All practice questions

How many integer values of $n$ satisfy the inequality $|n^2 - 9| \leq 2n + 6$ ?

Show answer & explanation

Answer: 6

Answer

There are 6 integer values of

n

that satisfy the given inequality.

Because

|n^2 - 9| \geq 0

for all real numbers,

2n + 6

must be non-negative, requiring

n \geq -3

. At

n = -3

, the inequality reduces to

0 \leq 0

, which is valid. For

n > -3

, the term

n + 3

is positive, so factoring

|n^2 - 9|

into

|n - 3|(n + 3)

and dividing by

n + 3

simplifies the inequality to

|n - 3| \leq 2

. This yields the integer range

1 \leq n \leq 5

, containing 5 integers. Including

n = -3

gives a total of 6 integer solutions.

Step-by-Step Solution

1

Determine the domain constraint from the non-negativity of the absolute value expression.

2n + 6 \geq 0 \implies n \geq -3

The absolute value expression

|n^2 - 9|

is non-negative for all real

n

, so the right-hand side must also be non-negative.

2

Evaluate the boundary case

n = -3

.

|(-3)^2 - 9| = 0

and

2(-3) + 6 = 0 \implies 0 \leq 0

(True)

When

n = -3

, both sides equal 0, making

n = -3

a valid integer solution.

3

Factor the quadratic inside the absolute value for

n > -3

and simplify.

|n - 3||n + 3| \leq 2(n + 3) \implies |n - 3| \leq 2

Since

n > -3

, the term

n + 3

is strictly positive, allowing division of both sides by

n + 3

without reversing the inequality.

4

Solve the linear absolute value inequality

|n - 3| \leq 2

.

-2 \leq n - 3 \leq 2 \implies 1 \leq n \leq 5

Removing the absolute value creates a compound inequality bounded between

-2

and

2

.

5

List all valid integer solutions and count them.

n \in \{-3, 1, 2, 3, 4, 5\} \implies 6

integer solutions

Combining the boundary root

n = -3

with the five consecutive integers from

1

to

5

gives 6 integer solutions.

Key Concept

Absolute Value Inequalities with Variable RHS and Boundary Factors

Question 42Question →

If $n$ is an integer such that $1 \le n \le 50$ , how many values of $n$ satisfy the condition that $n^2 + 3n$ is an odd integer?

Show answer & explanation

Answer: 0

Answer

0

For any integer

n

, the expression

n^2 + 3n

can be rewritten as

n(n + 3)

. Since

n

and

n + 3

differ by 3 (an odd number), one factor must be even and the other must be odd. The product of an even integer and an odd integer is always even. Therefore,

n^2 + 3n

is even for all integer values of

n

, meaning there are exactly 0 values of

n

in the given range for which the expression is odd.

Step-by-Step Solution

1

Analyze the algebraic expression

n^2 + 3n

by testing parity cases for any integer

n

.

The parity of the expression depends on whether

n

is even or odd.

Every integer is either even or odd.

2

Evaluate

n^2 + 3n

assuming

n

is even.

n^2

is even and

3n

is even; their sum is even.

Even

\times

Even = Even, and Even

+

Even = Even.

3

Evaluate

n^2 + 3n

assuming

n

is odd.

n^2

is odd and

3n

is odd; their sum is even.

Odd

\times

Odd = Odd, and Odd

+

Odd = Even.

4

Count the number of integers

n

in

1 \le n \le 50

for which

n^2 + 3n

is odd.

Because

n^2 + 3n

is even for all integer values of

n

, zero values of

n

make the expression odd.

No integer

n

satisfies the condition.

Key Concept

Odd and Even Integers (Parity)

Estimated Time:1m 0s

Question 43Question →

For how many positive integers $n$ less than or equal to $120$ is the expression $n^3 - n + 3^n$ an even integer?

Show answer & explanation

Answer: 0

Answer

0

Factoring

n^3 - n

into

(n-1)n(n+1)

shows that this component is the product of three consecutive integers and is therefore always even for any integer

n

. Since 3 is odd,

3^n

is always odd for any positive integer exponent

n

. The sum of an even integer and an odd integer is always odd. Consequently,

n^3 - n + 3^n

is odd for all positive integers

n

, meaning 0 values of

n \le 120

yield an even integer.

