Number Properties and Arithmetic Practice Questions — GMAT — Page 11

Question 201Question →

Let $n$ be the smallest positive integer that is a multiple of 15 and leaves a remainder equivalent to $-2$ when divided by 7. What is the value of $n$ ?

30

75

45

105

60

Show answer & explanation

Answer: 75

Answer

75

The correct value is 75. Since

n

is a multiple of 15,

n = 15k

. A remainder of

-2

when divided by 7 means

n \equiv 5 \pmod 7

. Substituting

n = 15k

gives

15k \equiv 5 \pmod 7

, which simplifies to

k \equiv 5 \pmod 7

since

15 \equiv 1 \pmod 7

. The smallest positive integer value for

k

is 5, giving

n = 15 \times 5 = 75

.

Step-by-Step Solution

1

Express

n

as a multiple of 15

n = 15k

for some positive integer

k

Since

n

is a multiple of 15, it must be of the form

15k

.

2

Convert the negative remainder condition into a standard positive remainder modulo 7

n \equiv -2 \equiv 5 \pmod 7

A remainder of

-2

modulo 7 is equivalent to

7 + (-2) = 5

.

3

Solve the modular congruence for

k

k = 5

Substituting

n = 15k

into

15k \equiv 5 \pmod 7

gives

1k \equiv 5 \pmod 7

because

15 \equiv 1 \pmod 7

. The smallest positive integer solution is

k = 5

.

4

Calculate the value of

n

n = 15 \times 5 = 75

Multiplying 15 by

k = 5

yields 75.

Key Concept

Modular Arithmetic and Remainder Properties

Estimated Time:1m 30s

Question 202Question →

For any positive integer $n$ , let $f(n)$ denote the remainder when $13^n + 17^n$ is divided by $10$ . What is the sum of all possible distinct values of $f(n)$ ?

0

8

10

18

20

Show answer & explanation

Answer: $10$

Answer

The sum of all possible distinct values is 10.

Finding the remainder when a number is divided by

10

is equivalent to finding its units digit. The units digit of

13^n

follows the repeating pattern

[3, 9, 7, 1]

, and the units digit of

17^n

follows the repeating pattern

[7, 9, 3, 1]

. Adding these term-by-term yields units digits of

(3+7) \rightarrow 0

,

(9+9) \rightarrow 8

,

(7+3) \rightarrow 0

, and

(1+1) \rightarrow 2

. The set of distinct remainders is

\{0, 8, 2\}

, and their sum is

0 + 8 + 2 = 10

.

Step-by-Step Solution

1

Relate division by 10 to units digits

The remainder when any positive integer is divided by

10

is equal to its units digit. Therefore,

f(n)

is the units digit of

13^n + 17^n

, which depends only on the units digits of

3^n

and

7^n

.

Dividing by 10 isolates the ones place of an integer.

2

Determine the units digit pattern for

3^n

and

7^n

The units digits of

3^n

follow a 4-term repeating cycle:

[3, 9, 7, 1]

. The units digits of

7^n

follow a 4-term repeating cycle:

[7, 9, 3, 1]

.

Powers of integers exhibit periodic patterns in their units digits.

3

Evaluate

f(n)

for each term in the 4-step pattern

For

n=1

: units digit of

3 + 7 = 10

is

0

, so

f(1) = 0

.
For

n=2

: units digit of

9 + 9 = 18

is

8

, so

f(2) = 8

.
For

n=3

: units digit of

7 + 3 = 10

is

0

, so

f(3) = 0

.
For

n=4

: units digit of

1 + 1 = 2

is

2

, so

f(4) = 2

.

Test one full period of length 4 to find all possible outputs.

4

Find the sum of all distinct values

The distinct values of

f(n)

are

0

,

8

, and

2

. Their sum is

0 + 8 + 2 = 10

.

The question asks for the sum of distinct possible remainders.

Key Concept

Units Digit Cyclicity and Modular Arithmetic

Estimated Time:1m 45s

Question 203Question →

For any positive integer $k$ , what is the units digit of the expression $2^{4k+2} + 3^{4k+3} + 7^{4k+1}$ ?

8

4

6

2

0

Show answer & explanation

Answer: 8

Answer

8

To find the units digit of

2^{4k+2} + 3^{4k+3} + 7^{4k+1}

, analyze the units digit cyclicity of each base. Powers of 2 have a units digit cycle of [2, 4, 8, 6]. Since

4k+2

leaves a remainder of 2 when divided by 4,

2^{4k+2}

ends in 4. Powers of 3 have a units digit cycle of [3, 9, 7, 1]. Since

4k+3

leaves a remainder of 3 when divided by 4,

3^{4k+3}

ends in 7. Powers of 7 have a units digit cycle of [7, 9, 3, 1]. Since

4k+1

leaves a remainder of 1 when divided by 4,

7^{4k+1}

ends in 7. Adding these units digits gives

4 + 7 + 7 = 18

, so the final units digit is 8.

Step-by-Step Solution

1

Determine the units digit cyclicity pattern for base 2

The units digits of powers of 2 repeat in a 4-step cycle: 2, 4, 8, 6. For

2^{4k+2}

, the exponent leaves a remainder of 2 when divided by 4, so its units digit is 4.

Units digits of powers follow periodic cycles modulo 10.

2

Determine the units digit cyclicity pattern for base 3

The units digits of powers of 3 repeat in a 4-step cycle: 3, 9, 7, 1. For

3^{4k+3}

, the exponent leaves a remainder of 3 when divided by 4, so its units digit is 7.

The exponent

4k+3

corresponds to the 3rd position in the 4-step cycle.

3

Determine the units digit cyclicity pattern for base 7

The units digits of powers of 7 repeat in a 4-step cycle: 7, 9, 3, 1. For

7^{4k+1}

, the exponent leaves a remainder of 1 when divided by 4, so its units digit is 7.

The exponent

4k+1

corresponds to the 1st position in the 4-step cycle.

4

Sum the units digits and take the units digit of the result

Sum =

4 + 7 + 7 = 18

, which has a units digit of 8.

The units digit of a sum of integers depends only on the sum of their individual units digits.

Key Concept

Units Digit Cyclicity of Exponents

Question 204Question →

A positive integer $n$ has exactly three distinct prime factors. If $n^2$ has exactly $147$ positive divisors, what is the minimum possible value of $n$ ?

Show answer & explanation

Answer: 1080

Answer

The minimum possible value of

n

is 1080.

By prime factorizing

n = p^a q^b r^c

, the number of divisors of

n^2

is given by

(2a+1)(2b+1)(2c+1) = 147

. Decomposing 147 into three factors greater than 1 gives

7 \times 7 \times 3

, which determines the exponents

a=3

,

b=3

, and

c=1

. Minimizing

n

by assigning the smallest primes (2, 3, and 5) to these exponents yields

2^3 \times 3^3 \times 5^1 = 1080

.

Step-by-Step Solution

1

Set up the prime factorization of

n

and

n^2

Let

n = p^a q^b r^c

, where

p < q < r

are prime numbers and

a, b, c \ge 1

. Then

n^2 = p^{2a} q^{2b} r^{2c}

.

Any positive integer can be uniquely represented as a product of prime powers.

2

Formulate the divisor counting equation for

n^2

The total number of positive divisors of

n^2

is

(2a+1)(2b+1)(2c+1) = 147

.

If an integer has prime factorization

\prod p_i^{e_i}

, its number of positive divisors is

\prod (e_i + 1)

.

3

Factor 147 into three odd factors greater than 1

The prime factorization of 147 is

3 \times 7^2

. The only factorization into three integers greater than 1 is

7 \times 7 \times 3

.

Since

a, b, c \ge 1

, we have

2a+1 \ge 3

,

2b+1 \ge 3

, and

2c+1 \ge 3

.

4

Solve for the exponents

a, b, c

Setting

2a+1=7

,

2b+1=7

, and

2c+1=3

yields

a=3

,

b=3

, and

c=1

.

Equating the factor values allows us to determine the required exponents of the prime factors of

n

.

5

Minimize

n

by choosing prime bases and assigning exponents

Choose the three smallest primes

p=2, q=3, r=5

and assign exponents

3, 3, 1

to get

n = 2^3 \times 3^3 \times 5^1 = 1080

.

To make the product as small as possible, smaller prime bases must be paired with larger exponents.

Key Concept

Determining the prime factorization and number of divisors of prime power expressions

Question 205Question →

For any positive integer $n$ , what is the units digit of the expression $S = 7^{4n+1} + 9^{2n} + 3^{4n+3}$ ?

0

3

5

7

9

Show answer & explanation

Answer: 5

Answer

The units digit of the expression is 5.

Each base has a predictable units digit cyclicity. The pattern for powers of

7

is

7, 9, 3, 1

(period 4), giving a units digit of

7

for exponent

4n+1

. The pattern for powers of

9

is

9, 1

(period 2), giving a units digit of

1

for even exponent

2n

. The pattern for powers of

3

is

3, 9, 7, 1

(period 4), giving a units digit of

7

for exponent

4n+3

. Adding these units digits gives

7 + 1 + 7 = 15

, whose units digit is

5

.

Step-by-Step Solution

1

Determine the units digit of

7^{4n+1}

The units digit of powers of

7

follows a repeating pattern of length 4:

7, 9, 3, 1

. Since the exponent

4n+1

has a remainder of

1

when divided by

4

, the units digit is

7

.

Units digit cyclicity of base

7

repeats every 4 powers.

2

Determine the units digit of

9^{2n}

The units digit of powers of

9

follows a repeating pattern of length 2:

9, 1

. For any positive integer

n

, the exponent

2n

is even, so the units digit is

1

.

Even powers of

9

always end in

1

.

3

Determine the units digit of

3^{4n+3}

The units digit of powers of

3

follows a repeating pattern of length 4:

3, 9, 7, 1

. Since the exponent

4n+3

has a remainder of

3

when divided by

4

, the units digit is

7

.

Units digit cyclicity of base

3

repeats every 4 powers.

4

Sum the units digits and take the final units digit

7 + 1 + 7 = 15

, which has a units digit of

5

.

The units digit of a sum of numbers equals the units digit of the sum of their individual units digits.

Key Concept

Units digit cyclicity of integer powers

Estimated Time:1m 30s

Question 206Question →

Set $S$ consists of $n$ consecutive integers, where $n > 1$ . The sum of all the elements in Set $S$ except the greatest element is $360$ , and the sum of all the elements in Set $S$ except the least element is $440$ . What is the median of the elements in Set $S$ ?

0

4

5

8

10

Show answer & explanation

Answer: $5$

Answer

The median of the elements in Set

S

is

5

.

Subtracting the given partial sums gives the difference between the largest and smallest elements:

(T - a_1) - (T - a_n) = 440 - 360 = 80

. For a set of

n

consecutive integers,

a_n - a_1 = n - 1

, so

n = 81

. In any set of consecutive integers, the arithmetic mean equals the median,

m

. Therefore, the total sum of all

81

elements is

81m

. The middle term is the 41st element (

m

), which means the 81st element is

m + 40

. Substituting these into

T - a_n = 360

yields

81m - (m + 40) = 360

, simplifying to

80m = 400

, which gives

m = 5

.

Step-by-Step Solution

1

Set up equations for the total sum

T

of Set

S

.

Let

a_1

be the least element and

a_n

be the greatest element.

T - a_n = 360

and

T - a_1 = 440

.

Subtracting the greatest element leaves

360

, and subtracting the least element leaves

440

.

2

Find the difference between the greatest and least elements

a_n - a_1

.

(T - a_1) - (T - a_n) = 440 - 360 \implies a_n - a_1 = 80

.

Subtracting the two sum equations eliminates the total sum

T

.

3

Determine the number of elements

n

in Set

S

.

For consecutive integers,

a_n - a_1 = n - 1

. Thus,

n - 1 = 80 \implies n = 81

.

The difference between the

n

-th and 1st term of consecutive integers is

n - 1

.

4

Relate the total sum

T

and the greatest element

a_n

to the median

m

.

Since

n = 81

is odd, the mean equals the median

m

. Total sum

T = 81m

. The greatest element is

a_{81} = m + 40

.

In an evenly spaced set, total sum is

n \times m

, and the last term is

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

.

5

Solve for the median

m

.

T - a_n = 360 \implies 81m - (m + 40) = 360 \implies 80m - 40 = 360 \implies 80m = 400 \implies m = 5

.

Substituting

T = 81m

and

a_n = m + 40

into the first equation allows solving for

m

directly.

Key Concept

Mean-Median Equivalence and Counting Terms in Consecutive Integer Sets

Question 207Question →

What is the sum of all the distinct prime factors of the integer $6^8 - 6^6$ ?

Show answer & explanation

Answer: 17

Answer

The sum of the distinct prime factors of

6^8 - 6^6

is 17.

Factoring out

6^6

yields

6^6(6^2 - 1) = 6^6 \times 35

. The prime factors of

6^6

are 2 and 3, while the prime factors of 35 are 5 and 7. The distinct prime factors are 2, 3, 5, and 7. Summing these prime numbers gives

2 + 3 + 5 + 7 = 17

.

Step-by-Step Solution

1

Factor out the common term

6^6

6^8 - 6^6 = 6^6(6^2 - 1) = 6^6 \times 35

Factoring simplifies the large exponent expression into a product of smaller integers.

2

Break down each base into its prime factors

6^6 = (2 \times 3)^6 = 2^6 \times 3^6

and

35 = 5 \times 7

Prime factorization requires expressing all bases as prime numbers.

3

List the distinct prime bases and compute their sum

Distinct prime factors: 2, 3, 5, 7. Sum =

2 + 3 + 5 + 7 = 17

.

Exponents do not affect which prime numbers are factors, only how many times they divide the number.

Key Concept

Prime Factorization of Difference of Exponents

Question 208Question →

If $n = 18^4 \times 35^3 \times 22^2$ , what is the total number of prime factors of $n$ , counting multiplicities (the sum of the exponents in its prime factorization)?

5

9

18

22

23

Show answer & explanation

Answer: 22

Answer

The total number of prime factors of

n

, counting multiplicities, is 22.

The correct answer is obtained by rewriting each base into prime factors (

18 = 2 \times 3^2

,

35 = 5 \times 7

,

22 = 2 \times 11

), combining powers of equal bases to obtain

n = 2^6 \times 3^8 \times 5^3 \times 7^3 \times 11^2

, and summing the resulting exponents (

6 + 8 + 3 + 3 + 2 = 22

).

Step-by-Step Solution

1

Express each composite base in terms of its prime factors.

18 = 2 \times 3^2

,

35 = 5 \times 7

, and

22 = 2 \times 11

.

Prime factorization requires writing every base strictly as a product of prime numbers.

2

Substitute the prime factorizations into the expression for

n

and apply exponent rules.

n = (2 \times 3^2)^4 \times (5 \times 7)^3 \times (2 \times 11)^2 = (2^4 \times 3^8) \times (5^3 \times 7^3) \times (2^2 \times 11^2)

.

Distribute exponents over multiplication using

(a \times b)^k = a^k \times b^k

and

(a^m)^n = a^{m \cdot n}

.

3

Combine like prime bases by adding their exponents.

n = 2^{4+2} \times 3^8 \times 5^3 \times 7^3 \times 11^2 = 2^6 \times 3^8 \times 5^3 \times 7^3 \times 11^2

.

Combine terms with identical bases according to exponent rules (

a^m \times a^n = a^{m+n}

).

4

Sum the exponents of all prime factors to find the total count including multiplicities.

6 + 8 + 3 + 3 + 2 = 22

.

The total number of prime factors counted with multiplicity is given by the sum of exponents in the canonical prime factorization.

Key Concept

Prime Factorization and Exponent Rules

Estimated Time:1m 30s

Question 209Question →

Set $S$ consists of $n$ consecutive odd integers. The sum of all elements in Set $S$ is $195$ . If the largest element in Set $S$ is $9$ times the smallest element in Set $S$ , what is the value of $n$ ?

3

9

12

13

15

Show answer & explanation

Answer: $13$

Answer

The total number of elements in Set

S

is

13

.

For any set of consecutive odd integers, the terms are spaced

2

units apart. Expressing the largest term as

a + 2(n-1) = 9a

yields

n = 4a + 1

. Using the average formula for evenly spaced sets,

\text{Sum} = n \times \frac{\text{first} + \text{last}}{2} = n(5a) = 195

, giving

an = 39

. Substituting

n = 4a + 1

gives

4a^2 + a - 39 = 0

, which factors as

(4a + 13)(a - 3) = 0

, so

a = 3

. Substituting

a = 3

into

n = 4a + 1

gives

n = 13

.

Step-by-Step Solution

1

Express the largest term in terms of the smallest term

a

and term count

n

.

The largest term is

a + 2(n - 1)

. Since the largest term is

9a

, we have

a + 2(n - 1) = 9a \implies 2(n - 1) = 8a \implies n - 1 = 4a \implies n = 4a + 1

.

Consecutive odd integers increase by increments of

2

.

2

Express the sum of the set using the arithmetic mean of an evenly spaced set.

\text{Average} = \frac{\text{Smallest} + \text{Largest}}{2} = \frac{a + 9a}{2} = 5a. \text{Sum} = n \times \text{Average} \implies 195 = n(5a) \implies an = 39.

For any set of consecutive odd integers, the average is the mean of the first and last terms.

3

Substitute

n = 4a + 1

into

an = 39

and solve for

a

.

a(4a + 1) = 39 \implies 4a^2 + a - 39 = 0 \implies (4a + 13)(a - 3) = 0. Since

a

must be a positive integer,

a = 3

.

The smallest element of a set of positive odd integers must be a positive odd integer.

4

Calculate

n

using

a = 3

.

n = 4(3) + 1 = 13.

Substitute the value of

a

back into the formula derived in Step 1.

Key Concept

Consecutive Integers and Number Sets

Question 210Question →

What is the value of $\sqrt{16^5 + 16^5 + 16^5 + 16^5}$ ?

2^{10}

2^{11}

2^{12}

2^{15}

2^{20}

Show answer & explanation

Answer: $2^{11}$

Answer

2^{11}

Combining the four identical terms inside the radical gives

4 \cdot 16^5

. Expressing both factors with base 2 yields

2^2 \cdot (2^4)^5 = 2^2 \cdot 2^{20} = 2^{22}

. Taking the square root gives

\sqrt{2^{22}} = 2^{11}

, which matches the value

2^{11}

.

Step-by-Step Solution

1

Factor out the common term inside the square root.

16^5 + 16^5 + 16^5 + 16^5 = 4 \cdot 16^5

Adding four identical terms is equivalent to multiplying the term by 4.

2

Convert both numbers to powers of 2.

4 = 2^2

and

16^5 = (2^4)^5 = 2^{20}

, so

4 \cdot 16^5 = 2^2 \cdot 2^{20} = 2^{22}

Expressing terms with a common prime base allows exponent rules to be applied.

3

Apply the square root to the simplified exponential expression.

\sqrt{2^{22}} = (2^{22})^{1/2} = 2^{11}

Taking the square root of a base raised to a power is equivalent to dividing the exponent by 2.

Key Concept

Combining like terms with exponents and applying prime factorization exponent laws under radical signs.

Estimated Time:1m 30s

Question 211Question →

For how many positive integer values of $n$ is $2^8 + 2^{11} + 2^n$ equal to the square of an integer?

Zero

One

Two

Three

Four

Show answer & explanation

Answer: One

Answer

There is exactly 1 positive integer value of

n

(specifically,

n = 12

).

The option stating 'One' is correct because factoring out

2^8

yields

2^8(9 + 2^{n-8})

. For this product to be a perfect square,

9 + 2^{n-8}

must equal

k^2

for some integer

k

. Rewriting this as

(k-3)(k+3) = 2^{n-8}

forces both factors to be powers of 2 whose difference is 6. The unique solution to

2^b - 2^a = 6

is

a = 1

and

b = 3

, which gives

n - 8 = 4

, so

n = 12

. No other positive integers

n \le 8

produce a square.

Step-by-Step Solution

1

Analyze the expression for

n > 8

by factoring out

2^8

.

2^8 + 2^{11} + 2^n = 2^8 (1 + 2^3 + 2^{n-8}) = 2^8 (9 + 2^{n-8})

.

Since

2^8 = (2^4)^2

is already a perfect square, the entire expression is a square if and only if

9 + 2^{n-8}

is a perfect square.

2

Set

9 + 2^{n-8} = k^2

for some integer

k > 3

and factor using prime power properties.

k^2 - 9 = 2^{n-8} \implies (k - 3)(k + 3) = 2^{n-8}

.

The difference of squares allows us to express the product of two integers as a power of 2.

3

Solve for the prime factors of the terms

(k-3)

and

(k+3)

.

Let

k - 3 = 2^a

and

k + 3 = 2^b

where

a + b = n - 8

and

b > a

. Subtracting the two equations gives

(k + 3) - (k - 3) = 2^b - 2^a = 6 \implies 2^a(2^{b-a} - 1) = 6 = 2^1 \cdot 3

.

The prime factorization of 6 uniquely dictates that

2^a = 2^1 \implies a = 1

, and

2^{b-a} - 1 = 3 \implies 2^{b-1} = 4 \implies b = 3

.

4

Determine

n

and test values of

n \le 8

.

Since

a = 1

and

b = 3

,

n - 8 = 1 + 3 = 4 \implies n = 12

. Testing

n \le 8

reveals no other squares (e.g., for

n=8

,

2^8(1+8+1)=10 \cdot 2^8

, not a square; for

n=3

,

2312 = 2^3 \cdot 17^2

, not a square).

This confirms that

n = 12

is the unique positive integer solution.

Key Concept

Prime factorization of differences of squares and prime power analysis

Estimated Time:2m 0s

Question 212Question →

A positive integer $n$ can be expressed in the form $p^3 q^2$ , where $p$ and $q$ are distinct prime numbers. If $n$ is divisible by 45 and is a factor of 4,050, what is the value of $n$ ?

Show answer & explanation

Answer: 675

Answer

675

Prime factorizing

45 = 3^2 \cdot 5

and

4,050 = 2 \cdot 3^4 \cdot 5^2

shows that

n

must be composed of the prime factors 3 and 5. The two candidate values for

p^3 q^2

are

3^3 \cdot 5^2 = 675

and

5^3 \cdot 3^2 = 1,125

. Both are multiples of 45, but only 675 is a factor of 4,050 because its power of 5 does not exceed

5^2

.

Step-by-Step Solution

1

Determine the prime factorizations of 45 and 4,050

45 = 3^2 \cdot 5^1

and

4,050 = 2^1 \cdot 3^4 \cdot 5^2

Decomposing the given numbers into prime factorizations determines the prime building blocks for

n

.

2

Identify the distinct prime factors

p

and

q

The primes

p

and

q

must be 3 and 5

Since

n

is divisible by 45, its prime factorization must contain at least

3^2

and

5^1

. Since

n = p^3 q^2

has exactly two distinct prime factors, those factors must be 3 and 5.

3

Evaluate both possible assignments for

p

and

q

n = 3^3 \cdot 5^2 = 675

or

n = 5^3 \cdot 3^2 = 1,125

Assigning 3 to

p

and 5 to

q

yields 675, while assigning 5 to

p

and 3 to

q

yields 1,125.

4

Filter using the constraint that

n

is a factor of 4,050

675 is a factor of 4,050; 1,125 is not a factor of 4,050

For a number to be a factor of

4,050 = 2^1 \cdot 3^4 \cdot 5^2

, the exponent of prime 5 cannot exceed 2. In 1,125, the exponent of 5 is 3, which makes 1,125 invalid.

Key Concept

Prime Factorization, Divisibility Rules, and Factors of Integers

Question 213Question →

Set $S$ consists of $n$ consecutive odd integers, ordered from least to greatest. The arithmetic mean of the integers in set $S$ is $42$ . If the sum of the smallest integer and the largest integer in set $S$ divided by $n$ equals $7$ , what is the largest integer in set $S$ ?

31

48

53

55

61

Show answer & explanation

Answer: 53

Answer

The largest integer in set

S

is 53.

For an evenly spaced set of numbers, the arithmetic mean is the average of the first and last terms. Therefore,

(a_1 + a_n) / 2 = 42

, which means

a_1 + a_n = 84

. Using the given condition

(a_1 + a_n) / n = 7

, we substitute

84 / n = 7

to find

n = 12

. Since consecutive odd integers increase by 2, the difference between the 12th term and the 1st term is

(12 - 1) \times 2 = 22

. Solving the system

a_n + a_1 = 84

and

a_n - a_1 = 22

yields

2a_n = 106

, so the largest integer

a_n

is 53.

Step-by-Step Solution

1

Express the sum of the smallest and largest integers using the mean.

Sum of smallest and largest integers

= 84

In any set of consecutive integers (an arithmetic progression), the mean is equal to the average of the first term

a_1

and the last term

a_n

:

\frac{a_1 + a_n}{2} = 42 \implies a_1 + a_n = 84

.

2

Determine the number of elements

n

in set

S

.

n = 12

The stem specifies that

\frac{a_1 + a_n}{n} = 7

. Substituting

a_1 + a_n = 84

gives

\frac{84}{n} = 7 \implies n = 12

.

3

Calculate the difference between the largest and smallest integers.

a_n - a_1 = 22

In a sequence of

n

consecutive odd integers, the common difference between terms is

2

. The number of gaps between the 1st term and the

n

-th term is

n - 1 = 11

. Thus,

a_n - a_1 = 11 \times 2 = 22

.

4

Solve for the largest integer

a_n

.

a_n = 53

Adding the equations

a_n + a_1 = 84

and

a_n - a_1 = 22

yields

2a_n = 106 \implies a_n = 53

.

Key Concept

Properties of consecutive odd integers and evenly spaced sets (mean-median equivalence and indexing)

Estimated Time:1m 30s

Question 214Question →

Let $K$ be a positive integer with prime factorization $K = 2^a \times 5^b \times 11^c$ , where $a$ , $b$ , and $c$ are positive integers. If $10K$ has 36 more positive divisors than $K$ , and $11K$ has 12 more positive divisors than $K$ , what is the value of $a + b + c$ ?

Show answer & explanation

Answer: 9

Answer

The value of

a + b + c

is 9.

For

K = 2^a 5^b 11^c

, the number of divisors is given by

(a+1)(b+1)(c+1)

. Multiplying by 11 increases the exponent of 11 by 1, yielding

(a+1)(b+1)

additional divisors. Since this difference is 12,

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

. Multiplying by 10 increases the exponents of both 2 and 5 by 1, yielding an additional

(c+1)(a+b+3)

divisors. Since this difference is 36,

(c+1)(a+b+3) = 36

. For positive integers

a

and

b

, the only factor pair of 12 that allows

36 / (a+b+3)

to be an integer is when

a+b = 6

. Consequently,

c+1 = 4

, so

c = 3

. Therefore,

a+b+c = 6+3 = 9

.

Step-by-Step Solution

1

Write the divisor count formulas for

K

,

10K

, and

11K

.

d(K) = (a+1)(b+1)(c+1)

,

d(11K) = (a+1)(b+1)(c+2)

, and

d(10K) = (a+2)(b+2)(c+1)

.

The number of positive divisors of a number

2^x 5^y 11^z

is

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

.

2

Use

d(11K) - d(K) = 12

to solve for

(a+1)(b+1)

.

(a+1)(b+1)(c+2) - (a+1)(b+1)(c+1) = 12 \implies (a+1)(b+1) = 12

.

Factoring out

(a+1)(b+1)

simplifies the equation directly.

3

Use

d(10K) - d(K) = 36

to find the relationship between

a, b,

and

c

.

(c+1)[(a+2)(b+2) - (a+1)(b+1)] = 36 \implies (c+1)(a+b+3) = 36

.

Expanding

(a+2)(b+2) - (a+1)(b+1) = ab + 2a + 2b + 4 - (ab + a + b + 1) = a + b + 3

.

4

Determine

a+b

and

c

using positive integer constraints.

a+b = 6

and

c = 3

.

Since

a, b \ge 1

, the pairs for

(a+1, b+1)

yielding 12 are

(2,6)

or

(3,4)

. If

(a+1,b+1)=(2,6)

,

a+b=6

, making

a+b+3=9

and

c+1=36/9=4 \implies c=3

. If

(3,4)

,

a+b=5

, so

a+b+3=8

, but

36/8

is not an integer.

5

Sum

a

,

b

, and

c

.

a + b + c = 6 + 3 = 9

.

Adding the derived sum

a+b=6

and

c=3

gives 9.

Key Concept

Divisor Count Function from Prime Factorization

Estimated Time:2m 0s

Question 215Question →

For real numbers $x$ , $y$ , and $z$ , none of which is equal to zero, suppose that $x y^3 z > 0$ , $\frac{x^2 z}{y} < 0$ , and $x + y > 0$ . Which of the following inequalities MUST be true?

\frac{x + z}{y} < 0

\frac{z}{x} < 0

y + z > 0

x^2 > y^2

x y z^2 > 0

Show answer & explanation

Answer: $\frac{x + z}{y} < 0$

Answer

\frac{x + z}{y} < 0

The condition

x y^3 z > 0

reduces to

x y z > 0

because

y^2 > 0

. The condition

\frac{x^2 z}{y} < 0

reduces to

\frac{z}{y} < 0

because

x^2 > 0

, implying

y

and

z

have opposite signs (

y z < 0

). Substituting

y z < 0

into

x (y z) > 0

forces

x < 0

. Furthermore,

x + y > 0

with

x < 0

requires

y > 0

. Then

y > 0

and

y z < 0

force

z < 0

. Consequently,

x

and

z

are both negative numbers, so

x + z

is negative. Dividing this negative sum by the positive number

y

guarantees that

\frac{x + z}{y} < 0

MUST be true.

Step-by-Step Solution

1

Analyze the first given inequality to establish a sign relationship among the variables.

Since

y \neq 0

,

y^2 > 0

. Dividing

x y^3 z > 0

by

y^2

gives

x y z > 0

.

Dividing an inequality by a strictly positive quantity preserves the inequality sign.

2

Analyze the second given inequality to determine the relationship between

y

and

z

.

Since

x \neq 0

,

x^2 > 0

. Dividing

\frac{x^2 z}{y} < 0

by

x^2

gives

\frac{z}{y} < 0

, which means

z

and

y

must have opposite signs (i.e.,

y z < 0

).

A quotient of two non-zero numbers is negative if and only if the numerator and denominator have opposite signs.

3

Deduce the individual signs of

x

,

y

, and

z

using the results of Steps 1 and 2 and the condition

x + y > 0

.

From

x(yz) > 0

and

yz < 0

, we deduce

x < 0

. Since

x < 0

and

x + y > 0

,

y

must be positive (

y > 0

). Because

y > 0

and

yz < 0

,

z

must be negative (

z < 0

).

Product of two negative numbers is positive, and adding a positive number larger in magnitude than a negative number yields a positive sum.

4

Evaluate the expression

\frac{x + z}{y}

using the established signs.

Since

x < 0

and

z < 0

, their sum

x + z < 0

. Dividing the negative sum by positive

y

gives

\frac{x + z}{y} < 0

.

The sum of two negative numbers is negative, and dividing a negative number by a positive number gives a negative quotient.

Key Concept

Deduction of positive and negative variable signs from products, quotients, and sums in inequalities.

Estimated Time:2m 0s

Question 216Question →

For how many positive two-digit integers $n$ is the sum $n + (n + 2)$ equal to the product of exactly two distinct prime numbers?

20

21

22

24

25

Show answer & explanation

Answer: 21

Answer

There are 21 positive two-digit integers that satisfy the given condition.

Simplifying

n + (n + 2)

gives

2(n + 1)

. For

2(n + 1)

to equal the product of two distinct prime numbers,

n + 1

must be an odd prime number

p

(so that 2 and

p

are distinct). Because

n

is a two-digit positive integer (

10 \le n \le 99

),

p = n + 1

must satisfy

11 \le p \le 100

. There are 25 prime numbers under 100, and exactly 4 of them (2, 3, 5, 7) are less than 11. Therefore, there are

25 - 4 = 21

suitable prime numbers, yielding 21 valid values of

n

.

Step-by-Step Solution

1

Algebraically simplify the given sum.

n + (n + 2) = 2n + 2 = 2(n + 1)

Combining like terms isolates the common factor of 2.

2

Analyze the prime factorization structure.

The expression

2(n + 1)

has 2 as a prime factor. For it to be the product of exactly two distinct prime numbers,

n + 1

must be a prime number

p

distinct from 2 (i.e., an odd prime).

If

n + 1

were composite or equal to 2, the total number of distinct prime factors would not be exactly two distinct primes.

3

Determine the valid range for the prime

p = n + 1

.

Since

n

is a two-digit positive integer,

10 \le n \le 99

. Adding 1 to all parts gives

11 \le n + 1 \le 100

, so

11 \le p \le 100

.

The constraints on

n

dictate the bounds for the prime number

p

.

4

Count the number of prime numbers in the range

[11, 100]

.

There are 25 prime numbers less than 100. The primes less than 11 are 2, 3, 5, and 7 (4 primes). Thus, there are

25 - 4 = 21

primes in the range

[11, 100]

.

Subtracting primes outside the valid range yields the count of valid values for

n

.

Key Concept

Prime Factorization and Prime Number Properties

Estimated Time:1m 45s

Question 217Question →

Let $x$ , $y$ , and $z$ be integers such that $-5 \le x \le 5$ , $-5 \le y \le 5$ , and $-5 \le z \le 5$ . If these integers satisfy all of the following conditions:
1. $x^3 y z^2 < 0$
2. $\frac{x}{z} < -1$
3. $x + y \le 1$

What is the maximum possible value of the expression $x + 2y - z$ ?

