k+1) = 210

. Factoring

210

into two consecutive integers gives

14 \times 15

, so

k = 14

.

2

Calculate the sum of all negative integers in set

S

The sum of all negative integers is

-136

The total sum of set

S

is the sum of its negative integers plus

0

plus the sum of its positive integers:

-31 = S_{\text{neg}} + 0 + 105 \implies S_{\text{neg}} = -136

.

3

Determine the count

p

of negative integers

The number of negative integers is

16

The negative integers are

-1, -2, \dots, -p

. Their sum is

-\frac{p(p+1)}{2} = -136 \implies p(p+1) = 272

. Solving

p(p+1) = 272

gives

p = 16

because

16 \times 17 = 272

.

Key Concept

Consecutive integer set properties and partitioning sets into positive and negative components using arithmetic series formulas.

Question 134Question →

A positive integer $N$ has the prime factorization $N = 3^a \cdot 5^b$ , where $a$ and $b$ are positive integers. If $N^2$ has exactly 35 positive integer divisors, what is the number of positive integer divisors of $N^3$ ?

Show answer & explanation

Answer: 70

Answer

70

Since

N = 3^a \cdot 5^b

,

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

. The number of positive divisors of

N^2

is

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

. The unique integer factor pair of 35 greater than 1 is

5 \times 7

, so the exponents

a

and

b

must be 2 and 3 (in some order). For

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

, the exponents are 6 and 9. Therefore, the number of positive divisors of

N^3

is

(6+1)(9+1) = 70

.

Step-by-Step Solution

1

Set up the formula for the number of positive divisors of

N^2

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

For an integer with prime factorization

p_1^{e_1} p_2^{e_2}

, the total number of divisors is

(e_1 + 1)(e_2 + 1)

.

2

Determine the positive integer values of

a

and

b

One exponent is 2 and the other exponent is 3

35 factors into

5 \times 7

. Solving

2a + 1 = 5

gives

a = 2

, and

2b + 1 = 7

gives

b = 3

.

3

Calculate the number of divisors of

N^3

(3(2) + 1)(3(3) + 1) = 7 \times 10 = 70

Exponents of

N^3

are

3a = 6

and

3b = 9

, so total divisors equal

(6 + 1)(9 + 1) = 70

.

Key Concept

Number of positive integer divisors from prime factorization

Estimated Time:1m 30s

Question 135Question →

Let $n = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. If the integer $\frac{n}{6}$ has exactly $32$ positive divisors and the integer $10n$ has exactly $90$ positive divisors, what is the value of $a + b + c$ ?

8

9

10

11

12

Show answer & explanation

Answer: 9

Answer

The value of

a + b + c

is

9

.

Writing the prime factorizations gives

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

with

a \cdot b \cdot (c+1) = 32

, and

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

with

(a+2)(b+1)(c+2) = 90

. Solving for positive integers

a

,

b

, and

c

yields

a = 4

,

b = 2

, and

c = 3

. Thus, the sum

a + b + c = 9

.

Step-by-Step Solution

1

Express the prime factorization of

\frac{n}{6}

and write its total divisor count equation.

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

. The number of positive divisors is

a \cdot b \cdot (c + 1) = 32

.

Dividing

n

by

6 = 2 \cdot 3

decreases the exponents of

2

and

3

by

1

each. The formula for the total number of divisors of a number

p_1^{x} p_2^{y} p_3^{z}

is

(x+1)(y+1)(z+1)

.

2

Express the prime factorization of

10n

and write its total divisor count equation.

10n = (2 \cdot 5) \cdot (2^a \cdot 3^b \cdot 5^c) = 2^{a+1} \cdot 3^b \cdot 5^{c+1}

. The number of positive divisors is

(a + 2) \cdot (b + 1) \cdot (c + 2) = 90

.

Multiplying by

10 = 2 \cdot 5

increases the exponents of

2

and

5

by

1

each.

3

Solve the system of equations for the positive integer exponents

a

,

b

, and

c

.

From

a \cdot b \cdot (c + 1) = 32

, test integer factors of

32

. Setting

c + 1 = 4 \implies c = 3

, we get

a \cdot b = 8

. Substituting

c = 3

into

(a + 2)(b + 1)(c + 2) = 90

yields

(a + 2)(b + 1)(5) = 90 \implies (a + 2)(b + 1) = 18

. Testing pairs where

a \cdot b = 8

: if

a = 4

and

b = 2

, then

(4 + 2)(2 + 1) = 6 \cdot 3 = 18

, which satisfies both equations.

Because

a, b, c

are positive integers, factor analysis uniquely pinpoints

a = 4

,

b = 2

, and

c = 3

.