Step-by-Step Solution

1

Analyze the algebraic expression by factoring

n^3 - n

.

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

, which represents the product of three consecutive integers.

The product of consecutive integers contains at least one even number, making

n^3 - n

even for all integers

n

.

2

Analyze the exponential term

3^n

.

Since 3 is odd,

3^n

is odd for every positive integer

n

.

An odd integer raised to any positive integer exponent yields an odd integer.

3

Determine the overall parity of the expression

n^3 - n + 3^n

.

\text{even} + \text{odd} = \text{odd}

. The expression is odd for all positive integers

n

.

Adding an odd integer to an even integer results in an odd integer.

4

Count the number of positive integers

n \le 120

for which the expression is even.

There are 0 such integers.

Because the expression is odd for every positive integer

n

, it is never even.

Key Concept

Parity rules of consecutive integer products, odd exponents, and addition of even/odd integers.

Question 44Question →

For how many integers $n$ satisfying $0 \le n \le 100$ is the expression $n^4 + 2^n + n$ an odd integer?

Show answer & explanation

Answer: 1

Answer

There is exactly 1 value of

n

(specifically

n = 0

) in the range

0 \le n \le 100

for which the expression

n^4 + 2^n + n

is an odd integer.

For any integer

n \ge 1

,

n^4

and

n

share the exact same parity (both even or both odd), making

n^4 + n

always even. Additionally, for

n \ge 1

,

2^n

is an even integer. The sum of two even integers is always even, so

n^4 + 2^n + n

is even for all

1 \le n \le 100

. When

n = 0

, however,

2^0 = 1

is odd and

0^4 + 0 = 0

is even. The sum

0 + 1 = 1

is odd. Thus,

n = 0

is the single value that satisfies the condition.

Step-by-Step Solution

1

Analyze parity of

n^4 + n

n^4 + n

is even for all integers

n

An integer

n

and its power

n^4

always have the same parity. The sum of two integers of the same parity is always even.

2

Evaluate

2^n

parity for

n \ge 1

2^n

is even for all

n \ge 1

Any positive integer power of 2 contains 2 as a prime factor and is therefore even.

3

Evaluate the expression for

n = 0

0^4 + 2^0 + 0 = 1

, which is odd

Zero is a non-negative integer within

0 \le n \le 100

. By definition of exponents,

2^0 = 1

, which is odd.

4

Combine results and count valid integers

Total count = 1

For

n \ge 1

, even + even = even. For

n = 0

, even + odd = odd. Hence, only

n = 0

produces an odd integer.

Key Concept

Parity Rules for Exponent Base 2 Edge Cases and Algebraic Expressions

Question 45Question →

Let $f(n) = n^5 + 4n^3 + 3n + 1$ for any positive integer $n$ . For how many integer values of $n$ in the range $1 \le n \le 150$ is the expression $E(n) = (f(n))^n + n^{f(n)}$ an even integer?

Show answer & explanation

Answer: 75

Answer

The total number of integer values of

n

in the given range for which the expression is even is 75.

Analyzing

f(n) \pmod 2

reveals that

n^5 \equiv n \pmod 2

and

3n \equiv n \pmod 2

, making

f(n) \equiv n + 0 + n + 1 \equiv 2n + 1 \equiv 1 \pmod 2

. Since

f(n)

is unconditionally odd,

(f(n))^n

is an odd number raised to a positive integer power, which remains odd. The second term

n^{f(n)}

has positive exponent

f(n) \ge 1

, so its parity matches the base

n

. Therefore,

E(n) = \text{Odd} + n

is even precisely when

n

is odd. In the range

1 \le n \le 150

, exactly half of the 150 integers are odd, yielding 75.

Step-by-Step Solution

1

Evaluate the parity of

f(n)

f(n) = n^5 + 4n^3 + 3n + 1 \equiv n + 0 + n + 1 \equiv 2n + 1 \equiv 1 \pmod 2

, so

f(n)

is always odd.

Powers of an integer retain the base parity (

n^5 \equiv n \pmod 2

), and coefficients modulo 2 simplify terms.

