All practice questions

What is the remainder when $3^{42}$ is divided by $5$ ?

Show answer & explanation

Answer: 4

Answer

The remainder when

3^{42}

is divided by 5 is 4.

The remainders of powers of 3 divided by 5 follow a repeating pattern of length 4: (3, 4, 2, 1). To find the remainder of

3^{42} \div 5

, divide the exponent 42 by 4. Since

42 = 4 \times 10 + 2

, the remainder of the exponent is 2. The 2nd term in the repeating sequence is 4, so

3^{42} \pmod 5 = 4

.

Step-by-Step Solution

1

Determine the remainder pattern for consecutive powers of 3 divided by 5.

The remainders for

3^1, 3^2, 3^3, 3^4, 3^5

are

3, 4, 2, 1, 3

, establishing a cycle of length 4.

Powers of integers divided by a positive integer yield repeating sequence patterns (cyclicity).

2

Divide the target exponent by the cycle length.

Dividing 42 by 4 gives a quotient of 10 and a remainder of 2.

The remainder indicates the specific term within the 4-step cycle.

3

Evaluate the value corresponding to the 2nd position in the cycle.

The 2nd term in the cycle (3, 4, 2, 1) is 4.

A remainder of 2 in the exponent position corresponds to the same remainder as

3^2

divided by 5.

Key Concept

Remainders and Units Digit Cyclicity

Question 642Question →

Three automated signal beacons emit flashes at regular time intervals of $x$ , $y$ , and $z$ seconds, respectively, where $x$ , $y$ , and $z$ are positive integers. If $\text{gcd}(x, y) = 18$ , $\text{gcd}(y, z) = 24$ , and $\text{gcd}(x, z) = 30$ , what is the minimum possible value of $\text{lcm}(x, y, z)$ ?

180

360

720

1{}080

12{}960

Show answer & explanation

Answer: $360$

Answer

360

By breaking each given greatest common divisor into its prime factorization, we establish minimum requirements for the exponents of prime factors

2

,

3

, and

5

across

x

,

y

, and

z

. Specifically,

\text{gcd}(y, z) = 24 = 2^3 \times 3

dictates that both

y

and

z

must contain

2^3

, making the maximum exponent of

2

equal to

3

. Similarly,

\text{gcd}(x, y) = 18 = 2 \times 3^2

forces the maximum exponent of

3

to be

2

, and

\text{gcd}(x, z) = 30 = 2 \times 3 \times 5

forces the maximum exponent of

5

to be

1

. Multiplying these prime powers together gives

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

.

Step-by-Step Solution

1

Express the given pairwise greatest common divisors in prime factorized form.

\text{gcd}(x, y) = 18 = 2^1 \times 3^2

,

\text{gcd}(y, z) = 24 = 2^3 \times 3^1

,

\text{gcd}(x, z) = 30 = 2^1 \times 3^1 \times 5^1

.

GCD conditions determine the minimum prime factor exponents across pairs of variables.

2

Analyze the exponents for prime factor

2

.

\min(v_2(x), v_2(y)) = 1

,

\min(v_2(y), v_2(z)) = 3

, and

\min(v_2(x), v_2(z)) = 1

. This implies

v_2(y) \ge 3

and

v_2(z) \ge 3

, while

v_2(x) = 1

. To minimize the LCM, set

v_2(x) = 1

,

v_2(y) = 3

,

v_2(z) = 3

. Thus,

\max(v_2(x), v_2(y), v_2(z)) = 3

.

The exponent of

2

in

\text{lcm}(x, y, z)

is the maximum of the individual exponents of

2

.

3

Analyze the exponents for prime factor

3

.

\min(v_3(x), v_3(y)) = 2

,

\min(v_3(y), v_3(z)) = 1

, and

\min(v_3(x), v_3(z)) = 1

. This implies

v_3(x) \ge 2

and

v_3(y) \ge 2

, while

v_3(z) = 1

. To minimize the LCM, set

v_3(x) = 2

,

v_3(y) = 2

,

v_3(z) = 1

. Thus,

\max(v_3(x), v_3(y), v_3(z)) = 2

.

The exponent of

3

in

\text{lcm}(x, y, z)

is the maximum of the individual exponents of

3

.

