All practice questions

Question 161Question →

What is the smallest positive integer greater than $1$ that leaves a remainder of $1$ when divided by each of $4$ , $6$ , and $8$ ?

Show answer & explanation

Answer: 25

Answer

The correct answer is 25.

The least common multiple of 4, 6, and 8 is 24, which is the smallest positive integer divisible by all three numbers. To leave a remainder of 1 upon division by each, we add 1 to 24, resulting in 25.

Step-by-Step Solution

1

Calculate the least common multiple (LCM) of 4, 6, and 8.

\text{LCM}(4, 6, 8) = 24

Any positive integer divisible by 4, 6, and 8 must be a multiple of their least common multiple.

2

Add the desired remainder of 1 to the LCM.

24 + 1 = 25

Adding 1 to a common multiple guarantees that division by 4, 6, or 8 yields a remainder of 1.

Key Concept

Least Common Multiple (LCM) and Remainder Properties

Question 162Question →

Two positive integers $a$ and $b$ are in the ratio $3 : 8$ . If the least common multiple (LCM) of $a$ and $b$ is $360$ , what is the greatest common divisor (GCD) of $a$ and $b$ ?

Show answer & explanation

Answer: 15

Answer

The greatest common divisor of

a

and

b

is 15.

Any two positive integers in the ratio

3 : 8

can be written as

3g

and

8g

, where

g

is their greatest common divisor. Because

3

and

8

are coprime (their GCD is

1

), the least common multiple of

3g

and

8g

is

3 \times 8 \times g = 24g

. Setting

24g = 360

and dividing by

24

yields

g = 15

.

Step-by-Step Solution

1

Represent the two integers using their ratio and their greatest common divisor.

Let

g = \gcd(a, b)

. Then

a = 3g

and

b = 8g

, where

3

and

8

share no common factors other than

1

.

When two numbers are in reduced ratio

p : q

, dividing both by their GCD leaves coprime factors

p

and

q

.

2

Express the LCM of

a

and

b

in terms of

g

.

\text{LCM}(a, b) = 3 \times 8 \times g = 24g.

The LCM of two numbers

p \cdot g

and

q \cdot g

with

\gcd(p, q) = 1

is

p \cdot q \cdot g

.

3

Solve for

g

using the given LCM value of

360

.

24g = 360 \implies g = 15.

Dividing the given LCM by the product of the coprime ratio components yields the GCD.

Key Concept

Relationship between GCD, LCM, and coprime factor ratios of two positive integers

Estimated Time:1m 15s

Question 163Question →

A positive integer $n$ has only two distinct prime factors, $2$ and $3$ . If $n$ is a multiple of $12$ , is not divisible by $8$ , and has exactly $12$ positive divisors, what is the value of $n$ ?

Show answer & explanation

Answer: 108

Answer

The value of

n

is 108.

Because

n

has only

2

and

3

as prime factors, its prime factorization is

n = 2^a \times 3^b

. Divisibility by

12 = 2^2 \times 3^1

requires

a \ge 2

, and non-divisibility by

8 = 2^3

requires

a < 3

, which forces

a = 2

. Using the formula for total positive divisors,

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

, which simplifies to

3(b+1) = 12

and gives

b = 3

. Calculating

n = 2^2 \times 3^3

yields

108

.

Step-by-Step Solution

1

Set up the prime factorization of

n

n = 2^a \times 3^b

where

a \ge 1

and

b \ge 1

The problem states that

2

and

3

are the only distinct prime factors of

n

.

2

Determine the exact value of exponent

a

a = 2

n

is divisible by

12 = 2^2 \times 3^1

(so

a \ge 2

) but not by

8 = 2^3

(so

a < 3

).

3

Determine the exact value of exponent

b

b = 3

The total number of positive divisors is

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

, which solves to

b = 3

.

4

Calculate the value of

n

n = 108

n = 2^2 \times 3^3 = 4 \times 27 = 108

.

Key Concept

Determining integer values using prime factorization and the number of divisors formula

Question 164Question →

What is the sum of the distinct prime factors of $60$ ?

Show answer & explanation

Answer: 10

Answer

The sum of the distinct prime factors of 60 is 10.

The prime factorization of

60

is

2^2 \times 3 \times 5

. The unique prime factors are

2

,

3

, and

5

. Adding these values together gives

2 + 3 + 5 = 10

.

