All practice questions

Question 221Question →

Set $S$ consists of $n$ consecutive integers. The sum of the smallest three integers in set $S$ is $-33$ , and the sum of the largest three integers in set $S$ is $69$ . What is the value of $n$ ?

Show answer & explanation

Answer: 37

Answer

The total number of consecutive integers in set

S

is 37.

Solving

3x + 3 = -33

yields a first term of

-12

, and solving

3y - 3 = 69

yields a last term of

24

. The total count of consecutive integers in an inclusive range is

\text{last} - \text{first} + 1

, giving

24 - (-12) + 1 = 37

.

Step-by-Step Solution

1

Find the smallest integer in set S

The smallest integer is -12

Let the smallest integer be

x

. The sum of the smallest three consecutive integers is

x + (x + 1) + (x + 2) = 3x + 3

. Setting

3x + 3 = -33

yields

3x = -36

, so

x = -12

.

2

Find the largest integer in set S

The largest integer is 24

Let the largest integer be

y

. The sum of the largest three consecutive integers is

(y - 2) + (y - 1) + y = 3y - 3

. Setting

3y - 3 = 69

yields

3y = 72

, so

y = 24

.

3

Calculate the total number of elements n in set S

n = 37

For an inclusive set of consecutive integers bounded by first term

a

and last term

b

, the total number of terms is

n = b - a + 1

. Here,

n = 24 - (-12) + 1 = 37

.

Key Concept

Counting inclusive terms in a sequence of consecutive integers

Question 222Question →

A set $S$ consists of $n$ consecutive integers. The sum of all the integers in set $S$ is $675$ , and the product of the smallest integer and the largest integer in set $S$ is $1{}976$ . What is the value of $n$ ?

Show answer & explanation

Answer: 15

Answer

15

For any set of

n

consecutive integers, the median

m

equals the arithmetic mean

\frac{675}{n}

. The smallest and largest elements can be written as

m - \frac{n-1}{2}

and

m + \frac{n-1}{2}

, respectively. Their product is

m^2 - \left(\frac{n-1}{2}\right)^2 = 1976

. Substituting

n = 15

gives

m = 45

, leading to

45^2 - 7^2 = 2025 - 49 = 1976

, which satisfies all conditions.

Step-by-Step Solution

1

Relate the sum of the set to its mean and number of terms

The mean (arithmetic average) of

n

consecutive integers is equal to the median

m

, so

\text{Sum} = n \cdot m = 675

, which implies

m = \frac{675}{n}

.

For any evenly spaced set, the sum equals the number of terms times the mean.

2

Express the smallest and largest elements in terms of the median

m

and number of terms

n

The smallest element is

a = m - \frac{n-1}{2}

and the largest element is

b = m + \frac{n-1}{2}

.

In a set of

n

consecutive integers, the distance from the median to either endpoint is

\frac{n-1}{2}

.

3

Formulate the product equation using the difference of squares

The product of the smallest and largest elements is

a \cdot b = \left(m - \frac{n-1}{2}\right)\left(m + \frac{n-1}{2}\right) = m^2 - \left(\frac{n-1}{2}\right)^2 = 1976

.

Applying the difference of squares identity

(x-y)(x+y) = x^2 - y^2

simplifies the expression.

4

Substitute

m = \frac{675}{n}

and solve for

n

Substituting

m

yields

\left(\frac{675}{n}\right)^2 - \left(\frac{n-1}{2}\right)^2 = 1976

. Testing odd integer factors

n

of

675

:
- If

n = 9

:

m = 75

, product

= 75^2 - 4^2 = 5625 - 16 = 5609 \neq 1976

.
- If

n = 15

:

m = 45

, product

= 45^2 - 7^2 = 2025 - 49 = 1976

.
Hence,

n = 15

.

Since

n

must be a positive integer factor of

675

, checking candidate factors narrows down the unique solution.

Key Concept

Properties of consecutive integer sets: mean-median equivalence and difference of squares decomposition for endpoints.

Alternative Method

Let the set be

\{a, a+1, \dots, a+n-1\}

. The sum is

\frac{n(2a + n - 1)}{2} = 675 \implies n(2a + n - 1) = 1350

. The product of endpoints is

a(a + n - 1) = 1976

. Solving the system of equations for integer values of

a

and

n

yields

a = 38

and

n = 15

.

Estimated Time:2m 30s

Question 223Question →

Let $k$ be a positive integer with the prime factorization $k = 2^x \cdot 3^y \cdot 5^z$ , where $x$ , $y$ , and $z$ are positive integers. If $k$ is divisible by both $18$ and $75$ , and $k$ has exactly $36$ positive integer divisors, what is the maximum possible value of $x + y + z$ ?

7

8

9

11

12

Show answer & explanation

Answer: 8

Answer

The maximum possible value of

x + y + z

is 8.

The correct answer is 8. The prime factorization of

k

requires

x \ge 1

,

y \ge 2

, and

z \ge 2

because

k

is a multiple of

18 = 2^1 \cdot 3^2

and

75 = 3^1 \cdot 5^2

. The number of positive divisors is given by

(x+1)(y+1)(z+1) = 36

. Under the constraints

x+1 \ge 2

,

y+1 \ge 3

, and

z+1 \ge 3

, the factorizations of 36 into three factors yield the sums

x+y+z = 8

(from factors

2, 3, 6

) and

x+y+z = 7

(from factors

4, 3, 3

). Therefore, the maximum possible value is 8.