4

Analyze the exponents for prime factor

5

.

\min(v_5(x), v_5(y)) = 0

,

\min(v_5(y), v_5(z)) = 0

, and

\min(v_5(x), v_5(z)) = 1

. This requires

v_5(x) \ge 1

and

v_5(z) \ge 1

, while

v_5(y) = 0

. To minimize the LCM, set

v_5(x) = 1

,

v_5(y) = 0

,

v_5(z) = 1

. Thus,

\max(v_5(x), v_5(y), v_5(z)) = 1

.

The exponent of

5

in

\text{lcm}(x, y, z)

is the maximum of the individual exponents of

5

.

5

Calculate the minimum value of

\text{lcm}(x, y, z)

.

\text{lcm}(x, y, z) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360

.

No additional prime factors are required to satisfy the conditions, yielding the minimum possible LCM.

Key Concept

Prime Exponent Constraints on Pairwise GCD and LCM

Estimated Time:2m 0s

Question 643Question →

A real estate developer purchased two tracts of land. Tract 1 consists of $15$ acres purchased at an average cost of $\$24,000$ per acre. Tract 2 consists of $35$ acres purchased at an average cost of $\$40,000$ per acre. What is the average cost per acre, in thousands of dollars, for the combined land purchase?

Show answer & explanation

Answer: 35.2

Answer

The average cost per acre for the combined land purchase is

35.2

thousand dollars (or

\$35,200

).

The overall weighted average is calculated by dividing the sum of all individual costs by the total number of acres across both tracts. Since

15

acres cost

\$24,000

per acre and

35

acres cost

\$40,000

per acre, the total cost is

15(24,000) + 35(40,000) = 360,000 + 1,400,000 = 1,760,000

dollars. Dividing by the total of

50

acres yields

\$35,200

per acre, or

35.2

thousand dollars.

Step-by-Step Solution

1

Determine the total financial expenditure for each tract

Tract 1 cost =

15 \times 24,000 = \$360,000

; Tract 2 cost =

35 \times 40,000 = \$1,400,000

Total value of a set equals the number of items multiplied by the mean of the items.

2

Calculate total combined cost and total combined acreage

Total cost =

\$360,000 + \$1,400,000 = \$1,760,000

; Total acres =

15 + 35 = 50

Combined sets require summing all total values and all individual counts.

3

Compute the weighted average cost per acre

Weighted Mean =

\frac{1,760,000}{50} = 35,200

dollars, which is

35.2

thousand dollars.

Divide total combined cost by total combined number of units.

Key Concept

Weighted Average of Combined Sets

Question 644Question →

What is the greatest prime factor of $2^5 - 2^3$ ?

1

2

3

4

8

Show answer & explanation

Answer: 3

Answer

3

Evaluating

2^5 - 2^3

yields

32 - 8 = 24

. The prime factorization of 24 is

2^3 \times 3

. The distinct prime factors are 2 and 3, making 3 the greatest prime factor.

Step-by-Step Solution

1

Evaluate the arithmetic expression

2^5 - 2^3 = 32 - 8 = 24

Calculate the numerical value of the expression before finding prime factors.

2

Find the prime factorization of 24

24 = 2^3 \times 3

Break down 24 into a product of prime numbers.

3

Identify the greatest prime factor

The prime factors are 2 and 3. The greatest prime factor is 3.

Compare the distinct prime factors (2 and 3) to find the largest one.

Key Concept

Prime Factorization

Estimated Time:45s

Question 645Question →

If the sum of five consecutive integers is $115$ , what is the value of the largest integer in this set?

21

23

24

25

26

Show answer & explanation

Answer: 25

Answer

The largest integer in the set is 25.

For any evenly spaced set with an odd number of elements, the arithmetic mean equals the median. Dividing the sum of

115

by

5

gives a median of

23

. The five consecutive integers are

21, 22, 23, 24, 25

, so the largest integer is

25

.

Step-by-Step Solution

1

Find the average (and median) of the five consecutive integers.

Dividing the sum

115

by

5

yields a median of

23

.

For an odd number of consecutive integers, the average of the set is equal to its median (the middle term).

2

Determine the five consecutive integers centered around the median.

The integers are

21, 22, 23, 24, 25

.

Consecutive integers differ by

1

, so the two terms above the median

23

are

24

and

25

.

3

Identify the largest integer.

The largest integer is

25

.

The question asks specifically for the maximum value in the set.

Key Concept

Average and Median Equivalence in Consecutive Integer Sets

Estimated Time:45s

Question 646Question →

In number theory, the units digit of a positive integer raised to successive positive integer powers follows a repeating cyclic pattern. What is the units digit of $4^{25}$ ?

Show answer & explanation

Answer: 4

Answer

The units digit of

4^{25}

is 4.

The units digit of powers of 4 alternates between 4 (for odd powers) and 6 (for even powers). Because 25 is an odd number,

4^{25}

has a units digit of 4.

Step-by-Step Solution

1

Identify the units digit pattern for powers of 4

The units digits cycle between 4 (for odd exponents) and 6 (for even exponents), giving a cycle length of 2.

Units digits of positive integer powers follow a periodic pattern determined by the base digit.

2

Determine the parity of the exponent 25

25 is an odd integer (remainder 1 when divided by 2).

The exponent's remainder modulo 2 determines which position in the 2-element cycle [4, 6] the number falls into.

3

Select the corresponding units digit from the cycle

Since 25 is odd, the units digit is 4.

Odd powers of 4 always have a units digit of 4.

Key Concept

Units Digit Cyclicity

Question 647Question →

If $n$ is a positive integer with exactly two distinct prime factors such that $n$ is a multiple of $12$ and $n^2$ has exactly $27$ positive integer divisors, what is the value of $n$ ?

36

48

72

144

162

Show answer & explanation

Answer: 48

Answer

48

Since

n

is a multiple of

12 = 2^2 \cdot 3

and has only two prime factors,

n = 2^a \cdot 3^b

with

a \ge 2

and

b \ge 1

. The square

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

has

(2a+1)(2b+1) = 27

divisors. The only factor pair of 27 satisfying

2a+1 \ge 5

is

9 \times 3

, yielding

2a+1=9 \implies a=4

and

2b+1=3 \implies b=1

. Thus,

n = 2^4 \cdot 3^1 = 48

.

Step-by-Step Solution

1

Express

n

in terms of its prime factorization.

Since

n

is a multiple of

12 = 2^2 \cdot 3

and has exactly two distinct prime factors, its prime factorization must be

n = 2^a \cdot 3^b

, where

a \ge 2

and

b \ge 1

.

Divisibility by 12 requires at least two factors of 2 and at least one factor of 3.

2

Formulate the expression for the number of positive divisors of

n^2

.

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

. The number of positive divisors is

(2a + 1)(2b + 1) = 27

.

The total number of positive integer divisors of a prime power product

p^x q^y

is

(x+1)(y+1)

.

3

Solve for integer values of

a

and

b

using factor pairs of 27.

The factor pairs of 27 into two integers greater than 1 are

9 \times 3

. Since

a \ge 2

,

2a + 1 \ge 5

. Thus,

2a + 1 = 9

and

2b + 1 = 3

, giving

a = 4

and

b = 1

.

Setting

2a + 1 = 3

yields

a = 1

, which violates the requirement that

a \ge 2

for divisibility by 12.

4

Calculate

n

.

n = 2^4 \cdot 3^1 = 16 \cdot 3 = 48

.

Substituting

a=4

and

b=1

into the prime factorization of

n

gives 48.

Key Concept

Divisibility rules and prime factorization count of positive divisors

Question 648Question →

If $3(x - 2) + 4 = 2x + 11$ , what is the value of $x$ ?

Show answer & explanation

Answer: 13

Answer

The value of

x

is 13.

Expanding the left side yields

3x - 6 + 4 = 3x - 2

. Equating this to the right side gives

3x - 2 = 2x + 11

. Subtracting

2x

from both sides gives

x - 2 = 11

, and adding

2

to both sides yields

x = 13

.

Step-by-Step Solution

1

Expand the left side of the equation

3x - 6 + 4 = 2x + 11

Apply the distributive property

3(x - 2) = 3x - 6

.

2

Combine like constant terms on the left side

3x - 2 = 2x + 11

Combine

-6

and

+4

to get

-2

.

3

Subtract

2x

from both sides

x - 2 = 11

Move variable terms to one side of the equation.

4

Add 2 to both sides

x = 13

Isolate the variable

x

.

Key Concept

Solving Linear Equations in One Variable

Question 649Question →

At a biotechnology facility, a solution tank initially contains a mixture of ethanol, water, and stabilizer liquid. By volume, $0.36$ of the initial mixture is ethanol and $\frac{2}{5}$ is water, with the remainder consisting of stabilizer liquid. During a purification process, $25\%$ of the ethanol and $\frac{1}{8}$ of the water evaporate, while the volume of the stabilizer liquid remains unchanged. If the volume of the remaining stabilizer liquid is $132$ liters, by what percentage did the total volume of the liquid mixture in the tank decrease?

14%

16.28%

18.75%

37.50%

24.00%

Show answer & explanation

Answer: 14%

Answer

The total volume of the liquid mixture in the tank decreased by 14%.

The initial mixture consists of 36% ethanol, 40% water, and 24% stabilizer by volume. Ethanol loses 25% of its volume, which corresponds to

0.25 \times 36\% = 9\%

of the total initial mixture volume. Water loses

\frac{1}{8}

of its volume, corresponding to

\frac{1}{8} \times 40\% = 5\%

of the total initial mixture volume. Combining these losses gives

9\% + 5\% = 14\%

net decrease relative to the initial total volume.

Step-by-Step Solution

1

Express initial component proportions as unified decimals or fractions.

Ethanol fraction =

0.36 = \frac{9}{25}

. Water fraction =

\frac{2}{5} = 0.40 = \frac{10}{25}

. Stabilizer fraction =

1 - (0.36 + 0.40) = 0.24 = \frac{6}{25}

.

Establishing all initial parts relative to the total initial volume allows step-by-step proportion tracking.

2

Calculate the fraction of total volume lost due to evaporation for each component.

Ethanol loss =

25\% \times 0.36 = 0.25 \times 0.36 = 0.09

of initial total volume. Water loss =

\frac{1}{8} \times 0.40 = 0.125 \times 0.40 = 0.05

of initial total volume.

Applying the component-specific percentage decrease to each component's fraction yields its contribution to the overall volume loss.

3

Sum the component volume losses to find the total percentage decrease.

Total volume loss fraction =

0.09 + 0.05 = 0.14

, which equals

14\%

of the total initial volume.

The stabilizer experience no loss, so the net change in total volume is the direct sum of the losses in ethanol and water relative to the original base.

Key Concept

Multi-step arithmetic operations combining decimals, fractions, and weighted percent changes using a consistent base value.

Estimated Time:1m 50s

Question 650Question →

A positive integer $n$ is a multiple of $6$ , but is neither a multiple of $4$ nor a multiple of $9$ . If $n$ has exactly $12$ positive integer divisors and $n^2$ has exactly $45$ positive integer divisors, how many positive integer divisors of $n^2$ are divisible by $n$ ?

Show answer & explanation

Answer: 12

Answer

The number of positive integer divisors of

n^2

that are divisible by

n

is

12

.

By analyzing the prime factorization

n = 2^1 \cdot 3^1 \cdot p^2

, we find

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

. For a divisor

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

of

n^2

to be divisible by

n

, the exponents must satisfy

x \in \{1, 2\}

,

y \in \{1, 2\}

, and

z \in \{2, 3, 4\}

, yielding

2 \times 2 \times 3 = 12

valid divisors.

Step-by-Step Solution

1

Analyze the prime factorization of

n

for prime factors

2

and

3

.

The exponents of

2

and

3

in

n

are both

1

.

Since

n

is a multiple of

6

, it must contain at least one factor of

2

and one factor of

3

. Because

n

is not a multiple of

4

, the exponent of

2

cannot exceed

1

. Because

n

is not a multiple of

9

, the exponent of

3

cannot exceed

1

.

2

Determine the remaining prime factors of

n

using the total divisor count of

n

.

n = 2^1 \cdot 3^1 \cdot p^2

for some prime

p > 3

.

The number of positive divisors is given by

d(n) = (1+1)(1+1)\prod(c_i+1) = 12

, which simplifies to

4\prod(c_i+1) = 12

, so

\prod(c_i+1) = 3

. Since

3

is prime, there is exactly one additional prime factor

p

with exponent

c = 2

.

3

Verify with the divisor count of

n^2

.

d(n^2) = (2(1)+1)(2(1)+1)(2(2)+1) = 3 \cdot 3 \cdot 5 = 45

.

Squaring

n

doubles all prime exponents, so

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

, which has

(2+1)(2+1)(4+1) = 45

positive divisors, consistent with the given information.

4

Calculate the number of divisors of

n^2

that are multiples of

n

.

12

Any divisor of

n^2

has the form

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

with

0 \le x \le 2

,

0 \le y \le 2

, and

0 \le z \le 4

. For this divisor to be a multiple of

n = 2^1 \cdot 3^1 \cdot p^2

, the exponents must satisfy

1 \le x \le 2

(

2

choices),

1 \le y \le 2

(

2

choices), and

2 \le z \le 4

(

3

choices). Multiplying the choices gives

2 \cdot 2 \cdot 3 = 12

.

Key Concept

Divisor counting formula and prime factor exponent constraints

Estimated Time:2m 0s

Question 651Question →

If $m$ and $n$ are positive two-digit integers such that $m < n$ , $\text{GCD}(m, n) = 12$ , and $\text{LCM}(m, n) = 420$ , what is the value of $n - m$ ?

24

36

72

144

408

Show answer & explanation

Answer: $24$

Answer

The value of

n - m

is

24

.

Writing

m = 12x

and

n = 12y

for coprime integers

x < y

gives

\text{LCM}(m, n) = 12xy = 420

, which simplifies to

xy = 35

. The coprime factor pairs of

35

are

(1, 35)

and

(5, 7)

. The pair

(1, 35)

yields

m = 12

and

n = 420

, but

420

is not a two-digit integer. The pair

(5, 7)

gives two-digit integers

m = 60

and

n = 84

. The difference is

84 - 60 = 24

.

Step-by-Step Solution

1

Express

m

and

n

in terms of their GCD.

Let

m = 12x

and

n = 12y

, where

x

and

y

are coprime positive integers with

x < y

.

Factoring out the greatest common divisor

\text{GCD}(m, n) = 12

leaves coprime factors

x

and

y

.

2

Relate GCD and LCM to determine

x \cdot y

.

\text{LCM}(m, n) = 12 \cdot x \cdot y = 420 \implies x \cdot y = 35

.

For integers written as

12x

and

12y

with

\text{GCD}(x,y)=1

, the least common multiple is

12xy

.

3

Determine the valid pair

(x, y)

that satisfies the two-digit constraint.

The coprime factor pairs of

35

with

x < y

are

(1, 35)

and

(5, 7)

. For

(1, 35)

,

m = 12

and

n = 420

(a three-digit number). For

(5, 7)

,

m = 60

and

n = 84

(both are two-digit numbers). Thus,

(x, y) = (5, 7)

.

The problem specifies that both

m

and

n

are two-digit positive integers.

4

Compute

n - m

.

n - m = 84 - 60 = 24

.

Subtracting

m = 60

from

n = 84

gives

24

.

Key Concept

Properties of Greatest Common Divisor (GCD) and Least Common Multiple (LCM)

Question 652Question →

A total of 360 identical notebooks are to be distributed equally among $n$ students, where $n > 1$ , such that each student receives an integer number of notebooks strictly greater than 1. If $n$ must be a multiple of 4, how many different values of $n$ are possible?

Show answer & explanation

Answer: 11

Answer

The total number of possible values for

n

is 11.

To find the number of valid values of

n

, prime factorize

360 = 2^3 \times 3^2 \times 5^1

. For

n

to be a factor of 360 and a multiple of 4, the exponent of 2 must be 2 or 3 (2 choices), the exponent of 3 can be 0, 1, or 2 (3 choices), and the exponent of 5 can be 0 or 1 (2 choices), yielding

2 \times 3 \times 2 = 12

total factors of 360 that are multiples of 4. Excluding

n = 360

, which results in 1 notebook per student, leaves

12 - 1 = 11

valid values.

Step-by-Step Solution

1

Prime factorize the total number of notebooks, 360.

360 = 2^3 \times 3^2 \times 5^1

Prime factorization allows us to systematically count factors meeting specific divisibility conditions.

2

Calculate the number of factors of 360 that are multiples of 4.

2 choices for the exponent of 2 (

2^2

or

2^3

), 3 choices for the exponent of 3 (

3^0, 3^1, 3^2

), and 2 choices for the exponent of 5 (

5^0, 5^1

). Total factors =

2 \times 3 \times 2 = 12

.

A factor is a multiple of 4 if and only if its prime factorization contains at least two factors of 2.

3

Apply the constraint that each student receives strictly more than 1 notebook.

If

n = 360

, each student receives

\frac{360}{360} = 1

notebook, which violates the condition. Subtracting this case yields

12 - 1 = 11

valid values.

The problem requires each student to receive an integer number of notebooks strictly greater than 1.

Key Concept

Counting constrained factors using prime factorization

Question 653Question →

What is the value of $\sqrt{2^6 + 2^6 + 2^6 + 2^6}$ ?

8

16

32

64

128

Show answer & explanation

Answer: 16

Answer

16

The four identical terms

2^6 + 2^6 + 2^6 + 2^6

sum to

4 \times 2^6

. Since

4 = 2^2

, the expression inside the radical becomes

2^2 \times 2^6 = 2^8

. Taking the square root yields

\sqrt{2^8} = 2^4 = 16

.

Step-by-Step Solution

1

Express the sum of the four identical terms as a product.

2^6 + 2^6 + 2^6 + 2^6 = 4 \times 2^6

Adding a number to itself 4 times is equivalent to multiplying that number by 4.

2

Convert 4 to a power of 2 and combine exponents.

4 \times 2^6 = 2^2 \times 2^6 = 2^{2+6} = 2^8

When multiplying powers with the same base, add their exponents.

3

Simplify the square root of

2^8

.

\sqrt{2^8} = (2^8)^{1/2} = 2^4 = 16

Taking the square root of a power means dividing the exponent by 2.

Key Concept

Factoring identical exponential terms and applying exponent laws to simplify radicals.

Question 654Question →

If $3^x \cdot 9^2 = 3^{10}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 6

Answer

The value of

x

is 6.

Rewriting

9^2

as

(3^2)^2 = 3^4

converts the expression to a common base of

3

. Applying the product rule gives

3^x \cdot 3^4 = 3^{x+4}

. Setting the exponent

x+4

equal to

10

yields

x = 6

.

Step-by-Step Solution

1

Express all powers using the common base 3

9^2 = (3^2)^2 = 3^4

Converting terms to a common base allows application of exponent rules.

2

Apply the product rule for exponents

3^x \cdot 3^4 = 3^{x+4}

When multiplying terms with identical bases, add the exponents.

3

Equate the exponents and solve for x

x + 4 = 10 \implies x = 6

If

b^m = b^n

for

b > 0

and

b \neq 1

, then

m = n

.

Key Concept

Exponent rules with base conversion and product of powers

Question 655Question →

If the sum of a set of $7$ consecutive odd integers is $105$ , what is the median integer of the set?

Show answer & explanation

Answer: 15

Answer

The median integer of the set is 15.

For any set of evenly spaced numbers, such as consecutive odd integers, the arithmetic mean is equal to the median. Since the sum of the

7

integers is

105

, the mean is

105 \div 7 = 15

. Therefore, the median of the set is

15

.

Step-by-Step Solution

1

Apply the property of evenly spaced sets.

For consecutive odd integers, the mean of the set is equal to its median.

The numbers in an arithmetic sequence are symmetrically distributed around the middle value.

2

Calculate the arithmetic mean.

Mean =

\frac{\text{Sum}}{\text{Number of terms}} = \frac{105}{7} = 15

.

Dividing the sum of the terms by the count gives the average value.

3

Determine the median.

Median =

15

.

Because mean equals median for evenly spaced sets, the median must be 15.