Step-by-Step Solution

1

Find the prime factorization of 60

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

Decompose 60 into prime factor powers.

2

List the distinct prime factors

The distinct prime factors are 2, 3, and 5

Focus only on the prime bases, ignoring exponents.

3

Sum the distinct prime factors

2 + 3 + 5 = 10

Add the unique prime numbers found in the factorization.

Key Concept

Prime Factorization and Distinct Prime Factors

Question 165Question →

For a positive integer $N$ , the greatest common divisor of $N$ and $360$ is $120$ , and the least common multiple of $N$ and $450$ is $9000$ . What is the value of $N$ ?

Show answer & explanation

Answer: 3000

Answer

The value of

N

is

3000

.

Prime factorizing

360

,

120

,

450

, and

9000

converts the GCD and LCM requirements into a system of min/max equations for the exponents of

2

,

3

, and

5

. The condition

\gcd(N, 360) = 120

dictates that the exponent of

3

in

N

must be exactly

1

, while the exponent of

2

is at least

3

. The condition

\text{lcm}(N, 450) = 9000

dictates that the exponent of

2

must be exactly

3

and the exponent of

5

must be exactly

3

. Combining these constraints gives

N = 2^3 \times 3^1 \times 5^3 = 3000

.

Step-by-Step Solution

1

Find the prime factorizations of all known numbers

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

,

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

,

450 = 2^1 \times 3^2 \times 5^2

, and

9000 = 2^3 \times 3^2 \times 5^3

.

Prime factorization allows us to analyze GCD and LCM conditions using exponent minimums and maximums.

2

Apply the GCD exponent rule

\min(\text{exp}_N(p), \text{exp}_A(p)) = \text{exp}_{GCD}(p)

For factor

2

:

\min(a, 3) = 3 \implies a \ge 3

; For factor

3

:

\min(b, 2) = 1 \implies b = 1

; For factor

5

:

\min(c, 1) = 1 \implies c \ge 1

.

The GCD of two numbers takes the minimum exponent for each prime factor.

3

Apply the LCM exponent rule

\max(\text{exp}_N(p), \text{exp}_B(p)) = \text{exp}_{LCM}(p)

For factor

2

:

\max(a, 1) = 3 \implies a = 3

; For factor

5

:

\max(c, 2) = 3 \implies c = 3

. No prime factors greater than

5

exist in

N

.

The LCM of two numbers takes the maximum exponent for each prime factor.

4

Synthesize the exponents and compute

N

N = 2^3 \times 3^1 \times 5^3 = 8 \times 3 \times 125 = 3000

.

Multiplying out the uniquely determined prime factors gives the value of

N

.

Key Concept

Simultaneous prime exponent analysis using GCD (minimum exponents) and LCM (maximum exponents) rules.

Question 166Question →

For two positive integers $x$ and $y$ with $x < y$ , the greatest common divisor is $\text{GCD}(x, y) = 15$ and the least common multiple is $\text{LCM}(x, y) = 9000$ . If $x$ is a multiple of $8$ but not a multiple of $9$ , what is the value of $y - x$ ?

Show answer & explanation

Answer: 1005

Answer

1005

By writing

x = 15a

and

y = 15b

with

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

, we derive

a \cdot b = \frac{9000}{15} = 600

. Prime factorizing

600 = 8 \times 3 \times 25

yields four coprime pairs

(a, b)

with

a < b

:

(1, 600)

,

(3, 200)

,

(8, 75)

, and

(24, 25)

. These yield candidate values for

x

of

15

,

45

,

120

, and

360

, respectively. Checking the divisibility conditions,

15

and

45

are not multiples of

8

, while

360

is a multiple of

9

. The only value of

x

that is a multiple of

8

and not a multiple of

9

is

120

(corresponding to

a = 8, b = 75

). Hence

y = 15 \times 75 = 1125

, and

y - x = 1125 - 120 = 1005

.

Step-by-Step Solution

1

Set up the algebraic representation using the GCD

x = 15a

and

y = 15b

, where

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

and

a < b

Any two integers can be written as the product of their GCD and coprime factor multipliers.

2

Relate LCM and GCD to find the product of multipliers

a \times b

a \times b = \frac{\text{LCM}(x, y)}{\text{GCD}(x, y)} = \frac{9000}{15} = 600

The product of GCD and LCM equals the product of the numbers:

\text{GCD}(x,y) \times \text{LCM}(x,y) = x \times y = 15a \times 15b = 225ab

.

3

Decompose 600 into coprime component blocks

600 = 2^3 \times 3^1 \times 5^2 = 8 \times 3 \times 25

Since

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

, prime powers cannot be split between

a

and

b

.

4

Form all valid candidate pairs

(a, b)

with

a < b

(1, 600)

,

(3, 200)

,

(8, 75)

, and

(24, 25)

There are

2^{3-1} = 4

ways to partition the 3 prime factor blocks into two coprime factors where

a < b

.

5

Apply divisibility constraints to isolate

x

and

y

x = 120

and

y = 1125

Only

x = 15 \times 8 = 120

satisfies being a multiple of 8 without being a multiple of 9.

6

Calculate the target difference

y - x

1125 - 120 = 1005

Subtracting

x

from

y

yields the required value.

Key Concept

Partitioning prime factor powers of LCM/GCD to identify coprime factor multipliers

Question 167Question →

A positive integer $n$ is not divisible by $3$ . If $n$ has exactly $15$ positive divisors and $2n$ has exactly $20$ positive divisors, how many positive divisors does $3n$ have?

Show answer & explanation

Answer: 30

Answer

30

The number of positive divisors of an integer with prime factorization

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

is

(e_1 + 1)(e_2 + 1) \dots

. Given

d(n) = 15

,

n

can be

p^{14}

or

p^4 q^2

. Given

d(2n) = 20

, 2 must be a prime factor of

n

with exponent 2, making

n = 2^2 p^4

. Since

n

is not divisible by 3,

p \neq 3

. Multiplying

n

by 3 introduces

3^1

into the prime factorization, giving

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

. The total number of positive divisors is

(2+1)(1+1)(4+1) = 30

.

Step-by-Step Solution

1

Analyze the number of positive divisors of

n

.

n

is either

p^{14}

or

p^4 q^2

for distinct primes

p

and

q

.

The number of positive divisors of

n = p_1^{e_1} p_2^{e_2} \dots

is given by

(e_1 + 1)(e_2 + 1) \dots

. Since

15 = 15 \times 1 = 5 \times 3

, the exponent structure must be 14 or 4 and 2.

2

Analyze the number of positive divisors of

2n

.

n = 2^2 p^4

, where

p

is a prime other than 2 and 3.

If

n = p^4 q^2

, multiplying by 2 increases the exponent of 2 by 1. If

q = 2

, then

n = 2^2 p^4

and

2n = 2^3 p^4

, yielding

(3 + 1)(4 + 1) = 20

divisors. Any other prime structure yields a different count.

3

Determine the prime factorization of

3n

and count its positive divisors.

The number of divisors of

3n

is 30.

Since

n

is not divisible by 3,

p \neq 3

. Thus

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

, which has

(2 + 1)(1 + 1)(4 + 1) = 30

positive divisors.

Key Concept

Divisor Count Formula and Prime Factorization

Question 168Question →

A store owner purchases a batch of books for $\$150$ each and sells them at a price that is $40\%$ higher than the purchase price. During a clearance sale, the selling price is reduced by $10\%$ . What is the final sale price, in dollars, of one book?

Show answer & explanation

Answer: 189

Answer

The final sale price of one book is

\$189

.

First, calculate the price after the

40\%

markup:

\$150 \times 1.40 = \$210

. Then apply the

10\%

discount to this new amount:

\$210 \times 0.90 = \$189

. The final price is

\$189

.

Step-by-Step Solution

1

Calculate the price after a 40% markup on the base price of $150.

$210

A 40% markup means multiplying the original purchase price by 1.40.

2

Apply the 10% clearance discount to the marked-up price of $210.

$189

A 10% discount means multiplying the new price by 0.90.

Key Concept

Successive Percentage Change

Estimated Time:45s

Question 169Question →

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 170Question →

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 171Question →

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 172Question →

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 173Question →

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 174Question →

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 175Question →

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 176Question →

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 177Question →

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 178Question →

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 179Question →

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 180Question →

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

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