Step-by-Step Solution

1

Determine the lower bounds for the exponents

x

,

y

, and

z

based on divisibility conditions.

Since

k

is divisible by

18 = 2^1 \cdot 3^2

, we must have

x \ge 1

and

y \ge 2

. Since

k

is divisible by

75 = 3^1 \cdot 5^2

, we must have

y \ge 2

and

z \ge 2

.

For a prime factorization to be divisible by another number, each prime factor's exponent in

k

must be at least as large as its corresponding exponent in the divisor.

2

Set up the equation for the total number of positive integer divisors of

k

.

(x + 1)(y + 1)(z + 1) = 36

, with constraints

x + 1 \ge 2

,

y + 1 \ge 3

, and

z + 1 \ge 3

.

The total number of positive divisors of

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

is given by

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

.

3

Find all valid integer factor triples

(A, B, C) = (x+1, y+1, z+1)

multiplying to 36 under the given constraints.

The valid triples

(A, B, C)

with

A \ge 2, B \ge 3, C \ge 3

are

(2, 3, 6)

,

(2, 6, 3)

, and

(4, 3, 3)

.

Testing factorizations of

36

:

2 \cdot 3 \cdot 6 = 36

,

2 \cdot 6 \cdot 3 = 36

, and

4 \cdot 3 \cdot 3 = 36

all meet the inequality bounds for each term.

4

Calculate the sum

x + y + z = (A + B + C) - 3

for each valid triple and identify the maximum.

For

(2, 3, 6)

:

x+y+z = (2-1) + (3-1) + (6-1) = 1 + 2 + 5 = 8

.
For

(2, 6, 3)

:

x+y+z = (2-1) + (6-1) + (3-1) = 1 + 5 + 2 = 8

.
For

(4, 3, 3)

:

x+y+z = (4-1) + (3-1) + (3-1) = 3 + 2 + 2 = 7

.
The maximum possible value is

8

.

Comparing all valid scenarios yields

8

as the maximum sum of exponents.

Key Concept

Calculating total positive integer divisors from prime factorizations and analyzing exponent constraints derived from divisibility conditions.

Estimated Time:2m 0s

Question 224Question →

What is the smallest positive integer $n$ such that $n$ is a multiple of $45$ , $n$ is not divisible by $8$ , and $n$ has exactly $18$ positive integer divisors?

Show answer & explanation

Answer: 180

Answer

180

The smallest positive integer satisfying all conditions is 180 because 180 = 2^2 * 3^2 * 5^1, which has (2+1)(2+1)(1+1) = 18 positive divisors, is a multiple of 45, and is not divisible by 8.

Step-by-Step Solution

1

Determine the prime factorization constraints for n

n must be of the form 2^a * 3^b * 5^c * ..., where b >= 2, c >= 1, and a <= 2.

n is a multiple of 45 = 3^2 * 5^1, requiring at least 3^2 and 5^1. Since n is not divisible by 8 = 2^3, the exponent of 2 cannot exceed 2.

2

Analyze the total number of divisors constraint

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

The total number of positive divisors is calculated by adding 1 to each exponent in the prime factorization and multiplying them together.

3

Find the optimal prime exponent configuration to minimize n

Factor 18 as 3 * 3 * 2, corresponding to exponents 2, 2, and 1.

Using three prime factors (2, 3, and 5) with smaller exponents yields a smaller total integer than using fewer prime factors with larger exponents.

4

Assign exponents to prime bases to yield the minimum integer value

n = 2^2 * 3^2 * 5^1 = 180

Assigning exponent 2 to base 2, exponent 2 to base 3, and exponent 1 to base 5 satisfies all constraints (b = 2 >= 2, c = 1 >= 1, a = 2 < 3) and minimizes the result.

Key Concept

Divisor count formula and prime factorization constraints

Question 225Question →

If $n$ is a positive integer such that the units digit of $7^n$ is $3$ , what is the remainder when $3^{n+2} + 8^{n+1}$ is divided by $5$ ?

0

1

2

3

4

Show answer & explanation

Answer: $4$

Answer

The correct remainder is

4

.

The cyclicity pattern of the units digit of

7^n

repeats every

4

terms:

7, 9, 3, 1

. For the units digit to be

3

,

n

must leave a remainder of

3

when divided by

4

(

n \equiv 3 \pmod 4

). Using this form,

n+2 \equiv 5 \equiv 1 \pmod 4

, which means

3^{n+2} \equiv 3^1 \equiv 3 \pmod 5

. Similarly,

n+1 \equiv 4 \equiv 0 \pmod 4

, and since

8 \equiv 3 \pmod 5

, we have

8^{n+1} \equiv 3^0 \equiv 1 \pmod 5

. Adding these values gives

3 + 1 = 4

, so the remainder when divided by

5

is

4

.

Step-by-Step Solution

1

Determine the remainder of

n

when divided by

4

using the units digit cyclicity of

7^n

.

n \equiv 3 \pmod 4

.

The units digits of powers of

7

follow a repeating pattern of period

4

:

7^1 \rightarrow 7

,

7^2 \rightarrow 9

,

7^3 \rightarrow 3

,

7^4 \rightarrow 1

. Since the units digit of

7^n

is

3

,

n

must be of the form

4k + 3

.

2

Evaluate

3^{n+2} \pmod 5

.

3^{n+2} \equiv 3 \pmod 5

.

Substitute

n = 4k + 3

into the exponent:

n + 2 = 4k + 5 = 4(k+1) + 1

. The powers of

3 \pmod 5

repeat every

4

powers (

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

). Thus,

3^{4k+5} \equiv 3^1 \equiv 3 \pmod 5

.

3

Evaluate

8^{n+1} \pmod 5

.

8^{n+1} \equiv 1 \pmod 5

.

First simplify the base:

8 \equiv 3 \pmod 5

, so

8^{n+1} \equiv 3^{n+1} \pmod 5

. Substitute

n = 4k + 3

:

n + 1 = 4k + 4 = 4(k+1)

. Since the exponent is a multiple of

4

,

3^{4(k+1)} \equiv 3^4 \equiv 1 \pmod 5

.

4

Sum the modular results to find the final remainder modulo

5

.

(3 + 1) \pmod 5 = 4

.

By properties of modular addition,

(3^{n+2} + 8^{n+1}) \pmod 5 \equiv (3 + 1) \pmod 5 = 4

.

Key Concept

Units Digit Cyclicity and Modular Arithmetic Exponent Rules

Estimated Time:2m 0s

Question 226Question →

Let $S = 2^{2026} - 3^{2026} + 4^{2026} - 7^{2026}$ . What is the remainder when $S$ is divided by $10$ ?

0

2

4

6

8

Show answer & explanation

Answer: 2

Answer

The remainder when

S

is divided by

10

is

2

.

Finding the remainder when an expression is divided by 10 is equivalent to finding the units digit of that expression. The units digits of powers of 2, 3, 4, and 7 repeat in periodic cycles of length 4, 4, 2, and 4, respectively. Since

2026 \equiv 2 \pmod 4

, the units digits correspond to the 2nd term of each cycle:

2^2 \rightarrow 4

,

3^2 \rightarrow 9

,

4^2 \rightarrow 6

, and

7^2 \rightarrow 9

. Evaluating the expression yields

4 - 9 + 6 - 9 = -8

. In modular arithmetic, a negative remainder

-8 \pmod{10}

is equivalent to

-8 + 10 = 2

. Therefore, the value representing 2 is correct.

Step-by-Step Solution

1

Determine the remainder of each term divided by 10 by finding the units digit cyclicity.

Powers of 2 cycle with period 4 (2, 4, 8, 6). Since

2026 = 4 \times 506 + 2

,

2^{2026} \equiv 2^2 \equiv 4 \pmod{10}

.

Dividing an integer by 10 yields a remainder equal to its units digit.

2

Evaluate the units digits for the remaining terms

3^{2026}

,

4^{2026}

, and

7^{2026}

.

Powers of 3 cycle with period 4 (3, 9, 7, 1);

3^{2026} \equiv 3^2 \equiv 9 \pmod{10}

. Powers of 4 cycle with period 2 (4, 6);

4^{2026} \equiv 4^2 \equiv 6 \pmod{10}

. Powers of 7 cycle with period 4 (7, 9, 3, 1);

7^{2026} \equiv 7^2 \equiv 9 \pmod{10}

.

Each base follows a repeating pattern of units digits when raised to successive positive integer powers.

3

Substitute the congruent remainder values back into the expression for

S

.

S \equiv 4 - 9 + 6 - 9 = -8 \pmod{10}

.

Modular arithmetic operations preserve addition and subtraction equivalences.

4

Convert the negative result to a standard non-negative remainder.

-8 + 10 = 2

. Thus, the remainder is

2

.

By definition, the remainder

r

when an integer is divided by

d

must satisfy

0 \leq r < d

.

Key Concept

Units Digit Cyclicity and Negative Remainder Rules

Estimated Time:2m 30s

Question 227Question →

A bakery sells blueberry muffins for $3 each and chocolate chip muffins for$ 4 each. On Saturday, the bakery sold a total of 50 muffins for $170. How many chocolate chip muffins were sold on Saturday?

Show answer & explanation

Answer: 20

Answer

The bakery sold 20 chocolate chip muffins on Saturday.

Let

b

represent the number of blueberry muffins and

c

represent the number of chocolate chip muffins sold. Based on the problem text, we construct the system of equations:

b + c = 50

and

3b + 4c = 170

. From the first equation,

b = 50 - c

. Substituting this into the second equation yields

3(50 - c) + 4c = 170

, which expands to

150 - 3c + 4c = 170

. Simplifying gives

150 + c = 170

, so

c = 20

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations.

b + c = 50

and

3b + 4c = 170

The total number of muffins sold is 50, and the total revenue from selling blueberry muffins at

3 each and chocolate chip muffins at

4 each is $170.

2

Substitute

b = 50 - c

into the cost equation.

3(50 - c) + 4c = 170 \implies 150 + c = 170

Substituting

b

in terms of

c

creates a single-variable linear equation for

c

.

3

Solve for

c

.

c = 20

Subtracting 150 from both sides gives the exact number of chocolate chip muffins sold.

Key Concept

Solving Systems of Linear Equations by Substitution or Elimination

Question 228Question →

What is the sum of all real solutions to the absolute value equation $|2x - 3| + |x + 5| = 12 - x$ ?

-\frac{9}{2}

\frac{5}{2}

-\frac{1}{2}

\frac{9}{2}

\frac{19}{2}

Show answer & explanation

Answer: $-\frac{9}{2}$

Answer

The sum of all real solutions to the equation is

-\frac{9}{2}

.

By splitting the real number line into three intervals based on the critical points

x = -5

and

x = \frac{3}{2}

, we find two valid solutions:

x = \frac{5}{2}

(from

x \ge \frac{3}{2}

) and

x = -7

(from

x < -5

). Their sum is

\frac{5}{2} + (-7) = -\frac{9}{2}

.

Step-by-Step Solution

1

Identify critical points for the absolute value expressions

The critical points are

x = \frac{3}{2}

and

x = -5

.

The expressions inside the absolute values,

2x - 3

and

x + 5

, change signs at

x = \frac{3}{2}

and

x = -5

respectively.

2

Evaluate Region 1 (

x \ge \frac{3}{2}

)

(2x - 3) + (x + 5) = 12 - x \implies 3x + 2 = 12 - x \implies 4x = 10 \implies x = \frac{5}{2}

.

In this region, both

2x - 3 \ge 0

and

x + 5 > 0

, so absolute value bars can be removed directly. Since

\frac{5}{2} \ge \frac{3}{2}

,

x = \frac{5}{2}

is a valid solution.

3

Evaluate Region 2 (

-5 \le x < \frac{3}{2}

)

-(2x - 3) + (x + 5) = 12 - x \implies -x + 8 = 12 - x \implies 8 = 12

(No solution).

In this interval,

2x - 3 < 0

while

x + 5 \ge 0

. The resulting equation produces a contradiction, so there are no solutions in this interval.

4

Evaluate Region 3 (

x < -5

)

-(2x - 3) - (x + 5) = 12 - x \implies -3x - 2 = 12 - x \implies -2x = 14 \implies x = -7

.

In this region, both

2x - 3 < 0

and

x + 5 < 0

. Since

-7 < -5

,

x = -7

is a valid solution.

5

Calculate the sum of all valid solutions

\frac{5}{2} + (-7) = \frac{5}{2} - \frac{14}{2} = -\frac{9}{2}

.

Combining the valid roots from Region 1 and Region 3 gives the final requested sum.

Key Concept

Solving piecewise linear equations involving multiple absolute value terms.

Estimated Time:2m 30s

Question 229Question →

Which of the following is equivalent to $\sqrt{\frac{8^{10} + 4^{15}}{8^4 + 4^6}}$ ?

2^8

2^9

2^{10}

2^{12}

2^{18}

Show answer & explanation

Answer: $2^9$

Answer

The expression is equivalent to

2^9

.

By converting all terms to base 2,

8^{10} + 4^{15} = (2^3)^{10} + (2^2)^{15} = 2^{30} + 2^{30} = 2^{31}

in the numerator, and

8^4 + 4^6 = (2^3)^4 + (2^2)^6 = 2^{12} + 2^{12} = 2^{13}

in the denominator. The expression simplifies to

\sqrt{\frac{2^{31}}{2^{13}}} = \sqrt{2^{18}} = 2^9

.

Step-by-Step Solution

1

Convert all terms in the numerator and denominator to a common base of 2.

Numerator:

8^{10} + 4^{15} = (2^3)^{10} + (2^2)^{15} = 2^{30} + 2^{30}

. Denominator:

8^4 + 4^6 = (2^3)^4 + (2^2)^6 = 2^{12} + 2^{12}

.

Expressing terms with prime bases enables simplification using exponent rules.

2

Factor out common terms to simplify addition in numerator and denominator.

Numerator:

2^{30} + 2^{30} = 2 \cdot 2^{30} = 2^{31}

. Denominator:

2^{12} + 2^{12} = 2 \cdot 2^{12} = 2^{13}

.

Adding two equal quantities

x + x

equals

2x

, increasing the power of 2 by 1.

3

Simplify the fraction inside the square root.

\frac{2^{31}}{2^{13}} = 2^{31 - 13} = 2^{18}.

Apply the quotient rule of exponents:

\frac{a^m}{a^n} = a^{m-n}

.

4

Evaluate the radical expression.

\sqrt{2^{18}} = (2^{18})^{1/2} = 2^{18/2} = 2^9.

Taking the square root of a power halves its exponent.

Key Concept

Exponents, Roots, and Powers of Integers

Estimated Time:2m 0s

Question 230Question →

An event planner ordered a total of $30$ gift baskets for a corporate conference. Standard gift baskets cost $\$25$ each, and Deluxe gift baskets cost $\$40$ each. If the total cost of all $30$ gift baskets was $\$900$ , how many Deluxe gift baskets were ordered?