Key Concept

Average and Median Equivalence in Evenly Spaced Sets

Estimated Time:45s

Question 656Question →

Two positive integers $m$ and $n$ satisfy $\text{GCD}(m, n) = 60$ and $\text{LCM}(m, n) = 75{}600$ . Given that $m$ is divisible by $63$ but not by $189$ , and $n$ is divisible by $400$ , what is the value of $m$ ?

Show answer & explanation

Answer: 1260

Answer

1260

By finding the prime factorizations of the GCD (

2^2 \cdot 3^1 \cdot 5^1

) and LCM (

2^4 \cdot 3^3 \cdot 5^2 \cdot 7^1

), we determine the prime powers for

m

and

n

using

\min

and

\max

rules. The constraint that

n

is divisible by

400 = 2^4 \cdot 5^2

fixes

e_2(n)=4

and

e_5(n)=2

, which forces

e_2(m)=2

and

e_5(m)=1

. The constraint that

m

is divisible by

63 = 3^2 \cdot 7^1

but not

189 = 3^3 \cdot 7^1

fixes

e_3(m)=2

and

e_7(m)=1

. Evaluating

m = 2^2 \cdot 3^2 \cdot 5^1 \cdot 7^1

gives

1{}260

.

Step-by-Step Solution

1

Prime factorize the GCD and LCM

GCD = 2^2 * 3^1 * 5^1 * 7^0 and LCM = 2^4 * 3^3 * 5^2 * 7^1

GCD represents the minimum exponent of each prime shared by m and n, whereas LCM represents the maximum exponent.

2

Determine the prime exponents for m using divisibility conditions

e_2(m) = 2, e_3(m) = 2, e_5(m) = 1, e_7(m) = 1

Divisibility of n by 400 forces e_2(n)=4 and e_5(n)=2, leaving e_2(m)=2 and e_5(m)=1. Divisibility of m by 63 but not 189 fixes e_3(m)=2 and e_7(m)=1.

3

Compute the product of prime powers for m

m = 4 * 9 * 5 * 7 = 1260

Multiplying the determined prime factors yields the exact value of integer m.

Key Concept

Greatest Common Divisor (GCD) and Least Common Multiple (LCM) Prime Exponent Rules

Estimated Time:2m 0s

Question 657Question →

Two positive integers $x$ and $y$ satisfy $x < y$ , $\text{GCD}(x, y) = 15$ , and $\text{LCM}(x, y) = 1,800$ . If $x$ is not a multiple of $9$ and $y$ is not a multiple of $25$ , what is the value of $y - x$ ?

Show answer & explanation

Answer: 285

Answer

The value of

y - x

is 285.

Expressing

x

and

y

as

15a

and

15b

yields

ab = 120

. Testing coprime pairs

(a,b)

shows that only

(5, 24)

gives values

x = 75

and

y = 360

that satisfy both constraints (75 is not a multiple of 9, and 360 is not a multiple of 25). The difference is

360 - 75 = 285

.

Step-by-Step Solution

1

Set up coprime factor representation

x = 15a

and

y = 15b

with

\text{GCD}(a, b) = 1

and

a < b

Any two numbers can be expressed as their GCD multiplied by coprime integer quotients.

2

Solve for the product of coprime quotient factors

a \cdot b

a \cdot b = 120

The LCM of two numbers divided by their GCD equals the product of their coprime quotients.

3

Identify candidate coprime pairs

(a, b)

(1, 120), (3, 40), (5, 24), (8, 15)

Since

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

, all prime factor powers must be assigned entirely to either

a

or

b

to maintain

\text{GCD}(a, b) = 1

.

4

Filter pairs using divisibility constraints on

x

and

y

Only

(75, 360)

satisfies

x \not\equiv 0 \pmod 9

and

y \not\equiv 0 \pmod{25}

75 = 3 \cdot 25

(not divisible by 9) and

360 = 8 \cdot 9 \cdot 5

(not divisible by 25).

5

Calculate the target difference

y - x

285

360 - 75 = 285

.

Key Concept

Relationship between GCD, LCM, and prime factorization distribution in coprime quotients.

Question 658Question →

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 659Question →

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 660Question →

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

Answer

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