4

Calculate

a + b + c

.

a + b + c = 4 + 2 + 3 = 9

.

Summing the derived values of the positive integer exponents.

Key Concept

Prime Factorization and Total Number of Divisors

Estimated Time:2m 0s

Question 136Question →

If $m$ and $n$ are positive integers such that $3^m \cdot 5^n$ is a factor of $15!$ , what is the maximum possible value of $m + n$ ?

8

9

10

11

12

Show answer & explanation

Answer: 9

Answer

9

To find the maximum power of a prime

p

dividing

n!

, we sum

\lfloor n/p^k \rfloor

for all

k \ge 1

. For

p = 3

,

m = \lfloor 15/3 \rfloor + \lfloor 15/9 \rfloor = 5 + 1 = 6

. For

p = 5

,

n = \lfloor 15/5 \rfloor = 3

. Thus, the maximum value of

m + n

is

6 + 3 = 9

.

Step-by-Step Solution

1

Find the maximum exponent

m

of the prime factor 3 in

15!

using Legendre's formula.

m = \lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{3^2} \rfloor = 5 + 1 = 6

The prime factor 3 appears in multiples of 3 (3, 6, 9, 12, 15) and contributes an extra factor in 9 (

3^2

).

2

Find the maximum exponent

n

of the prime factor 5 in

15!

using Legendre's formula.

n = \lfloor \frac{15}{5} \rfloor = 3

The prime factor 5 appears in multiples of 5 (5, 10, 15).

3

Sum the maximum possible integer values of

m

and

n

.

m + n = 6 + 3 = 9

The maximum possible value of

m+n

is the sum of the maximum individual prime exponents.

Key Concept

Counting Prime Factors in a Factorial (Legendre's Formula)

Alternative Method

List out the prime factorizations of all numbers from 1 to 15: 3 contributing numbers are 3, 6 (

2 \cdot 3

), 9 (

3^2

), 12 (

2^2 \cdot 3

), 15 (

3 \cdot 5

). Total factors of 3 =

1 + 1 + 2 + 1 + 1 = 6

. 5 contributing numbers are 5, 10 (

2 \cdot 5

), 15 (

3 \cdot 5

). Total factors of 5 =

1 + 1 + 1 = 3

. Sum

6 + 3 = 9

.

Estimated Time:1m 30s

Question 137Question →

If $k$ is a positive integer such that $\frac{3^{k+2} + 3^{k+4}}{10} = 3^{2k-1}$ , what is the value of $k$ ?

1

2

3

4

5

Show answer & explanation

Answer: 3

Answer

The value of

k

is 3.

Factoring

3^{k+2}

from the numerator gives

3^{k+2}(1 + 3^2) = 3^{k+2}(10)

. Dividing by 10 simplifies the left side to

3^{k+2}

. Setting

3^{k+2} = 3^{2k-1}

requires

k + 2 = 2k - 1

, which yields

k = 3

.

Step-by-Step Solution

1

Factor out the common exponential term

3^{k+2}

from the numerator on the left-hand side.

3^{k+2} + 3^{k+4} = 3^{k+2}(1 + 3^2) = 3^{k+2}(1 + 9) = 10 \cdot 3^{k+2}

When adding terms with identical bases, factor out the term with the smallest exponent to simplify the sum.

2

Substitute the factored expression into the fraction and simplify.

\frac{10 \cdot 3^{k+2}}{10} = 3^{k+2}

Canceling the common factor of 10 in the numerator and denominator simplifies the left-hand side.

3

Set the simplified left-hand side equal to the right-hand side of the original equation.

3^{k+2} = 3^{2k-1}

Both sides now have the same base of 3.

4

Equate the exponents since the bases are equal and solve for

k

.

k + 2 = 2k - 1 \implies 2 + 1 = 2k - k \implies k = 3

For any non-zero, non-one base

b

,

b^x = b^y

implies

x = y

.

Key Concept

Factoring exponential expressions with identical bases and equating exponents.

Question 138Question →

Let $N = 2^4 \times 3^3 \times 5^2 \times 7$ . How many positive integer divisors of $N$ are even, divisible by 15, but not divisible by 9?

Show answer & explanation

Answer: 16

Answer

16

Any divisor of

N = 2^4 \times 3^3 \times 5^2 \times 7^1

takes the form

2^a \times 3^b \times 5^c \times 7^d

. The condition that the divisor is even requires

a \ge 1

, which gives 4 choices (

a \in \{1, 2, 3, 4\}

). The condition that the divisor is divisible by 15 requires

b \ge 1

and

c \ge 1

. The condition that it is not divisible by 9 requires

b < 2

. Together,

b \ge 1

and

b < 2

mean

b = 1

(1 choice). The condition

c \ge 1

allows

c \in \{1, 2\}

(2 choices). The exponent

d

can be 0 or 1 (2 choices). Multiplying these choices gives

4 \times 1 \times 2 \times 2 = 16

.