2

Evaluate the parity of

(f(n))^n

and

n^{f(n)}

(f(n))^n

is always odd because an odd base raised to any positive integer exponent is odd.

n^{f(n)}

has the same parity as

n

because

f(n) \ge 1

.

Exponent parity rules state that

\text{odd}^k = \text{odd}

and

\text{even}^k = \text{even}

for positive integer exponents

k

.

3

Formulate the condition for

E(n)

to be even

E(n) = \text{Odd} + n

. Thus,

E(n)

is even if and only if

n

is odd.

The sum of two integers is even if and only if both integers have the same parity (Odd + Odd = Even).

4

Count odd integers in the specified interval

There are 75 odd integers in

\{1, 2, 3, \dots, 150\}

.

Half of the integers in any consecutive sequence of even length starting at 1 are odd.

Key Concept

Odd and Even Integers (Parity)

Estimated Time:2m 0s

Question 46Question →

What is the total number of positive factors of the integer $36$ ?

Show answer & explanation

Answer: 9

Answer

The total number of positive factors of

36

is 9.

The prime factorization of

36

is

2^2 \times 3^2

. By the total factors formula, adding 1 to each exponent and multiplying gives

(2+1)(2+1) = 9

. The factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Step-by-Step Solution

1

Find the prime factorization of 36

36 = 4 \times 9 = 2^2 \times 3^2

Decomposing the number into prime factors allows the application of the factor count formula.

2

Apply the exponent formula for counting factors

(2 + 1)(2 + 1) = 3 \times 3 = 9

Each factor of

36

is of the form

2^x \times 3^y

, where

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

(3 choices) and

y \in \{0, 1, 2\}

(3 choices).

Key Concept

Calculating the total number of positive factors from prime factorization exponents.

Estimated Time:45s

Question 47Question →

A positive integer $N$ has no prime factors other than $2$ and $3$ . If $N$ is a multiple of $12$ and has exactly $18$ positive divisors, what is the sum of all possible values of $N$ ?

Show answer & explanation

Answer: 2028

Answer

2028

The correct numerical answer is 2028, obtained by determining all exponent combinations of 2 and 3 that yield 18 total factors while guaranteeing divisibility by 12.

Step-by-Step Solution

1

Set up the prime factorization of

N

with constraints.

N = 2^a \cdot 3^b

with

a \ge 2

and

b \ge 1

.

Since the only prime factors are

2

and

3

, and

12 = 2^2 \cdot 3^1

divides

N

, the exponents must satisfy

a \ge 2

and

b \ge 1

.

2

Apply the total number of divisors formula.

(a+1)(b+1) = 18

.

The number of positive divisors of

p_1^{e_1} p_2^{e_2}

is

(e_1 + 1)(e_2 + 1)

.

3

Identify all valid integer solution pairs for

(a+1, b+1)

.

Valid pairs are

(9,2)

,

(6,3)

, and

(3,6)

, corresponding to

(a,b) = (8,1), (5,2), (2,5)

.

We require

a+1 \ge 3

and

b+1 \ge 2

. Pairs

(18,1)

and

(2,9)

violate these lower bounds.

4

Compute the corresponding values of

N

and find their sum.

768 + 288 + 972 = 2028

.

Evaluating

2^8 \cdot 3^1 = 768

,

2^5 \cdot 3^2 = 288

, and

2^2 \cdot 3^5 = 972

yields a total sum of

2028

.

Key Concept

Divisor Count Formula & Multiplicativity Constraints

Question 48Question →

For how many integer values of $k$ in the range $1 \le k \le 60$ is the expression $3k^2 + 5k + 7$ an even integer?

Show answer & explanation

Answer: 0

Answer

0

The expression

3k^2 + 5k + 7

can be evaluated for parity by considering the cases for

k

.
If

k

is even,

3k^2

is even,

5k

is even, and

7

is odd. The sum of two even integers and an odd integer (even + even + odd) is always odd.
If

k

is odd,

3k^2

is odd,

5k

is odd, and

7

is odd. The sum of three odd integers (odd + odd + odd) is always odd.
Because the expression yields an odd integer for every integer

k

, there are no integer values of

k

in the specified range for which the expression is even. Therefore, the correct count is 0.

Step-by-Step Solution

1

Examine parity by testing even and odd cases for k

If k is even: 3(even)^2 + 5(even) + 7 = even + even + odd = odd. If k is odd: 3(odd)^2 + 5(odd) + 7 = odd + odd + odd = odd.

Covering both cases establishes the parity of the expression for all integer inputs.

2

Count the number of values of k in 1 <= k <= 60 that yield an even result

Since the expression is odd for all integer values of k, zero values yield an even integer.

The question specifically asks for the number of integer values of k that make the expression even.

Key Concept

Parity Rules for Addition, Multiplication, and Algebraic Expressions

Question 49Question →

Let $A = 12^6 \cdot 35^4$ and $B = 18^4 \cdot 14^6$ . If $d = \text{gcd}(A, B)$ , how many positive factors of $d^2$ are not factors of $d$ ?

Show answer & explanation

Answer: 2072

Answer

The number of positive factors of

d^2

that are not factors of

d

is 2072.

Decomposing

A

and

B

into prime factors yields

A = 2^{12} \cdot 3^6 \cdot 5^4 \cdot 7^4

and

B = 2^{10} \cdot 3^8 \cdot 7^6

. Taking the minimum power of each common prime gives

d = \text{gcd}(A, B) = 2^{10} \cdot 3^6 \cdot 7^4

, which has

(10+1)(6+1)(4+1) = 385

positive factors. For

d^2 = 2^{20} \cdot 3^{12} \cdot 7^8

, the total number of positive factors is

(20+1)(12+1)(8+1) = 2457

. Subtracting the factors of

d

yields

2457 - 385 = 2072

.

Step-by-Step Solution

1

Find the prime factorizations of

A

and

B

A = 2^{12} \cdot 3^6 \cdot 5^4 \cdot 7^4

and

B = 2^{10} \cdot 3^8 \cdot 7^6

Converting composite bases into prime factors allows determination of common divisor properties.

2

Determine the greatest common divisor

d = \text{gcd}(A, B)

d = 2^{10} \cdot 3^6 \cdot 7^4

The GCD takes the minimum exponent for each common prime factor between

A

and

B

.

3

Calculate the total number of positive factors of

d

385 positive factors

Adding 1 to each prime exponent of

d

and multiplying gives

(10+1)(6+1)(4+1) = 385

.

4

Find the prime factorization and number of positive factors of

d^2

d^2 = 2^{20} \cdot 3^{12} \cdot 7^8

, which has 2457 positive factors

Squaring

d

doubles all prime exponents. The number of factors is

(20+1)(12+1)(8+1) = 2457

.

5

Subtract the number of factors of

d

from the number of factors of

d^2

2457 - 385 = 2072

Since every factor of

d

is also a factor of

d^2

, the factors of

d^2

that are not factors of

d

equal the total factors of

d^2

minus the factors of

d

.

Key Concept

Prime Factorization, Greatest Common Divisor (GCD), and the Divisor Count Formula

Question 50Question →

For how many integers $x$ in the range $-50 \le x \le 50$ is the value of the expression $x^5 - 5x^3 + 4x + 3$ an even integer?

Show answer & explanation

Answer: 0

Answer

There are 0 integers in the specified range for which the expression evaluates to an even integer.

Factoring

x^5 - 5x^3 + 4x

yields

(x - 2)(x - 1)x(x + 1)(x + 2)

, which is the product of 5 consecutive integers. The product of consecutive integers is always even for any integer

x

. Adding 3 (an odd integer) to an even integer results in an odd integer for all values of

x

. Consequently, zero integers in the given range result in an even value.

Step-by-Step Solution

1

Factor the variable portion of the polynomial

x^5 - 5x^3 + 4x + 3

.

The expression rewrites as

(x - 2)(x - 1)x(x + 1)(x + 2) + 3

.

Factoring helps identify structural properties such as consecutive terms.

2

Analyze the parity of the product of 5 consecutive integers

(x - 2)(x - 1)x(x + 1)(x + 2)

.

The product is always an even integer for all integer values of

x

.

Any sequence of 5 consecutive integers includes multiple even numbers. Even when

x = 0

, the product equals 0, which is an even integer.

3

Determine the parity of the full expression by adding the constant term 3.

Even integer + 3 (odd integer) = odd integer.

The sum of an even integer and an odd integer is always an odd integer.

4

Count the number of integer values of

x

in

-50 \le x \le 50

that yield an even integer.

0 integers.

Since the expression evaluates to an odd integer for every integer

x

, no integer value can produce an even integer.

Key Concept

Parity of consecutive integer products and addition rules for even and odd integers.

Estimated Time:2m 0s

Question 51Question →

What is the least common multiple (LCM) of $12$ and $18$ ?

Show answer & explanation

Answer: 36

Answer

The least common multiple of

12

and

18

is

36

.

To find the least common multiple of

12

and

18

, determine the prime factorization of each number (

12 = 2^2 \times 3^1

and

18 = 2^1 \times 3^2

). The LCM is calculated by taking the maximum power of each prime factor present:

2^2 \times 3^2 = 4 \times 9 = 36

.

Step-by-Step Solution

1

Express both numbers in terms of their prime factorizations.

12 = 2^2 \times 3^1

and

18 = 2^1 \times 3^2

Prime factorization separates each number into basic prime components.

2

Select the highest power of each prime factor that appears in either factorization.

The highest power of

2

is

2^2

, and the highest power of

3

is

3^2

.

The least common multiple must contain enough factors to be divisible by both original numbers.

3

Calculate the product of these highest prime powers.

\text{LCM}(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36

Multiplying the chosen powers yields the minimum integer divisible by both

12

and

18

.

Key Concept

Finding the Least Common Multiple (LCM) via Prime Factorization

Estimated Time:45s

Question 52Question →

For how many ordered pairs of positive integers $(a, b)$ , where $1 \le a \le 15$ and $1 \le b \le 15$ , is the value of the expression $(a^2 + a + 1)(b^3 + b + 7)$ an even integer?

Show answer & explanation

Answer: 0

Answer

The expression evaluates to an odd integer for all positive integer pairs

(a, b)

, so there are 0 ordered pairs for which the value is an even integer.

The factor

(a^2 + a + 1)

simplifies to

a(a + 1) + 1

. Because

a(a + 1)

is the product of two consecutive integers, it is always even, making

a(a + 1) + 1

odd for every integer

a

. The factor

(b^3 + b + 7)

is always odd because

b^3

and

b

have matching parities (their sum is always even), so adding 7 yields an odd number. Since the product of two odd integers is always odd, the expression is never even, yielding exactly 0 ordered pairs.

Step-by-Step Solution

1

Examine the parity of the factor

a^2 + a + 1

a^2 + a + 1

is odd for all integer values of

a

The expression

a^2 + a = a(a + 1)

represents the product of two consecutive integers, which is always even. Adding 1 to an even integer results in an odd integer.

2

Examine the parity of the factor

b^3 + b + 7

b^3 + b + 7

is odd for all integer values of

b

Since

b^3

and

b

always share the same parity, their sum

b^3 + b

is always even. Adding 7 to an even integer results in an odd integer.

3

Determine the overall parity of the product

The product

(a^2 + a + 1)(b^3 + b + 7)

is odd for all inputs

The product of two odd integers is strictly an odd integer.

4

Count the number of pairs satisfying the even condition

0 pairs

Because the product is never even, zero pairs satisfy the requirement.

Key Concept

Parity rules of consecutive integer products and polynomial expressions

Estimated Time:1m 30s

Question 53Question →

What is the units digit of $7^{43}$ ?

Show answer & explanation

Answer: 3

Answer

The units digit of

7^{43}

is 3.

The units digits of powers of 7 repeat in a pattern of four terms: 7, 9, 3, 1. Dividing the exponent 43 by 4 yields a remainder of 3. The 3rd term in the repeating pattern is 3, so the units digit of

7^{43}

is 3.

Step-by-Step Solution

1

Determine the cyclicity pattern of the units digit of powers of 7.

The units digits for

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

follow the repeating sequence 7, 9, 3, 1 with a cycle length of 4.

Units digits of positive integer powers follow a periodic pattern.

2

Divide the exponent 43 by the pattern cycle length of 4.

43 \div 4 = 10

remainder 3.

The remainder determines which term in the repeating sequence gives the units digit.

3

Find the units digit corresponding to the 3rd term in the cyclicity sequence.

The 3rd digit in the sequence (7, 9, 3, 1) is 3.

A remainder of 3 corresponds to the 3rd power in the cycle,

7^3

.

Key Concept

Units Digit Cyclicity

Question 54Question →

What is the smallest prime factor of $143$ ?

Show answer & explanation

Answer: 11

Answer

The smallest prime factor of 143 is 11.

To find the smallest prime factor of 143, test prime numbers in increasing order starting from 2. Testing 2, 3, 5, and 7 shows none are factors of 143. Testing 11 yields

143 \div 11 = 13

, which is an integer. Thus,

143 = 11 \times 13

. Since 11 and 13 are both prime, 11 is the smallest prime factor of 143.

Step-by-Step Solution

1

Test divisibility of 143 by small prime numbers in ascending order

143 is not divisible by 2, 3, 5, or 7.

Basic divisibility rules eliminate 2 (odd), 3 (digit sum 8), and 5 (ends in 3). Direct division shows 143 = 7 × 20 + 3.

2

Test divisibility by the next prime number, 11

143 divided by 11 equals 13 exactly.

Since 143 = 11 × 13 and 11 is prime, 11 is a prime factor.

3

Identify the smallest prime factor

The prime factors of 143 are 11 and 13, making 11 the smallest.

No prime number smaller than 11 divides 143.

Key Concept

Prime Factorization

Question 55Question →

A consultancy allocated a total budget for a digital transformation project. In the first phase, $\frac{3}{8}$ of the initial budget was spent. In the second phase, $40\%$ of the remaining budget was spent. If the unspent amount after both phases is $\$27,000$ , what was the initial budget of the project, in dollars?

Show answer & explanation

Answer: 72000

Answer

72,000

To determine the initial budget, track the unspent fraction step by step. After the first phase,

\frac{5}{8}

of the budget remains. In the second phase,

40\%

(or

\frac{2}{5}

) of that remaining fraction is spent, which leaves

60\%

(or

\frac{3}{5}

) of the

\frac{5}{8}

unspent. Multiplying

\frac{3}{5} \times \frac{5}{8} = \frac{3}{8}

. Because

\frac{3}{8}

of the initial budget equals

\$27,000

, the full initial budget is

27,000 \times \frac{8}{3} = 72,000

.

Step-by-Step Solution

1

Find the fraction of the budget remaining after the first phase.

The fraction remaining after Phase 1 is

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

.

The first phase spends

\frac{3}{8}

of the initial total budget.

2

Find the fraction of the total initial budget spent in the second phase.

Phase 2 expenditure =

40\% \times \frac{5}{8} = \frac{2}{5} \times \frac{5}{8} = \frac{2}{8}

.

40\%

is equivalent to

\frac{2}{5}

, which applies to the remaining

\frac{5}{8}

of the budget.

3

Determine the remaining fraction of the total budget after both phases.

Remaining fraction =

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

.

Subtracting the fraction spent in Phase 2 from the fraction left after Phase 1 yields the unspent portion.

4

Calculate the initial total budget in dollars.

Initial Budget =

\$27,000 \times \frac{8}{3} = \$72,000

.

Setting

\frac{3}{8}

of the total budget equal to the unspent amount of

\$27,000

determines the total.

Key Concept

Sequential fractional and percentage reductions from a base quantity

Question 56Question →

Three automated security beacons flash at regular intervals of $18$ seconds, $24$ seconds, and $42$ seconds, respectively. If all three beacons flash simultaneously at 8:00:00 AM, how many times will all three beacons flash simultaneously between 8:01:00 AM and 9:00:00 AM, inclusive?

Show answer & explanation

Answer: 7

Answer

7

The least common multiple of 18, 24, and 42 is 504 seconds. The timeframe between 8:01:00 AM and 9:00:00 AM inclusive represents elapsed times from 60 seconds to 3600 seconds after 8:00:00 AM. Dividing 3600 by 504 gives 7 full periods (3528 seconds), all of which are greater than 60 seconds. Therefore, the beacons flash simultaneously 7 times within the specified period.

Step-by-Step Solution

1

Find the Least Common Multiple (LCM) of the three flash intervals.

LCM(18, 24, 42) = 504 seconds

The beacons flash simultaneously at time intervals that are integer multiples of the LCM of their individual cycle times.

2

Convert the target time interval (8:01:00 AM to 9:00:00 AM) into seconds after the initial flash at 8:00:00 AM (t = 0).

Range is from t = 60 seconds to t = 3600 seconds

8:01:00 AM is 60 seconds after 8:00:00 AM, and 1 hour (9:00:00 AM) equals 3600 seconds after 8:00:00 AM.

3

Determine the number of positive multiples of 504 within the interval [60, 3600].

7 multiples (504, 1008, 1512, 2016, 2520, 3024, 3528)

Dividing 3600 by 504 gives 7 with a remainder of 72. Since the first positive multiple (504) is greater than 60, all 7 multiples fall strictly within the specified window.

Key Concept

Application of Least Common Multiple (LCM) to periodic events

Question 57Question →

Two positive integers $m$ and $n$ satisfy $m < n$ . Their greatest common divisor is $18$ and their least common multiple is $1,080$ . If $m$ is not divisible by $27$ and $n$ is not divisible by $5$ , what is the value of $n - m$ ?

Show answer & explanation

Answer: 126

Answer

The value of

n - m

is

126

.

By writing

m = 18a

and

n = 18b

with

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

, we find

a \cdot b = 60

. Evaluating all coprime factor pairs

(a, b)

shows that only

(5, 12)

satisfies all constraints (

m = 90

is not a multiple of

27

, and

n = 216

is not a multiple of

5

). The difference is

216 - 90 = 126

.

Step-by-Step Solution

1

Relate GCD, LCM, and coprime factors.

Let

m = 18a

and

n = 18b

, where

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

and

a < b

. Then

a \cdot b = \frac{1080}{18} = 60

.

The product of two positive integers equals the product of their GCD and LCM.

2

Determine all coprime factor pairs of

60

.

The coprime pairs

(a, b)

with

a < b

such that

a \cdot b = 60

are

(1, 60)

,

(3, 20)

,

(4, 15)

, and

(5, 12)

.

Since

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

, prime factors of

60 = 2^2 \cdot 3 \cdot 5

cannot be shared between

a

and

b

.

3

Apply the given constraints to eliminate invalid pairs.

The pair

(5, 12)

is the only pair satisfying both conditions:

m = 90

is not divisible by

27

, and

n = 216

is not divisible by

5

.

Pairs

(1,60)

and

(4,15)

yield

n

divisible by

5

; pair

(3,20)

yields

m=54

which is divisible by

27

.

4

Compute the final difference

n - m

.

n - m = 216 - 90 = 126

.

Subtract

m

from

n

using the valid values found.

Key Concept

Properties of GCD and LCM, prime factorization, and coprime factor pairs

Estimated Time:2m 0s

Question 58Question →

A clothing store reduced the original price of a jacket by $20\%$ . During a promotional event, the store offered an additional $15\%$ discount off the reduced price. If the final price of the jacket was $\$102$ , what was the original price of the jacket, in dollars?

Show answer & explanation

Answer: 150

Answer

The original price of the jacket was $150.

To find the original price, express the final price as a decimal multiplier of the original price. A

20\%

discount leaves

80\%

of the original price (

0.80

). An additional

15\%

discount off the reduced price leaves

85\%

of that amount (

0.85

). The overall price multiplier is

0.80 \times 0.85 = 0.68

. Setting

0.68 \times \text{Original Price} = 102

gives an original price of

\frac{102}{0.68} = 150

dollars.

Step-by-Step Solution

1

Determine the combined price multiplier after both successive discounts.

The first discount leaves

80\%

(

0.80

) of the original price, and the second discount leaves

85\%

(

0.85

) of that reduced price. The combined multiplier is

0.80 \times 0.85 = 0.68

.

Successive percentage discounts are multiplicative rather than additive.

2

Set up an equation equating the final discounted price to

102

.

0.68 \times P = 102

, where

P

represents the original price.

The final price is

68\%

of the original base price.

3

Solve for the original price

P

.

P = \frac{102}{0.68} = \frac{10200}{68} = 150

.

Dividing the final amount by the net decimal multiplier yields the original value.

Key Concept

Successive percentage change calculation

Estimated Time:1m 0s

Question 59Question →

What is the remainder when the sum $S = 3^1 + 3^2 + 3^3 + \dots + 3^{2025}$ is divided by $100$ ?

Show answer & explanation

Answer: 63

Answer

The remainder when the sum

S

is divided by

100

is

63

.

Using the geometric series formula,

S = \frac{3(3^{2025} - 1)}{2}

. Computing

3^{2025} \pmod{200}

gives

43 \pmod{200}

because

3^{20} \equiv 1 \pmod{200}

and

2025 \equiv 5 \pmod{20}

. Substituting this into the numerator yields

3(43 - 1) = 126 \pmod{200}

. Dividing by

2

produces the remainder

63

modulo

100

.

Step-by-Step Solution

1

Sum the geometric series

S = 3^1 + 3^2 + \dots + 3^{2025}

S = \frac{3(3^{2025} - 1)}{2}

Applying the standard finite geometric series formula

a\frac{r^n - 1}{r - 1}

with

a = 3

and

r = 3

.

2

Set up modular equation for

S \pmod{100}

Evaluate

3(3^{2025} - 1) \pmod{200}

Dividing an integer congruence by

2

requires doubling the modulus to

200

to maintain exact divisibility.

3

Find the cyclicity period of

3^n \pmod{200}

3^{20} \equiv 1 \pmod{200}

3^5 = 243 \equiv 43

,

3^{10} \equiv 43^2 = 1849 \equiv 49

, and

3^{20} \equiv 49^2 = 2401 \equiv 1 \pmod{200}

.

4

Compute

3^{2025} \pmod{200}

3^{2025} \equiv 3^5 \equiv 43 \pmod{200}

Dividing the exponent

2025

by the period length

20

leaves a remainder of

5

.

5

Calculate the numerator

3(3^{2025} - 1) \pmod{200}

and divide by 2

3(43 - 1) = 126 \pmod{200}

, so

S \equiv 63 \pmod{100}

Simplifying

126 / 2 = 63

gives the unique remainder modulo

100

.

Key Concept

Geometric series summation combined with modular arithmetic and cyclicity mod 200

Question 60Question →

At a logistics distribution hub, a shipment of incoming packages was processed over two shifts. During the morning shift, $0.35$ of the total shipment was processed and dispatched. During the evening shift, $\frac{4}{13}$ of the remaining packages were processed. If $450$ packages remained unprocessed at the end of both shifts, what was the total number of packages in the initial shipment?

Show answer & explanation

Answer: 1000

Answer

1000 packages

To find the initial shipment size, first express

0.35

as the fraction

\frac{7}{20}

. Subtracting this from

1

leaves

\frac{13}{20}

of the initial total after the morning shift. Next, processing

\frac{4}{13}

of these remaining packages leaves

1 - \frac{4}{13} = \frac{9}{13}

of that remaining amount. Multiplying

\frac{9}{13} \times \frac{13}{20}

shows that

\frac{9}{20}

of the original total remains unprocessed. Finally, setting

\frac{9}{20}

of the total equal to

450

packages yields an initial total of

1000

packages.

Step-by-Step Solution

1

Convert decimal portion to fraction and find remaining fraction after morning shift

Remaining fraction after morning shift is

\frac{13}{20}

Since

0.35 = \frac{7}{20}

of the total shipment was processed,

1 - \frac{7}{20} = \frac{13}{20}

of the initial shipment remained.

2

Determine the remaining fraction of the shipment after the evening shift

Final remaining fraction of the initial shipment is

\frac{9}{20}

The evening shift processed

\frac{4}{13}

of the remaining packages, leaving

1 - \frac{4}{13} = \frac{9}{13}

of the remaining packages. Thus,

\frac{9}{13} \times \frac{13}{20} = \frac{9}{20}

of the initial shipment remained.

3

Solve for the total initial number of packages

Total initial packages =

1000

Setting

\frac{9}{20} N = 450

gives

N = 450 \times \frac{20}{9} = 1000

.

Key Concept

Combining decimal-to-fraction conversions with successive remaining fraction calculations

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