10

12

15

20

25

Show answer & explanation

Answer: $10$

Answer

The correct number of Deluxe gift baskets ordered is

10

.

By setting up the linear system

x + y = 30

and

25x + 40y = 900

, substituting

x = 30 - y

into the cost equation yields

750 + 15y = 900

, which solves directly to

y = 10

. Therefore,

10

Deluxe baskets were ordered.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

x

be the number of Standard baskets and

y

be the number of Deluxe baskets. Equation 1:

x + y = 30

. Equation 2:

25x + 40y = 900

.

The total number of baskets establishes a quantity relation, while the individual prices establish a cost relation.

2

Express

x

in terms of

y

using Equation 1

x = 30 - y

Isolating

x

allows direct substitution into the total cost equation to solve for

y

.

3

Substitute

x = 30 - y

into Equation 2 and solve for

y

25(30 - y) + 40y = 900 \implies 750 - 25y + 40y = 900 \implies 15y = 150 \implies y = 10

.

Simplifying the single-variable equation yields the exact number of Deluxe gift baskets.

Key Concept

Solving Systems of Linear Equations using Substitution or Elimination

Alternative Method

Use the elimination method: multiply the total quantity equation by

25

(

25x + 25y = 750

) and subtract it from the total cost equation (

25x + 40y = 900

) to get

15y = 150

, leading directly to

y = 10

.

Estimated Time:45s

Question 231Question →

An integer sequence is defined by $T_n = 7^n - (-2)^n$ for all positive integers $n$ . What is the remainder when $T_{40}$ is divided by $5$ ?

1

2

0

4

3

Show answer & explanation

Answer: 0

Answer

The remainder when

T_{40}

is divided by

5

is

0

.

Modulo

5

,

7 \equiv 2

, so

7^{40} \equiv 2^{40}

. Since

40

is an even exponent,

(-2)^{40} = 2^{40}

. Substituting these into the formula yields

T_{40} \equiv 2^{40} - 2^{40} = 0 \pmod 5

. Thus, the remainder is

0

.

Step-by-Step Solution

1

Reduce the base

7

modulo

5

7 \equiv 2 \pmod 5

, so

7^{40} \equiv 2^{40} \pmod 5

.

Simplifying the base makes modular exponentiation straightforward.

2

Evaluate the negative base term

(-2)^{40}

Since

40

is an even integer,

(-2)^{40} = 2^{40}

.

An even power of a negative number yields a positive result.

3

Compute

T_{40}

modulo

5

T_{40} = 7^{40} - (-2)^{40} \equiv 2^{40} - 2^{40} = 0 \pmod 5

.

Subtracting identical values yields

0

.

Key Concept

Modular Arithmetic and Exponent Parity

Estimated Time:1m 30s

Question 232Question →

When the positive integer $n$ is divided by $12$ , the remainder is $7$ . What is the units digit of $9^n + 4^{n+1} - 7^{n+2}$ ?

2

4

6

8

0

Show answer & explanation

Answer: 8

Answer

The units digit of the expression is 8.

The correct answer is 8 because evaluating each component using unit digit cyclicity gives

9^n ≡ 9 \pmod{10}

(since

n

is odd),

4^{n+1} ≡ 6 \pmod{10}

(since

n+1

is even), and

7^{n+2} ≡ 7 \pmod{10}

(since

n+2 ≡ 1 \pmod 4

). Combining these yields

(9 + 6 - 7) = 8

.

Step-by-Step Solution

1

Express

n

using division algorithm and determine its properties.

Since

n = 12k + 7

for some non-negative integer

k

,

n

is odd,

n+1

is even, and

n+2 = 12k + 9

.

Establishing the form of

n

determines the exponents for cyclicity calculations.

2

Find the units digit of

9^n

.

Units digit of

9^n

is 9.

Powers of 9 alternate units digits:

9^1 = 9, 9^2 = 1, 9^3 = 9...

Any odd power of 9 ends in 9. Since

n = 12k+7

is odd,

9^n

ends in 9.

3

Find the units digit of

4^{n+1}

.

Units digit of

4^{n+1}

is 6.

Powers of 4 alternate units digits:

4^1 = 4, 4^2 = 6, 4^3 = 4...

Any even power of 4 ends in 6. Since

n

is odd,

n+1

is even, so

4^{n+1}

ends in 6.

4

Find the units digit of

7^{n+2}

.

Units digit of

7^{n+2}

is 7.

Powers of 7 follow a 4-step cyclicity pattern: 7, 9, 3, 1. The exponent

n+2 = 12k + 9 = 4(3k+2) + 1 ≡ 1 \pmod 4

. Thus,

7^{n+2}

has the same units digit as

7^1

, which is 7.

5

Combine the units digits.

Units digit =

9 + 6 - 7 = 8

.

Adding and subtracting the respective units digits gives

15 - 7 = 8

.

Key Concept

Units Digit Cyclicity and Modular Arithmetic

Estimated Time:2m 0s

Question 233Question →

A courier service calculates its total delivery charge using a fixed base fee plus a constant per-mile rate. On Monday, a delivery of 15 miles received a 20% discount on the fixed base fee and a 25% surcharge on the per-mile rate, resulting in a total charge of $30.50. On Tuesday, a delivery of 20 miles incurred a 40% increase on the fixed base fee and received a 15% discount on the per-mile rate, resulting in a total charge of$ 34.40. What is the standard total delivery charge, in dollars, for a 25-mile delivery with no fee adjustments or rate changes?

Show answer & explanation

Answer: 40

Answer

The standard total delivery charge for a 25-mile delivery is 40 dollars.

Translating the scenario into linear equations gives

0.80B + 18.75r = 30.50

and

1.40B + 17.00r = 34.40

. Solving this linear system yields a standard base fee of

B = 10.00

dollars and a standard per-mile rate of

r = 1.20

dollars. Substituting these into the standard 25-mile cost expression

B + 25r

yields

10.00 + 25(1.20) = 40.00

dollars.

Step-by-Step Solution

1

Formulate linear equations from the word problem context.

System of equations:

0.80B + 18.75r = 30.50

and

1.40B + 17.00r = 34.40

.

Applying the percentage adjustments to the fixed base fee

B

and the rate per mile

r

for the given distances yields exact linear expressions.

2

Eliminate variable

B

to solve for

r

.

Multiply equations to equate coefficients of

B

:

5.60B + 131.25r = 213.50

and

5.60B + 68.00r = 137.60

. Subtracting gives

63.25r = 75.90

, so

r = 1.20

.

Finding the per-mile rate

r

allows determination of the standard mileage component.

3

Solve for base fee

B

using

r = 1.20

.

1.40B + 17(1.20) = 34.40 \implies 1.40B = 14.00 \implies B = 10.00

.

Substituting

r

into either linear equation gives the fixed base fee.

4

Calculate the target standard cost for 25 miles.

B + 25r = 10.00 + 25(1.20) = 40.00

.

Evaluating the standard pricing expression

B + 25r

with

B = 10

and

r = 1.20

gives the total cost.

Key Concept

Solving Systems of Two-Variable Linear Equations from Word Problems

Question 234Question →

What is the value of the positive integer $n$ if $n$ is a multiple of 18 and $n$ has exactly 9 positive integer divisors?

Show answer & explanation

Answer: 36

Answer

36

The prime factorization of 18 is

2^1 \cdot 3^2

. Any multiple

n

of 18 must take the form

n = 2^a \cdot 3^b \dots

where

a \ge 1

and

b \ge 2

. The total number of positive divisors of

n

is given by

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

. Given that

a+1 \ge 2

and

b+1 \ge 3

, the only product of integers equal to 9 is

3 \times 3

. This requires

a+1 = 3 \implies a = 2

and

b+1 = 3 \implies b = 2

, with no additional prime factors present. Therefore,

n = 2^2 \cdot 3^2 = 36

.

Step-by-Step Solution

1

Determine the prime factorization constraints imposed by 18.

18 = 2^1 \cdot 3^2

, so

n = 2^a \cdot 3^b \dots

with

a \ge 1

and

b \ge 2

.

Any multiple of 18 must contain at least one factor of 2 and two factors of 3.

2

Apply the divisor count formula to set up an equation.

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

The number of positive divisors of

2^a \cdot 3^b

is given by

(a+1)(b+1)

.

3

Solve for the exponents

a

and

b

.

a = 2

and

b = 2

Because 9 can only be factored as

3 \times 3

for integer components where

a+1 \ge 2

and

b+1 \ge 3

, both

a+1

and

b+1

must equal 3.

4

Compute the value of

n

.

n = 2^2 \cdot 3^2 = 36

Multiply the prime power factors together to find the value of

n

.

Key Concept

The total number of positive divisors of a positive integer

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

is given by

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

.

Question 235Question →

If $m$ and $n$ are positive integers such that $2^m - 2^n = 1920$ , what is the value of $m + n$ ?

Show answer & explanation

Answer: 18

Answer

18

To solve

2^m - 2^n = 1920

, factor out

2^n

to express the left side as

2^n(2^{m-n} - 1)

. Prime factorizing 1920 gives

2^7 \times 15

. Since

m

and

n

are positive integers with

m > n

, the term

(2^{m-n} - 1)

is an odd integer. Therefore, the power-of-2 term

2^n

must equal

2^7

, which implies

n = 7

. The odd term

(2^{m-n} - 1)

must equal

15

, leading to

2^{m-n} = 16 = 2^4

, so

m - n = 4

. Solving for

m

gives

m = 11

. Finally,

m + n = 11 + 7 = 18

.

Step-by-Step Solution

1

Factor the exponential expression

2^n(2^{m-n} - 1) = 1920

Factoring out the smaller power of 2 separates the expression into a power of 2 and an odd integer multiplier.

2

Determine the prime factorization of 1920

1920 = 2^7 \times 15

Prime factorization isolates the highest power of 2 (

2^7

) from the remaining odd factor (

15

).

3

Equate corresponding power-of-2 and odd factors

n = 7

and

m - n = 4

The even component

2^n

must equal

2^7

, giving

n = 7

. The odd component

2^{m-n} - 1

must equal

15

, so

2^{m-n} = 16 = 2^4

, giving

m - n = 4

.

4

Solve for

m

and compute

m + n

m = 11

and

m + n = 18

Adding

n = 7

to

m - n = 4

yields

m = 11

. The requested sum is

m + n = 11 + 7 = 18

.

Key Concept

Factoring difference of powers using fundamental exponent rules and equating even/odd prime components.

Question 236Question →

What is the smallest positive integer $n$ such that $n!$ is divisible by $10^{10}$ ?

Show answer & explanation

Answer: 45

Answer

The smallest positive integer

n

such that

n!

is divisible by

10^{10}

is 45.

To find the smallest integer

n

such that

n!

is divisible by

10^{10}

, we need

n!

to contain at least 10 prime factors of 5 (since 2s are abundant). Using Legendre's formula,

E_5(40!) = \lfloor 40/5 \rfloor + \lfloor 40/25 \rfloor = 8 + 1 = 9

, which is insufficient. For

n = 45

,

E_5(45!) = \lfloor 45/5 \rfloor + \lfloor 45/25 \rfloor = 9 + 1 = 10

, satisfying the requirement. Thus, 45 is the smallest positive integer.

Step-by-Step Solution

1

Determine the prime factorization requirement for divisibility by

10^{10}

.

10^{10} = 2^{10} \times 5^{10}

. Thus,

n!

must contain at least 10 factors of 5.

The power of 2 in any factorial

n!

(where

n \ge 5

) is always strictly greater than the power of 5, making 5 the limiting prime factor.

2

Apply Legendre's formula for the exponent of prime

p = 5

in

n!

.

E_5(n!) = \lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \dots \ge 10

.

Legendre's formula accounts for single multiples of 5, double multiples of 5 (25), etc.

3

Evaluate candidate values for

n

.

For

n = 40

,

E_5(40!) = 8 + 1 = 9

factors. For

n = 45

,

E_5(45!) = 9 + 1 = 10

factors.

Testing multiples of 5 systematically pinpoints the exact boundary where the total count of prime factor 5 reaches 10.

Key Concept

Finding the exponent of a prime factor in a factorial using Legendre's formula

Estimated Time:2m 0s

Question 237Question →

A set $S$ consists of $n$ consecutive positive integers, where $n > 1$ is an odd integer. If the sum of all elements in set $S$ is equal to $3^{10}$ , what is the minimum possible value of the median of set $S$ ?

Show answer & explanation

Answer: 243

Answer

The minimum possible value of the median of set

S

is 243.

For an odd number

n

of consecutive integers, the sum of the set equals

n \times m

, where

m

is the median. Given that

n \times m = 3^{10}

, both

n

and

m

must be powers of 3, so

n = 3^k

and

m = 3^{10-k}

. To ensure all terms in the set are positive, the smallest term

m - \frac{n-1}{2}

must be at least 1, which requires

2m > n

. Substituting the powers of 3 yields

2 \cdot 3^{10-k} > 3^k \implies 3^{2k} < 2 \cdot 3^{10} = 118,098

. The largest integer

k

satisfying this condition is

k = 5

(since

3^{10} = 59,049 < 118,098

while

3^{12} = 531,441 > 118,098

). Maximizing

k

minimizes the median

m = 3^{10-5} = 3^5 = 243

.

Step-by-Step Solution

1

Relate the sum of an evenly spaced set to its number of terms and median.

n \times m = 3^{10}

, where

n

is the number of terms and

m

is the median.

For any set of

n

consecutive integers where

n

is odd, the sum of the set equals the number of terms times the middle term (median).

2

Express

n

and

m

as powers of 3.

n = 3^k

and

m = 3^{10-k}

where

k \ge 1

.

Because

3^{10}

has only 3 as a prime factor, any integer factors

n

and

m

must be powers of 3.

3

Establish the positivity constraint for the terms in set

S

.

2m > n

The smallest term in the set is

m - \frac{n-1}{2}

. Requiring

m - \frac{n-1}{2} \ge 1

gives

2m \ge n + 1

, or strictly

2m > n

.

4

Solve the inequality

2m > n

in terms of

k

.

3^{2k} < 2 \cdot 3^{10} = 118,098

Substituting

n = 3^k

and

m = 3^{10-k}

yields

2 \cdot 3^{10-k} > 3^k

, which rearranges to

3^{2k} < 2 \cdot 3^{10}

.

5

Find the maximum valid integer value of

k

and calculate the corresponding minimum median

m

.

Maximum

k = 5

, giving minimum median

m = 3^{5} = 243

.

For

k = 5

,

3^{10} = 59,049 < 118,098

. For

k = 6

,

3^{12} = 531,441 > 118,098

. Thus

k=5

is the maximum integer

k

, which yields the minimum median

m = 3^{10-5} = 243

.

Key Concept

Sum of consecutive integers set formula and positivity constraints

Question 238Question →

When a positive integer $n$ is divided by $12$ , the remainder is $7$ . What is the remainder when $n^2 + 5n + 11$ is divided by $12$ ?

Show answer & explanation

Answer: 11

Answer

11

Since

n

leaves a remainder of

7

when divided by

12

, we can substitute

n \equiv 7 \pmod{12}

directly into

n^2 + 5n + 11

. Evaluating

7^2 + 5(7) + 11

gives

49 + 35 + 11 = 95

. Dividing

95

by

12

gives

7

with a remainder of

11

.

Step-by-Step Solution

1

Express

n

using modular arithmetic

n \equiv 7 \pmod{12}

A positive integer

n

that leaves a remainder of

7

when divided by

12

can be expressed as

n = 12k + 7

for some non-negative integer

k

.

2

Substitute

n \equiv 7 \pmod{12}

into the target expression

n^2 + 5n + 11 \equiv 7^2 + 5(7) + 11 \pmod{12}

By the algebraic properties of remainders (modular arithmetic), substituting the remainder

7

for

n

yields an equivalent remainder modulo

12

.

3

Evaluate the arithmetic sum

49 + 35 + 11 = 95

Computing

7^2 = 49

,

5 \times 7 = 35

, and adding

11

gives

95

.

4

Find the remainder of

95

modulo

12

95 = 12 \times 7 + 11 \implies 95 \equiv 11 \pmod{12}

Dividing

95

by

12

yields a quotient of

7

and a remainder of

11

.

Key Concept

Modular Arithmetic and Polynomial Remainders

Question 239Question →

If $3x + 2y = 18$ and $x + 4y = 16$ , what is the value of $x + y$ ?

3

4

7

10

17

Show answer & explanation

Answer: 7

Answer

7

Solving the system of linear equations

3x + 2y = 18

and

x + 4y = 16

yields

x = 4

and

y = 3

. Adding these values together produces

4 + 3 = 7

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the second equation

x = 16 - 4y

Isolating one variable allows straightforward substitution into the other equation.

2

Substitute

x = 16 - 4y

into the first equation

3(16 - 4y) + 2y = 18 \implies 48 - 12y + 2y = 18 \implies 48 - 10y = 18

This creates an equation containing only the variable

y

.

3

Solve for

y

-10y = 18 - 48 = -30 \implies y = 3

Isolating the numerical term isolates the value of

y

.

4

Substitute

y = 3

back to find

x

x = 16 - 4(3) = 16 - 12 = 4

Determines the specific value of

x

.

5

Calculate the requested sum

x + y

x + y = 4 + 3 = 7

Provides the final requested expression value.

Key Concept

Solving a system of linear equations by substitution or elimination to evaluate a combined linear expression.

Estimated Time:45s

Question 240Question →

If $n$ is a positive integer greater than $20$ such that $\sqrt{2^{28} + 2^{20} + 2^n}$ is an integer, what is the least possible value of $n$ ?

22

24

25

26

34

Show answer & explanation

Answer: 25

Answer

The least possible value of

n

is 25.

The expression inside the radical must be a perfect square. Matching

2^{28} + 2^{20} + 2^n

to

(2^{14} + 2^{10})^2 = 2^{28} + 2 \cdot 2^{14} \cdot 2^{10} + 2^{20} = 2^{28} + 2^{25} + 2^{20}

reveals that

2^n = 2^{25}

, so

n = 25

. Since 25 is greater than 20 and smaller than the other valid solution

n = 34

, it is the least possible value.

Step-by-Step Solution

1

Set up the perfect square structure for the radical expression.

For

\sqrt{2^{28} + 2^{20} + 2^n}

to be an integer, the expression

2^{28} + 2^{20} + 2^n

must equal

(2^a + 2^b)^2 = 2^{2a} + 2^{a+b+1} + 2^{2b}

for some positive integers

a > b

.

Expanding a binomial power of 2 generates three power-of-2 terms matching the three terms in the expression.

2

Analyze Case 1 where

2^{28}

corresponds to the leading term

2^{2a}

.

2a = 28 \implies a = 14

. The remaining terms

2^{20}

and

2^n

must correspond to

2^{2b}

and

2^{a+b+1}

.

Matching highest powers establishes the value of the parameter

a

.

3

Evaluate sub-cases for matching

2^{20}

.

Subcase 1: If

2b = 20 \implies b = 10

, then the middle term exponent is

a + b + 1 = 14 + 10 + 1 = 25

. Thus

n = 25

.
Subcase 2: If

a + b + 1 = 20 \implies 14 + b + 1 = 20 \implies b = 5

, then

2b = 10

. Thus

n = 10

.

Checking both term assignments determines possible integer values for

n

.

4

Analyze Case 2 where

2^{20}

corresponds to

2^{2b}

and

2^{28}

is the middle term.

If

2b = 20 \implies b = 10

and

a + b + 1 = 28 \implies a + 11 = 28 \implies a = 17

, then

2a = 34

. Thus

n = 34

.

Checking alternative middle term assignments finds all valid values for

n

.

5

Filter by constraint

n > 20

and select the minimum.

The valid values satisfying

n > 20

are

n = 25

and

n = 34

. The least possible value is 25.

The question asks specifically for the least possible value greater than 20.

Key Concept

Perfect Square Trinomial Expansion with Exponent Rules

Estimated Time:2m 0s

Answer

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