Step-by-Step Solution

1

Express the general prime factorization of a divisor

d = 2^a \times 3^b \times 5^c \times 7^d

with

0 \le a \le 4

,

0 \le b \le 3

,

0 \le c \le 2

,

0 \le d \le 1

Any divisor of

N

must consist only of the prime factors of

N

up to their respective maximum powers.

2

Apply the condition for even numbers

a \in \{1, 2, 3, 4\}

(4 choices)

An even integer must contain at least one factor of 2.

3

Apply the condition for divisibility by 15 and non-divisibility by 9

b = 1

(1 choice) and

c \in \{1, 2\}

(2 choices)

Divisibility by

15 = 3 \times 5

requires

b \ge 1

and

c \ge 1

. Non-divisibility by

9 = 3^2

requires

b < 2

. Thus

b

must be exactly 1.

4

Apply the condition for the exponent of 7

d \in \{0, 1\}

(2 choices)

There are no restrictions given for prime factor 7.

5

Calculate the product of choices

4 \times 1 \times 2 \times 2 = 16

By the fundamental counting principle, the total number of valid combinations of prime exponents is the product of the number of choices for each exponent.

Key Concept

Counting Divisors with Prime Factorization Restrictions

Question 139Question →

Set $S$ consists of a sequence of consecutive integers, ordered from least to greatest. If Set $S$ contains exactly $45$ integers and the sum of all the integers in Set $S$ is $405$ , what is the value of the smallest integer in Set $S$ ?

Show answer & explanation

Answer: -13

Answer

The smallest integer in Set

S

is

-13

.

The average (arithmetic mean) of the set is calculated by dividing the total sum (

405

) by the number of terms (

45

), which equals

9

. In an evenly spaced set with an odd number of elements, the mean is equal to the median (middle term). Because there are

45

terms, exactly

22

terms lie below the median. Subtracting

22

from

9

yields

-13

as the smallest integer in the set.

Step-by-Step Solution

1

Calculate the arithmetic mean and median of the set

Arithmetic mean = Median = 9

For an evenly spaced set, the arithmetic mean is equal to the median. Dividing the sum (

405

) by the number of terms (

45

) gives

9

.

2

Determine the number of terms preceding the median

22 terms precede the median

With

45

terms in total, the median is the

23\text{rd}

term, which leaves

\frac{45 - 1}{2} = 22

terms smaller than the median.

3

Calculate the smallest integer

Smallest integer =

-13

Subtracting

22

from the median gives

9 - 22 = -13

.

Key Concept

In any set of consecutive integers with

n

terms, the arithmetic mean equals the median. If

n

is odd, the smallest integer is given by

\text{Median} - \frac{n-1}{2}

.

Question 140Question →

Let $x$ and $y$ be non-zero integers such that $-6 \le x \le 6$ and $-6 \le y \le 6$ . If $x^3 y < 0$ and $\frac{x}{y^2} < -\frac{1}{4}$ , what is the minimum possible value of $x - y$ ?

Show answer & explanation

Answer: -10

Answer

The minimum possible value of

x - y

is

-10

.

Analyzing the signs shows x must be negative and y must be positive. Multiplying x/y² < -1/4 by 4y² gives y² < -4x. Testing x = -6 gives y² < 24, so the largest positive integer y is 4. The expression x - y reaches its minimum value of -6 - 4 = -10.

Step-by-Step Solution

1

Determine the signs of variables x and y

x < 0 and y > 0

Since y² > 0 for any non-zero integer y, x/y² < -1/4 requires x < 0. Furthermore, x³y < 0 requires x³ and y to have opposite signs; since x < 0 implies x³ < 0, y must be positive.

2

Transform the inequality without altering its sign direction

y² < -4x

Multiplying x/y² < -1/4 by 4y² > 0 yields 4x < -y², which rearranges to y² < -4x.

3

Evaluate candidate integer pairs to minimize x - y

The minimum value of x - y is -10 when x = -6 and y = 4

To minimize x - y, select the most negative integer x and the largest allowable positive integer y. Setting x = -6 gives y² < 24, making max integer y = 4. Therefore, x - y = -6 - 4 = -10.

Key Concept

Deducing signs from products and quotients, and manipulating inequalities involving non-zero variables.

Number Properties and Arithmetic

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Common Mistakes

Answer

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution