Number Properties and Arithmetic Practice Questions — GMAT — Page 10

Question 181Question →

A cloud data center processes an incoming raw data stream through sequential filtering stages. In Stage 1, $\frac{3}{8}$ of the incoming raw data volume is discarded as noise, and $0.20$ of the remaining data is flagged for long-term archiving. In Stage 2, $\frac{4}{5}$ of the data not flagged for long-term archiving is processed into active storage, while the rest is discarded. If the volume of data processed into active storage in Stage 2 is $50$ terabytes greater than the volume of data discarded in Stage 1, what was the initial volume of the raw data stream, in terabytes?

400

1,600

2,000

2,500

4,000

Show answer & explanation

Answer: 2,000

Answer

2,000 terabytes

Let the initial data volume be

X

. Discarded in Stage 1 is

\frac{3}{8}X

, leaving

\frac{5}{8}X

. Archiving takes

20\%

(

0.20 = \frac{1}{5}

) of this remainder, which is

\frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X

. The data entering Stage 2 is

\frac{5}{8}X - \frac{1}{8}X = \frac{1}{2}X

. Active storage in Stage 2 is

\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X

. Subtracting Stage 1 discarded volume from Stage 2 active storage volume gives

\frac{2}{5}X - \frac{3}{8}X = \frac{16}{40}X - \frac{15}{40}X = \frac{1}{40}X

. Setting

\frac{1}{40}X = 50

yields

X = 2,000

terabytes.

Step-by-Step Solution

1

Define the variable and compute Stage 1 discarded volume and remaining volume.

Discarded in Stage 1 =

\frac{3}{8}X

; Remaining after Stage 1 discard =

X - \frac{3}{8}X = \frac{5}{8}X

.

Establishing quantities in terms of the total initial volume

X

allows setting up a single-variable linear equation.

2

Calculate the volume flagged for archiving and the volume available for Stage 2.

Archived volume =

0.20 \times \frac{5}{8}X = \frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X

. Available for Stage 2 = \frac{5}{8}X - \frac{1}{8}X = \frac{4}{8}X = \frac{1}{2}X$.

The

0.20

decimal must be converted to a fraction (

\frac{1}{5}

) and applied to the remaining

\frac{5}{8}X

base.

3

Calculate the volume processed into active storage in Stage 2.

Active storage volume =

\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X

.

Stage 2 processes

\frac{4}{5}

of the data that entered Stage 2.

4

Set up the algebraic equation comparing active storage volume in Stage 2 to discarded volume in Stage 1.

\frac{2}{5}X - \frac{3}{8}X = 50

.

The problem states that active storage volume in Stage 2 is

50

terabytes greater than the Stage 1 discarded volume.

5

Find a common denominator and solve for

X

.

\left(\frac{16}{40} - \frac{15}{40}\right)X = 50 \implies \frac{1}{40}X = 50 \implies X = 2,000

.

Converting fractions to a denominator of

40

yields

\frac{1}{40}X = 50

, so multiplying by

40

gives

X = 2,000

terabytes.

Key Concept

Multi-step arithmetic operations involving sequential fractions, decimals, and percent bases

Estimated Time:2m 0s

Question 182Question →

Let $K = 7^{83} - 4^{42}$ . What is the remainder when $-K$ is divided by $5$ ?

1

2

3

4

0

Show answer & explanation

Answer: 3

Answer

3

To find the remainder of

-K

divided by 5, first calculate

K \pmod 5

. By analyzing powers of 7 modulo 5,

7^1 \equiv 2, 7^2 \equiv 4, 7^3 \equiv 3, 7^4 \equiv 1

, so

7^{83} = 7^{4(20)+3} \equiv 3 \pmod 5

. For powers of 4 modulo 5,

4^1 \equiv 4, 4^2 \equiv 1

, so

4^{42} \equiv 1 \pmod 5

. Therefore,

K \equiv 3 - 1 = 2 \pmod 5

. Consequently,

-K \equiv -2 \pmod 5

. Adding the divisor 5 gives a non-negative remainder of 3.

Step-by-Step Solution

1

Find the remainder of

7^{83}

when divided by

5

.

7^1 \equiv 2

,

7^2 \equiv 4

,

7^3 \equiv 3

,

7^4 \equiv 1 \pmod 5

. Since

83 = 4 \times 20 + 3

,

7^{83} \equiv 7^3 \equiv 3 \pmod 5

.

Powers of 7 follow a 4-term repeating remainder cycle modulo 5: (2, 4, 3, 1).

2

Find the remainder of

4^{42}

when divided by

5

.

4^1 \equiv 4

,

4^2 \equiv 1 \pmod 5

. Since 42 is even,

4^{42} \equiv 1 \pmod 5

.

Powers of 4 follow a 2-term repeating remainder cycle modulo 5: (4, 1).

3

Compute

K \pmod 5

.

K \equiv 3 - 1 = 2 \pmod 5

.

Modular subtraction allows evaluating terms independently.

4

Determine the positive remainder of

-K \pmod 5

.

-K \equiv -2 \equiv -2 + 5 = 3 \pmod 5

.

Standard division algorithm requires the remainder

r

when dividing by 5 to satisfy

0 \le r < 5

.

Key Concept

Modular cyclicity and negative remainder conversion

Estimated Time:2m 0s

Question 183Question →

If the sum of three consecutive even integers is $42$ , what is the smallest of these integers?

10

12

13

14

16

Show answer & explanation

Answer: $12$

Answer

12

The arithmetic mean (average) of three consecutive even integers is equal to the middle integer. Dividing the total sum

42

by

3

yields

14

, which is the middle integer. Subtracting

2

gives the smallest integer,

12

.

Step-by-Step Solution

1

Define the three consecutive even integers algebraically

Let the three consecutive even integers be

n

,

n + 2

, and

n + 4

, where

n

represents the smallest integer.

Consecutive even integers differ by

2

.

2

Set up the equation for their sum

n + (n + 2) + (n + 4) = 42 \implies 3n + 6 = 42

The problem states that the sum of the three integers equals

42

.

3

Solve for

n

3n = 36 \implies n = 12

Subtract

6

from both sides and divide by

3

to find the smallest integer.

Key Concept

Properties and Sums of Consecutive Even Integers

Estimated Time:45s

Question 184Question →

What is the remainder when $5^{99}$ is divided by $13$ ?

Show answer & explanation

Answer: 8

Answer

The remainder when

5^{99}

is divided by

13

is

8

.

To find the remainder of

5^{99}

divided by

13

, find the repeating pattern of remainders for powers of

5

modulo

13

. The first four powers yield remainders

5, 12, 8,

and

1

. Since

5^4 \equiv 1 \pmod{13}

, the remainders repeat every

4

powers. Dividing the exponent

99

by

4

gives

99 = 4 \times 24 + 3

. The remainder of

3

indicates that

5^{99}

has the exact same remainder modulo

13

as

5^3

, which is

8

.

Step-by-Step Solution

1

Analyze the remainders of powers of 5 modulo 13 to identify the repeating period length.

5^1 \equiv 5 \pmod{13}

,

5^2 \equiv 12 \pmod{13}

,

5^3 \equiv 8 \pmod{13}

, and

5^4 \equiv 1 \pmod{13}

. The sequence of remainders repeats every 4 powers.

Integral powers modulo a positive integer exhibit periodic behavior.

2

Compute the remainder of the exponent 99 divided by the cycle length 4.

99 \div 4 = 24

with a remainder of

3

.

The exponent's remainder modulo the cycle length determines which element in the periodic cycle gives the equivalent value.

3

Evaluate the 3rd term in the remainder cycle.

The 3rd element in the sequence

(5, 12, 8, 1)

is

8

.

An exponent congruent to

3 \pmod 4

yields the same remainder as

5^3 \pmod{13}

.

Key Concept

Modular Arithmetic and Cyclicity of Powers

Question 185Question →

At a chemical refining plant, a raw liquid compound containing Substance X, Substance Y, and an inert solvent is processed in two sequential purification phases. Initially, Substance X accounts for $0.25$ of the total weight of the compound, and Substance Y accounts for $\frac{2}{5}$ of the remaining weight, with the inert solvent comprising the rest. In Phase 1, $20\%$ of Substance X and $30\%$ of Substance Y are removed, while all of the inert solvent is retained. In Phase 2, a certain percentage $p\%$ of the inert solvent present after Phase 1 is removed, while no other substances are removed. If Substance X represents exactly $40\%$ of the total weight of the compound remaining after Phase 2, what is the value of $p$ ?

Show answer & explanation

Answer: 80

Answer

The value of

p

is 80.

Assuming a total initial weight of 100 units, Substance X is 25 units and the remaining weight is 75 units. Substance Y is

\frac{2}{5}

of 75, which equals 30 units, leaving 45 units of inert solvent. After Phase 1, 20 units of Substance X and 21 units of Substance Y remain, along with the full 45 units of solvent. In Phase 2, Substance X (20 units) becomes

40\%

of the total mixture, making the final total weight

\frac{20}{0.40} = 50

units. Since Substance X and Substance Y together account for

20 + 21 = 41

units, the remaining solvent after Phase 2 must be

50 - 41 = 9

units. Reducing solvent from 45 units down to 9 units requires removing

\frac{45 - 9}{45} = \frac{36}{45} = 80\%

of the solvent. Thus,

p = 80

.

Step-by-Step Solution

1

Determine initial component amounts using decimal and fractional breakdown.

In a 100-unit mixture, Substance X = 25 units, Substance Y = 30 units, and Inert Solvent = 45 units.

Substance X is

0.25

of the total (

25

units). Of the remaining

75

units, Substance Y is

\frac{2}{5} \times 75 = 30

units. The rest (

75 - 30 = 45

units) is inert solvent.

2

Calculate remaining component amounts after Phase 1 percentage reductions.

Substance X = 20 units, Substance Y = 21 units, Inert Solvent = 45 units.

Removing

20\%

of Substance X leaves

25 \times 0.80 = 20

units. Removing

30\%

of Substance Y leaves

30 \times 0.70 = 21

units. No solvent is removed in Phase 1.

3

Determine the final total mixture weight using the final percentage of Substance X.

Final total mixture weight = 50 units.

Substance X (20 units) represents

40\%

(

0.40

) of the final mixture after Phase 2, so the total weight is

\frac{20}{0.40} = 50

units.

4

Formulate and solve the linear equation for

p

.

p = 80

.

The total weight is the sum of all remaining components:

20 + 21 + 45\left(1 - \frac{p}{100}\right) = 50

. Solving

41 + 45\left(1 - \frac{p}{100}\right) = 50

yields

45\left(1 - \frac{p}{100}\right) = 9

, so

1 - \frac{p}{100} = 0.20

, giving

p = 80

.

Key Concept

Multi-step percentage change, fractional remaining parts, and algebraic mixture equations

Question 186Question →

An inventory manager is organizing items into storage boxes. When the items are packed into boxes of $24$ , $36$ , or $45$ items each, there are always $11$ items remaining unpacked. If the total number of items is between $700$ and $1,000$ , how many items are there in total?

Show answer & explanation

Answer: 731

Answer

731

The number of items

N

satisfies

N \equiv 11 \pmod{24}

,

N \equiv 11 \pmod{36}

, and

N \equiv 11 \pmod{45}

. Consequently,

N - 11

is a multiple of

\text{LCM}(24, 36, 45)

. The prime factorizations are

24 = 2^3 \times 3

,

36 = 2^2 \times 3^2

, and

45 = 3^2 \times 5

. The LCM is

2^3 \times 3^2 \times 5 = 360

. Thus,

N = 360k + 11

for some positive integer

k

. Testing values of

k

within the range

700 < N < 1,000

yields

k = 2

, giving

N = 360(2) + 11 = 731

.

Step-by-Step Solution

1

Find the Least Common Multiple (LCM) of the divisors 24, 36, and 45.

LCM(24, 36, 45) = 360

Any number that leaves a constant remainder when divided by a set of numbers must be a constant remainder greater than a common multiple of those numbers.

2

Formulate the general equation for the total number of items N.

N = 360k + 11

Adding the common remainder 11 to any multiple of the LCM yields a valid total number of items.

3

Evaluate integer values of k to satisfy the constraint 700 < N < 1,000.

k = 2 gives N = 731

k = 1 gives 371, which is too small, and k = 3 gives 1,091, which exceeds 1,000.

Key Concept

Least Common Multiple (LCM) with Constant Remainder

Question 187Question →

At the beginning of a fiscal year, a technology company allocated its total operating budget among three divisions: Research & Development, Marketing, and Operations. Research & Development received $0.45$ of the total budget, Marketing received $\frac{1}{3}$ of the total budget, and Operations received the remaining portion. Mid-year, the company adjusted the budget by increasing the Research & Development allocation by $20\%$ of its initial amount and reducing the Marketing allocation by $\frac{1}{4}$ of its initial amount, while the Operations allocation remained unchanged. If these mid-year adjustments resulted in a net increase of $\$30,000$ in the company's total operating budget, what was the company's total operating budget at the beginning of the fiscal year?

$187,500

$600,000

$2,700,000

$3,600,000

$4,500,000

Show answer & explanation

Answer: $4,500,000

Answer

$4,500,000

The option specifying

4,500,000 is correct. The increase in the Research & Development division is

20\%

of

0.45 B

, which equals

0.09 B

(or

\frac{9}{100} B

). The reduction in Marketing is

\frac{1}{4}

of

\frac{1}{3} B

, which equals

\frac{1}{12} B

. Combining these using a common denominator gives a net increase of

\frac{27}{300} B - \frac{25}{300} B = \frac{2}{300} B = \frac{1}{150} B

. Setting

\frac{1}{150} B = \

30,000

yields

B = \$4,500,000

.

Step-by-Step Solution

1

Express initial allocations as fractions of total budget B

Research & Development initial share =

0.45 B = \frac{9}{20} B

; Marketing initial share =

\frac{1}{3} B

Converting decimals to fractions simplifies exact combined calculations.

2

Calculate the monetary change in each division relative to total budget B

R&D increase =

20\% \times 0.45 B = 0.20 \times 0.45 B = 0.09 B = \frac{9}{100} B

. Marketing decrease =

\frac{1}{4} \times \frac{1}{3} B = \frac{1}{12} B

. Operations change =

0

.

Each change applies specifically to the division's initial allocation, not the total budget.

3

Find the net change in the total budget by taking the difference between the increase and the decrease

Net change =

\left(\frac{9}{100} - \frac{1}{12}\right) B = \left(\frac{27}{300} - \frac{25}{300}\right) B = \frac{2}{300} B = \frac{1}{150} B

Determining the net fraction allows direct equation setup with the given dollar increase.

4

Solve for the initial total operating budget B using the given net increase of $30,000

\frac{1}{150} B = 30,000 \implies B = 30,000 \times 150 = 4,500,000

Multiplying both sides by 150 yields the original total budget.

Key Concept

Combining decimal and fraction arithmetic to determine net percentage changes across component allocations.

Alternative Method

Convert all components to a common denominator out of 300 parts initially. Total budget = 300 parts. R&D = 135 parts (0.45). Marketing = 100 parts (1/3). Operations = 65 parts. R&D increase = 20% of 135 = +27 parts. Marketing decrease = 1/4 of 100 = -25 parts. Net change = +2 parts =

30,000. Therefore 1 part =

15,000, and 300 parts = $4,500,000.

Estimated Time:1m 45s

Question 188Question →

Let $N = 19^{43} + 11^{43} - 6^{43}$ . What is the remainder when $N$ is divided by $7$ ?

Show answer & explanation

Answer: 3

Answer

3

Reducing each base modulo 7 simplifies the expression significantly:

19 \equiv -2 \pmod{7}

,

11 \equiv 4 \pmod{7}

, and

6 \equiv -1 \pmod{7}

. Utilizing the cyclicity period of 3 for powers of 2 modulo 7 (

2^3 \equiv 1 \pmod{7}

),

19^{43} \equiv -(2^{43}) \equiv -2 \equiv 5 \pmod{7}

. Similarly,

11^{43} \equiv 4^{43} = 2^{86} = (2^3)^{28} \times 4 \equiv 4 \pmod{7}

. Lastly,

6^{43} \equiv (-1)^{43} = -1 \pmod{7}

. Combining these yields

N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}

.

Step-by-Step Solution

1

Reduce each base modulo 7

19 \equiv -2 \pmod{7}

,

11 \equiv 4 \pmod{7}

, and

6 \equiv -1 \pmod{7}

Simplifying bases using modular arithmetic reduces large power evaluations to smaller equivalents.

2

Evaluate

19^{43} \pmod{7}

using cyclicity

19^{43} \equiv -2 \equiv 5 \pmod{7}

Since

2^3 \equiv 1 \pmod{7}

and

43 = 3(14) + 1

,

2^{43} \equiv 2 \pmod{7}

, so

(-2)^{43} = -(2^{43}) \equiv -2 \pmod{7}

.

3

Evaluate

11^{43} \pmod{7}

using powers of 2

11^{43} \equiv 4 \pmod{7}

11^{43} \equiv 4^{43} = 2^{86} = (2^3)^{28} \times 2^2 \equiv 1^{28} \times 4 = 4 \pmod{7}

.

4

Evaluate

6^{43} \pmod{7}

6^{43} \equiv -1 \pmod{7}

Any odd power of

-1

equals

-1

.

5

Combine all components to determine the final remainder

N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}

Summing the individual remainders modulo 7 gives a final positive remainder of 3.

Key Concept

Modular Arithmetic and Cyclicity of Powers

Question 189Question →

A positive integer $n$ is divisible by $20$ but is not divisible by $8$ . If $n$ has exactly $15$ positive divisors and $10n$ has exactly $24$ positive divisors, what is the value of $n$ ?

Show answer & explanation

Answer: 2500

Answer

2500

To determine

n

, analyze its prime factorization. Divisibility by

20 = 2^2 \cdot 5^1

requires that

n

contains

2

raised to at least the power of

2

and

5

raised to at least the power of

1

. The condition that

n

is not divisible by

8 = 2^3

restricts the exponent of

2

to exactly

2

. Thus,

n = 2^2 \cdot 5^b \cdot K'

, where

b \ge 1

and

K'

is a product of powers of distinct prime factors other than

2

and

5

. The number of positive divisors of

n

is

d(n) = (2+1)(b+1)K = 3(b+1)K = 15

, which simplifies to

(b+1)K = 5

. Because

b \ge 1

, we have

b+1 \ge 2

. Since

5

is a prime number, its only divisor greater than or equal to

2

is

5

. Thus,

b+1 = 5

, giving

b = 4

, and

K = 1

, which means

n

has no prime factors other than

2

and

5

. Therefore,

n = 2^2 \cdot 5^4 = 4 \cdot 625 = 2500

. We verify that

10n = 2^3 \cdot 5^5

has

(3+1)(5+1) = 24

positive divisors, confirming the solution.

Step-by-Step Solution

1

Determine the power of 2 in the prime factorization of

n

The exponent of

2

in

n

is exactly

2

Because

n

is a multiple of

20 = 2^2 \cdot 5

but not a multiple of

8 = 2^3

,

2^2

divides

n

but

2^3

does not.

2

Set up the divisor counting formula for

n

d(n) = (2+1)(b+1)K = 15 \implies (b+1)K = 5

, where

b \ge 1

is the exponent of

5

and

K

represents the product of terms from any additional prime factors

The total number of positive divisors of an integer

N = p_1^{a_1} p_2^{a_2} \dots

is given by

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

3

Solve for the exponents and prime factors of

n

b = 4

and

K = 1

, giving

n = 2^2 \cdot 5^4

Since

n

is divisible by

20

,

5

is a prime factor of

n

, so

b \ge 1

, which implies

b+1 \ge 2

. Since

5

is prime, its only factor greater than

1

is

5

, forcing

b+1 = 5

and

K = 1

.

4

Verify

d(10n)

and evaluate

n

10n = 2^3 \cdot 5^5

has

(3+1)(5+1) = 24

divisors, and

n = 4 \cdot 625 = 2500

Multiplying

n

by

10 = 2 \cdot 5

increases the exponent of

2

from

2

to

3

and the exponent of

5

from

4

to

5

.

Key Concept

Divisor Count Formula and Prime Factorization Constraints

Question 190Question →

Let $S = \sum_{k=1}^{82} (-1)^k \cdot k \cdot 7^k$ . What is the remainder when $-S$ is divided by $10$ ?

1

3

7

9

0

Show answer & explanation

Answer: 9

Answer

The remainder when

-S

is divided by

10

is

9

.

Grouping the 82 terms into 20 blocks of 4 shows that each block contributes

6 \pmod{10}

, summing to

120 \equiv 0 \pmod{10}

. The 81st and 82nd terms contribute

3

and

8

respectively, giving

S \equiv 11 \equiv 1 \pmod{10}

. Therefore,

-S \equiv -1 \equiv 9 \pmod{10}

.

Step-by-Step Solution

1

Analyze the units digit cyclicity of

7^k \pmod{10}

.

The sequence of units digits for

7^k

repeats every 4 powers:

7^1 \equiv 7

,

7^2 \equiv 9

,

7^3 \equiv 3

,

7^4 \equiv 1 \pmod{10}

.

Units digits of powers of 7 follow a repeating cycle of length 4.

2

Evaluate the terms modulo 10 for a general block of 4 consecutive integers

k = 4m+1, 4m+2, 4m+3, 4m+4

.

The four terms evaluate to:

(-1)^{4m+1}(4m+1)7^{4m+1} \equiv -(4m+1)(7) \equiv 2m+3 \pmod{10}

;

(-1)^{4m+2}(4m+2)7^{4m+2} \equiv (4m+2)(9) \equiv 6m+8 \pmod{10}

;

(-1)^{4m+3}(4m+3)7^{4m+3} \equiv -(4m+3)(3) \equiv 8m+1 \pmod{10}

; and

(-1)^{4m+4}(4m+4)7^{4m+4} \equiv (4m+4)(1) \equiv 4m+4 \pmod{10}

.

Modular arithmetic simplifies each term according to its position in the 4-term cycle.

3

Sum the four terms in a single block of 4.

(2m+3) + (6m+8) + (8m+1) + (4m+4) = 20m + 16 \equiv 6 \pmod{10}.

Every complete block of 4 consecutive terms contributes exactly 6 modulo 10, independent of

m

.

4

Calculate the total sum

S \pmod{10}

for

k=1

to

82

.

82 terms contain 20 full blocks of 4 terms (

k=1

to

80

) plus 2 remaining terms (

k=81

and

k=82

). The 20 blocks contribute

20 \times 6 = 120 \equiv 0 \pmod{10}

. For

m=20

,

k=81

gives

2(20)+3 = 43 \equiv 3 \pmod{10}

, and

k=82

gives

6(20)+8 = 128 \equiv 8 \pmod{10}

. Thus

S \equiv 0 + 3 + 8 = 11 \equiv 1 \pmod{10}

.

Summing the contributions of all complete blocks and the remaining boundary terms gives

S \pmod{10}

.

5

Determine the positive remainder of

-S

when divided by 10.

Since

S \equiv 1 \pmod{10}

,

-S \equiv -1 \equiv 10 - 1 = 9 \pmod{10}

.

By definition of integer division, the remainder

r

must satisfy

0 \le r < 10

.

Key Concept

Units Digit Cyclicity and Modular Arithmetic with Negative Integer Remainders

Question 191Question →

Let $N = 2^3 \cdot 3^a \cdot 5^b$ , where $a$ and $b$ are positive integers. If $N$ has exactly 60 positive integer divisors that are multiples of 10 and exactly 40 positive integer divisors that are multiples of 15, how many positive integer divisors of $N$ are multiples of 12?

20

22

24

33

66

Show answer & explanation

Answer: 22

Answer

22

The correct answer is 22. From the problem statement, divisors divisible by 10 require

x \in \{1, 2, 3\}

,

y \in \{0, \dots, a\}

, and

z \in \{1, \dots, b\}

, leading to

3(a+1)b = 60

, or

(a+1)b = 20

. Divisors divisible by 15 require

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

,

y \in \{1, \dots, a\}

, and

z \in \{1, \dots, b\}

, leading to

4ab = 40

, or

ab = 10

. Substituting

ab = 10

into

ab + b = 20

yields

b = 10

and

a = 1

. To be a multiple of

12 = 2^2 \cdot 3^1

, a divisor must have

x \in \{2, 3\}

(2 choices),

y \in \{1\}

(1 choice), and

z \in \{0, 1, \dots, 10\}

(11 choices). The total number of such divisors is

2 \times 1 \times 11 = 22

.

Step-by-Step Solution

1

Set up the equation for positive divisors of

N

that are multiples of 10.

Multiples of 10 must be divisible by

2^1 \cdot 5^1

. Thus, for a divisor

2^x \cdot 3^y \cdot 5^z

, we need

1 \le x \le 3

(3 choices),

0 \le y \le a

(

a+1

choices), and

1 \le z \le b

(

b

choices). The total number of such divisors is

3(a + 1)b = 60

, which simplifies to

(a + 1)b = 20

.

Divisibility by 10 requires at least one factor of 2 and at least one factor of 5.

2

Set up the equation for positive divisors of

N

that are multiples of 15.

Multiples of 15 must be divisible by

3^1 \cdot 5^1

. Thus, we need

0 \le x \le 3

(4 choices),

1 \le y \le a

(

a

choices), and

1 \le z \le b

(

b

choices). The total number of such divisors is

4ab = 40

, which simplifies to

ab = 10

.

Divisibility by 15 requires at least one factor of 3 and at least one factor of 5.

3

Solve the system of equations for

a

and

b

.

Expanding

(a + 1)b = 20

yields

ab + b = 20

. Substituting

ab = 10

into this equation gives

10 + b = 20 \implies b = 10

. Since

ab = 10

, we find

a(10) = 10 \implies a = 1

.

Determining the exact exponents

a

and

b

allows us to fully specify

N = 2^3 \cdot 3^1 \cdot 5^{10}

.

4

Calculate the number of divisors of

N

that are multiples of 12.

Multiples of 12 must be divisible by

12 = 2^2 \cdot 3^1

. For a divisor

2^x \cdot 3^y \cdot 5^z

, we require

2 \le x \le 3

(2 choices:

2^2, 2^3

),

1 \le y \le a = 1

(1 choice:

3^1

), and

0 \le z \le b = 10

(

b + 1 = 11

choices:

5^0, 5^1, \dots, 5^{10}

). Total multiples of 12 =

2 \times 1 \times 11 = 22

.

Applying the fundamental counting principle to the constrained prime factor exponent ranges.

Key Concept

Divisor counting formula under prime factor constraint conditions

Question 192Question →

If $x$ is a real number such that $\sqrt[3]{\frac{9^{2x+1} \cdot 27^{x-1}}{3^{x+2}}} = 243$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

Converting all terms inside the radical to powers of 3 yields

\sqrt[3]{\frac{3^{4x+2} \cdot 3^{3x-3}}{3^{x+2}}} = \sqrt[3]{3^{6x-3}} = 3^{2x-1}

. Setting

3^{2x-1} = 243 = 3^5

gives

2x - 1 = 5

, which solves to

x = 3

.

Step-by-Step Solution

1

Express all exponential terms inside the radical using a common base of 3.

9^{2x+1} = 3^{4x+2}

and

27^{x-1} = 3^{3x-3}

.

Unifying bases allows the application of exponent multiplication and division rules.

2

Simplify the algebraic expression inside the cube root.

\frac{3^{4x+2} \cdot 3^{3x-3}}{3^{x+2}} = 3^{(4x+2) + (3x-3) - (x+2)} = 3^{6x-3}

.

Adding exponents for multiplication of like bases and subtracting for division.

3

Simplify the radical expression using fractional exponent rules.

\sqrt[3]{3^{6x-3}} = (3^{6x-3})^{1/3} = 3^{2x-1}

.

Taking the

n

-th root of a power corresponds to dividing the exponent by

n

.

4

Solve for

x

by equating the exponent to the prime factorized form of 243.

3^{2x-1} = 3^5 \implies 2x - 1 = 5 \implies x = 3

.

Since the bases are identical and non-zero, the exponents must be equal.

Key Concept

Exponent and radical rules with common base conversion

Question 193Question →

An investment fund allocated its total initial capital among three venture portfolios: Healthcare, Clean Tech, and Artificial Intelligence. Initially, $\frac{2}{5}$ of the capital was allocated to Healthcare and $0.35$ was allocated to Clean Tech, with the remaining fraction allocated to Artificial Intelligence. During a mid-year portfolio rebalancing, the amount in Healthcare was increased by $25\%$ , the amount in Clean Tech was decreased by $20\%$ , and the amount in Artificial Intelligence remained unchanged. If the total capital across all three portfolios after rebalancing was $\$5,150,000$ , what was the initial amount, in dollars, allocated to Artificial Intelligence?

$1,250,000

$1,287,500

$1,400,000

$1,750,000

$2,000,000

Show answer & explanation

Answer: $1,250,000

Answer

The initial amount allocated to Artificial Intelligence was $1,250,000 dollars.

The correct option of

1,250,000 is obtained by first calculating the initial proportion allocated to Artificial Intelligence as

1 - 0.40 - 0.35 = 0.25

. Applying the percentage adjustments yields Healthcare at

0.50C

, Clean Tech at

0.28C

, and Artificial Intelligence at

0.25C

. The sum of these updated allocations gives

1.03C = \

5,150,000

, which implies

C = \$5,000,000

. Taking

25\%

of

\$5,000,000

yields

\$1,250,000

.

Step-by-Step Solution

1

Determine the initial decimal fraction allocated to each portfolio.

Healthcare =

\frac{2}{5} = 0.40

; Clean Tech =

0.35

; Artificial Intelligence =

1 - (0.40 + 0.35) = 0.25

.

The sum of all initial portfolio fractions must equal

1

.

2

Express the post-rebalancing portfolio values in terms of the initial total capital

C

.

Healthcare =

0.40C \times (1 + 0.25) = 0.50C

; Clean Tech =

0.35C \times (1 - 0.20) = 0.28C

; Artificial Intelligence =

0.25C

.

A

25\%

increase scales a value by

1.25

, and a

20\%

decrease scales a value by

0.80

.

3

Sum the new portfolio values to set up an equation for total capital

C

.

0.50C + 0.28C + 0.25C = 1.03C = 5,150,000

.

The sum of all rebalanced portfolios equals the given final capital of

\$5,150,000

.

4

Solve for initial total capital

C

and calculate the Artificial Intelligence allocation.

C = \frac{5,150,000}{1.03} = 5,000,000

; Artificial Intelligence =

0.25 \times 5,000,000 = 1,250,000

.

Multiplying the initial total capital by

0.25

yields the original Artificial Intelligence allocation.

Key Concept

Combining sequential percentage changes and decimal/fraction conversions to solve multi-step financial allocation word problems.

Question 194Question →

For how many positive integers $n$ is the expression $\frac{n^2 + 7n + 120}{n + 3}$ equal to an integer?

6

9

10

11

12

Show answer & explanation

Answer: 9

Answer

9 positive integers

By dividing the numerator by the denominator, the expression simplifies to

n + 4 + \frac{108}{n + 3}

. For this to yield an integer,

(n + 3)

must be a positive divisor of 108. The number 108 has prime factorization

2^2 \cdot 3^3

, giving it

(2+1)(3+1) = 12

total positive divisors. Since

n

must be a positive integer (

n \ge 1

),

n + 3

must be at least 4. Eliminating the divisors 1, 2, and 3 leaves 9 valid values for

n + 3

, which correspond to 9 unique positive integers

n

.

Step-by-Step Solution

1

Perform polynomial long division or algebraic manipulation on the numerator

\frac{n^2 + 7n + 120}{n + 3} = \frac{(n^2 + 3n) + (4n + 12) + 108}{n + 3} = \frac{n(n+3) + 4(n+3) + 108}{n + 3} = n + 4 + \frac{108}{n + 3}

Separating the expression into an integer term

(n + 4)

and a proper fractional term

\frac{108}{n + 3}

isolates the divisibility condition.

2

Determine the condition for the fractional term to be an integer

The expression is an integer if and only if

(n + 3)

is a positive integer divisor of 108.

Since

n

is a positive integer,

(n + 4)

is always an integer, so

\frac{108}{n + 3}

must also be an integer.

3

Calculate the total number of positive divisors of 108

Prime factorization:

108 = 2^2 \cdot 3^3

. Total divisors =

(2+1)(3+1) = 3 \cdot 4 = 12

.

The formula for the number of positive divisors of

p_1^{a} p_2^{b}

is

(a+1)(b+1)

.

4

Apply the constraint that

n

is a positive integer (

n \ge 1

)

Since

n \ge 1

, we have

n + 3 \ge 4

. The divisors of 108 are {1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108}. The divisors less than 4 are {1, 2, 3} (3 divisors).

If

n + 3

were 1, 2, or 3,

n

would be

-2

,

-1

, or

0

, none of which are positive integers.

5

Subtract invalid divisors from total divisors

Valid divisors =

12 - 3 = 9

.

Each divisor

d \ge 4

yields exactly one unique positive integer

n = d - 3

.

Key Concept

Algebraic Divisibility and Prime Factorization Divisor Counting

Question 195Question →

A cargo logistics warehouse handles three types of freight: Standard, Express, and Fragile. At the start of the week, $\frac{2}{5}$ of the total shipments in the warehouse were Standard freight, $0.35$ of the total shipments were Express freight, and the remaining shipments were Fragile freight. During a midweek inventory adjustment, the number of Standard freight shipments increased by $25\%$ , the number of Express freight shipments decreased by $20\%$ , and the number of Fragile freight shipments remained unchanged at $125$ units. What was the total number of shipments in the warehouse after the midweek inventory adjustment?

Show answer & explanation

Answer: 515

Answer

515 shipments

Converting

\frac{2}{5}

to a decimal gives

0.40

. The initial proportion of Fragile freight is

1 - 0.40 - 0.35 = 0.25

. Since Fragile freight equals

125

shipments, the initial total number of shipments is

\frac{125}{0.25} = 500

. Prior to adjustment, the warehouse has

0.40 \times 500 = 200

Standard shipments and

0.35 \times 500 = 175

Express shipments. After the adjustment, Standard shipments increase to

200 \times 1.25 = 250

, Express shipments decrease to

175 \times 0.80 = 140

, and Fragile shipments remain

125

. Summing these post-adjustment counts yields

250 + 140 + 125 = 515

shipments.

Step-by-Step Solution

1

Determine the initial fraction representing Fragile freight.

Fragile freight constitutes

0.25

(or

25\%

) of the total initial shipments.

Converting

\frac{2}{5}

to a decimal yields

0.40

. Subtracting the Standard (

0.40

) and Express (

0.35

) proportions from the total (

1.00

) leaves

1.00 - 0.40 - 0.35 = 0.25

.

2

Calculate the total initial number of shipments in the warehouse.

Initial total shipments =

500

.

Since

25\%

of the total initial shipments equals

125

Fragile units, the initial total is

\frac{125}{0.25} = 500

.

3

Calculate the initial quantities of Standard and Express shipments.

Initial Standard shipments =

200

, Initial Express shipments =

175

.

Standard shipments =

0.40 \times 500 = 200

, and Express shipments =

0.35 \times 500 = 175

.

4

Apply the inventory percentage changes to each category.

New Standard shipments =

250

, New Express shipments =

140

, Fragile shipments =

125

.

A

25\%

increase in Standard shipments gives

200 \times 1.25 = 250

. A

20\%

decrease in Express shipments gives

175 \times 0.80 = 140

. Fragile shipments remain

125

.

5

Sum the adjusted shipment counts to determine the net total.

Final total shipments =

515

.

Adding the three updated categories gives

250 + 140 + 125 = 515

.

Key Concept

Fractions, Decimals, and Percents Arithmetic

Estimated Time:1m 30s

Question 196Question →

If $k = 2^3 \cdot 3^a \cdot 7^b$ , where $a$ and $b$ are positive integers, and $k$ has exactly 36 positive integer divisors, how many positive integer divisors does $3k$ have?

39

40

48

54

72

Show answer & explanation

Answer: 48

Answer

48

The number of positive integer divisors of a prime-factored number

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

is

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

. For

k = 2^3 \cdot 3^a \cdot 7^b

, the divisor count is

4(a+1)(b+1) = 36

, which simplifies to

(a+1)(b+1) = 9

. Given

a, b \ge 1

, the only integer solution is

a = 2

and

b = 2

. Multiplying

k

by 3 yields

3k = 2^3 \cdot 3^3 \cdot 7^2

, which has

(3+1)(3+1)(2+1) = 48

positive divisors.

Step-by-Step Solution

1

Use the formula for the number of positive divisors of an integer.

The number of positive divisors of

k = 2^3 \cdot 3^a \cdot 7^b

is given by

(3+1)(a+1)(b+1) = 4(a+1)(b+1)

.

If

N = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}

, the total number of positive integer divisors is

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

.

2

Set the divisor count equal to 36 and solve for

a

and

b

.

4(a+1)(b+1) = 36 \implies (a+1)(b+1) = 9

.

Dividing both sides by 4 isolates the term containing

a

and

b

.

3

Determine the values of

a

and

b

using the constraint that

a

and

b

are positive integers.

Since

a \ge 1

and

b \ge 1

, we have

a+1 \ge 2

and

b+1 \ge 2

. The only factor pair of 9 where both factors are at least 2 is

3 \times 3

. Thus,

a+1 = 3 \implies a = 2

and

b+1 = 3 \implies b = 2

.

Positive integers must be greater than or equal to 1.

4

Write the prime factorization of

3k

and compute its number of positive divisors.

3k = 3 \cdot (2^3 \cdot 3^2 \cdot 7^2) = 2^3 \cdot 3^3 \cdot 7^2

. The number of divisors is

(3+1)(3+1)(2+1) = 4 \cdot 4 \cdot 3 = 48

.

Multiplying

k

by 3 increases the exponent of 3 from 2 to 3.

Key Concept

Prime Factorization and Total Divisors Rule

Estimated Time:1m 30s

Question 197Question →

At an urban hydroponic farm, a monthly harvest consists of three types of crops: leafy greens, herbs, and microgreens. Leafy greens account for $\frac{5}{8}$ of the total harvest by weight, and herbs account for $0.25$ of the total harvest by weight. If the yield of microgreens increases by $40\%$ in the following month while the yields of leafy greens and herbs remain constant, what fraction of the new total harvest by weight will be microgreens?

\frac{7}{40}

\frac{1}{6}

\frac{1}{5}

\frac{7}{22}

\frac{7}{32}

Show answer & explanation

Answer: $\frac{1}{6}$

Answer

The fraction of the new total harvest that will be microgreens is

\frac{1}{6}

.

Initially, leafy greens and herbs account for

\frac{5}{8} + 0.25 = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}

of the total harvest weight, leaving microgreens with

1 - \frac{7}{8} = \frac{1}{8}

of the total weight. When microgreens increase by

40\%

, their new yield becomes

\frac{1}{8} \times 1.40 = \frac{7}{40}

of the original total harvest. The total harvest weight increases to

\frac{7}{8} + \frac{7}{40} = \frac{35}{40} + \frac{7}{40} = \frac{42}{40}

. The fraction of microgreens in this new total harvest is

\frac{7/40}{42/40} = \frac{7}{42} = \frac{1}{6}

.

Step-by-Step Solution

1

Convert all initial crop portions into consistent fractions to find the initial microgreens fraction.

Leafy greens =

\frac{5}{8}

. Herbs =

0.25 = \frac{1}{4} = \frac{2}{8}

. Microgreens fraction =

1 - \left(\frac{5}{8} + \frac{2}{8}\right) = 1 - \frac{7}{8} = \frac{1}{8}

.

The sum of all three crop fractions must equal

1

(the total harvest).

2

Calculate the updated relative weight of microgreens after a

40\%

increase.

New microgreens weight relative to original total =

\frac{1}{8} \times (1 + 0.40) = \frac{1}{8} \times \frac{7}{5} = \frac{7}{40}

.

Increasing a quantity by

40\%

is equivalent to multiplying by

1.40

or

\frac{7}{5}

.

3

Determine the new total harvest weight relative to the original harvest.

New total harvest = Leafy greens + Herbs + New microgreens =

\frac{5}{8} + \frac{2}{8} + \frac{7}{40} = \frac{35}{40} + \frac{7}{40} = \frac{42}{40} = \frac{21}{20}

.

Since the yields of leafy greens and herbs did not change, their combined relative weight remains

\frac{35}{40}

.

4

Compute the fraction of the new total harvest that consists of microgreens.

Fraction =

\frac{\frac{7}{40}}{\frac{42}{40}} = \frac{7}{42} = \frac{1}{6}

.

Dividing the new microgreens weight by the new total harvest weight gives the updated proportion.

Key Concept

Fractions, Decimals, and Percents Arithmetic

Estimated Time:1m 30s

Question 198Question →

Set $A$ consists of $k$ consecutive integers, where $k > 1$ . Set $B$ consists of $2k$ consecutive integers. The median of Set $A$ is equal to the least integer in Set $B$ , and the sum of all integers in Set $A$ is equal to the sum of all integers in Set $B$ . If the sum of the integers in Set $B$ is $-153$ , what is the greatest integer in Set $A$ ?

-21

-17

-13

-12

-9

Show answer & explanation

Answer: $-13$

Answer

The greatest integer in Set

A

is

-13

.

The problem establishes a relationship between two sets of consecutive integers using their medians and sums. Equating the sum formulas

S_A = k\left(a + \frac{k-1}{2}\right)

and

S_B = 2k\left(a + \frac{3k-2}{2}\right)

leads to

a = \frac{3-5k}{2}

. Substituting this into the sum expression gives

k - 2k^2 = -153

, yielding

k = 9

. Thus, the smallest element is

a = -21

, and the greatest element is

-21 + 9 - 1 = -13

.

Step-by-Step Solution

1

Express the sum of Set

A

in terms of

k

and its smallest element

a

.

The median of Set

A

is

M_A = a + \frac{k-1}{2}

, so the sum of Set

A

is

S_A = k \cdot M_A = k\left(a + \frac{k-1}{2}\right)

.

In any set of consecutive integers, the sum equals the number of terms multiplied by the average (median).

2

Express the sum of Set

B

using the given conditions.

The least integer in Set

B

is

M_A = a + \frac{k-1}{2}

. Since Set

B

has

2k

consecutive integers, its median is

M_B = \left(a + \frac{k-1}{2}\right) + \frac{2k-1}{2} = a + \frac{3k-2}{2}

. The sum of Set

B

is

S_B = 2k \cdot M_B = 2k\left(a + \frac{3k-2}{2}\right) = k(2a + 3k - 2)

.

Set

B

starts at the median of Set

A

and contains

2k

terms.

3

Equate

S_A

and

S_B

to solve for

a

in terms of

k

.

Since

S_A = S_B

and

k > 1

, we divide by

k

:

a + \frac{k-1}{2} = 2a + 3k - 2 \implies a = \frac{3 - 5k}{2}

.

The problem states that the sums of the two sets are equal.

4

Substitute

a = \frac{3 - 5k}{2}

into

S_B

and set it equal to

-153

.

S_B = k\left(2\left(\frac{3-5k}{2}\right) + 3k - 2\right) = k(3 - 5k + 3k - 2) = k(1 - 2k) = k - 2k^2

. Setting

k - 2k^2 = -153 \implies 2k^2 - k - 153 = 0

. Factoring gives

(2k + 17)(k - 9) = 0

. Since

k

must be a positive integer greater than 1,

k = 9

.

Finding

k

allows us to calculate the exact elements of Set

A

.

5

Calculate the greatest integer in Set

A

.

With

k = 9

, the smallest integer

a = \frac{3 - 5(9)}{2} = -21

. The greatest integer in Set

A

is

a + k - 1 = -21 + 9 - 1 = -13

.

The

k

-th term of a sequence of consecutive integers starting at

a

is

a + k - 1

.

Key Concept

Properties of consecutive integer sets: sum equals number of terms times median; indexing terms in evenly spaced sets.

Question 199Question →

What is the remainder when the integer $N = 3^{83} + 7^{83}$ is divided by $100$ ?

Show answer & explanation

Answer: 70

Answer

The remainder when

N = 3^{83} + 7^{83}

is divided by

100

is 70.

Finding the remainder when an integer is divided by

100

is equivalent to finding its last two digits. Powers of

3

repeat their last two digits in cycles of

20

, and

3^{20} \equiv 1 \pmod{100}

. Thus

3^{83} \equiv 3^3 = 27 \pmod{100}

. Powers of

7

repeat their last two digits in cycles of

4

, with

7^4 = 2401 \equiv 1 \pmod{100}

. Thus

7^{83} \equiv 7^3 = 343 \equiv 43 \pmod{100}

. Combining these gives

27 + 43 = 70

.

Step-by-Step Solution

1

Find the remainder of

3^{83}

modulo

100

by identifying the cyclicity pattern of powers of

3

.

3^{20} \equiv 1 \pmod{100}

, which implies

3^{83} = (3^{20})^4 \cdot 3^3 \equiv 1^4 \cdot 27 = 27 \pmod{100}

.

Since

3^{20} \equiv 1 \pmod{100}

, exponent

83

can be reduced to

83 \pmod{20} = 3

.

2

Find the remainder of

7^{83}

modulo

100

by identifying the cyclicity pattern of powers of

7

.

7^4 = 2401 \equiv 1 \pmod{100}

, which implies

7^{83} = (7^4)^{20} \cdot 7^3 \equiv 1^{20} \cdot 343 \equiv 43 \pmod{100}

.

Since

7^4 \equiv 1 \pmod{100}

, exponent

83

can be reduced to

83 \pmod{4} = 3

.

3

Sum the modular remainders to obtain the final remainder of

N

.

N = 3^{83} + 7^{83} \equiv 27 + 43 = 70 \pmod{100}

.

The remainder of a sum is equal to the sum of the individual remainders modulo

100

.

Key Concept

Modular Arithmetic and Cyclicity of Last Two Digits

Question 200Question →

What is the smallest positive integer $m$ such that the remainder when $6^m$ is divided by $11$ is equal to $8$ ?

Show answer & explanation

Answer: 7

Answer

The smallest positive integer

m

is 7.

Evaluating powers of 6 modulo 11 sequentially gives:

6^1 \equiv 6

,

6^2 \equiv 3

,

6^3 \equiv 7

,

6^4 \equiv 9

,

6^5 \equiv 10

,

6^6 \equiv 5

, and

6^7 \equiv 8 \pmod{11}

. Therefore,

m = 7

is the smallest positive integer satisfying the condition.

Step-by-Step Solution

1

Calculate the remainder of powers of 6 when divided by 11 starting from

m = 1

.

6^1 \equiv 6 \pmod{11}

,

6^2 \equiv 3 \pmod{11}

,

6^3 \equiv 7 \pmod{11}

,

6^4 \equiv 9 \pmod{11}

,

6^5 \equiv 10 \pmod{11}

,

6^6 \equiv 5 \pmod{11}

,

6^7 \equiv 8 \pmod{11}

.

Modular multiplication allows computing powers step-by-step by multiplying the previous remainder by 6 and finding the remainder modulo 11.

2

Identify the smallest exponent

m

that yields a remainder of 8.

m = 7

At

m = 7

,

6^7 = 279,936 = 11 \times 25,448 + 8

, which leaves a remainder of 8.

Key Concept

Modular Arithmetic and Cyclicity of Powers

Number Properties and Arithmetic

Answer

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