Show answer & explanation

Answer: 5

Answer

The maximum possible value of the expression

x + 2y - z

is 5.

By analyzing the given inequalities,

x

and

y

must have opposite signs, and

x

and

z

must have opposite signs, which means

y

and

z

must have the same sign. To maximize

x + 2y - z

, we examine the scenario where

x

is negative, while

y

and

z

are positive. Taking

z = 1

(the smallest positive integer),

y = 5

(the largest positive integer within the given range), and

x = -4

satisfies

x + y \le 1

,

\frac{x}{z} = -4 < -1

, and

x^3 y z^2 = -320 < 0

. This yields

x + 2y - z = -4 + 10 - 1 = 5

.

Step-by-Step Solution

1

Analyze product and quotient inequalities to deduce the relative signs of

x

,

y

, and

z

.

x

and

y

have opposite signs,

x

and

z

have opposite signs, and

|x| > |z| \ge 1

. Consequently,

y

and

z

share the same sign.

Because

z^2 > 0

for non-zero

z

,

x^3 y z^2 < 0

requires

x^3 y < 0

. Also

\frac{x}{z} < -1

implies

\frac{x}{z}

is negative and has magnitude greater than 1.

2

Evaluate the sign cases to determine which case yields a larger value for

x + 2y - z

.

Case A (

x < 0, y > 0, z > 0

) allows positive contributions from

2y

and

-z

, whereas Case B (

x > 0, y < 0, z < 0

) bounds the expression below 2.

In Case B,

y

and

z

are negative, so

2y \le -2

suppresses the sum.

3

Apply integer domain bounds and inequality constraints to maximize

x + 2y - z

in Case A.

The maximum value is 5, achieved when

x = -4

,

y = 5

, and

z = 1

.

Setting

z = 1

(smallest positive integer) and

y = 5

(largest positive integer) with

x = -4

satisfies

x + y \le 1

and all problem conditions.

Key Concept

Deduction of variable signs from inequality products and quotients, combined with integer range optimization.

Estimated Time:2m 0s

Question 218Question →

If $a$ , $b$ , and $c$ are non-zero real numbers such that $a^2 b^3 c < 0$ , $\frac{a}{b^5 c^2} > 0$ , and $a c > b c$ , which of the following expressions MUST be true?

a > b

a c > 0

\frac{b - a}{c} < 0

b + c > 0

\frac{a}{c} > 0

Show answer & explanation

Answer: $\frac{b - a}{c} < 0$

Answer

The expression $\frac{b - a}{c} < 0$ MUST be true.

The condition

a^2 b^3 c < 0

requires

b c < 0

because

a^2 > 0

. The condition

\frac{a}{b^5 c^2} > 0

requires

a b > 0

because

c^2 > 0

. Combining these shows that

a

and

b

share the same sign, whereas

c

has the opposite sign. Consequently,

a c < 0

. From

a c > b c

, subtracting

b c

yields

(a - b) c > 0

. Multiplying by

-1

gives

(b - a) c < 0

, and dividing by

c^2 > 0

produces

\frac{b - a}{c} < 0

, which MUST be true in all cases.

Step-by-Step Solution

1

Determine the relative signs of

b

and

c

using

a^2 b^3 c < 0

.

b c < 0

, meaning

b

and

c

have opposite signs.

Since

a \neq 0

,

a^2 > 0

always. Dividing

a^2 b^3 c < 0

by

a^2

yields

b^3 c < 0

. Because

b^3

has the same sign as

b

,

b c < 0

.

2

Determine the relative signs of

a

and

b

using

\frac{a}{b^5 c^2} > 0

.

a b > 0

, meaning

a

and

b

have the same sign.

Since

c \neq 0

,

c^2 > 0

always. Multiplying by

c^2

gives

\frac{a}{b^5} > 0

, which implies

a

and

b^5

have the same sign. Thus

a b > 0

.

3

Determine the relationship between

a

and

c

.

a c < 0

, meaning

a

and

c

have opposite signs.

Since

a

and

b

have the same sign (

a b > 0

) while

b

and

c

have opposite signs (

b c < 0

),

a

and

c

must have opposite signs.

4

Analyze the inequality

a c > b c

.

\frac{b - a}{c} < 0

Rearranging

a c > b c

gives

a c - b c > 0 \implies (a - b) c > 0

. Multiplying both sides by

-1

flips the inequality:

(b - a) c < 0

. Dividing by the strictly positive quantity

c^2

yields

\frac{(b - a) c}{c^2} < 0 \implies \frac{b - a}{c} < 0

.

Key Concept

Deducing sign relationships of variables in inequalities and algebraic transformations without assuming positive signs.

Estimated Time:2m 0s

Question 219Question →

A sequence consists of $k$ consecutive positive odd integers. The arithmetic mean of the $3$ largest integers in the sequence is $29$ . If the sum of all $k$ integers in the sequence is $207$ , what is the value of $k$ ?

7

9

11

15

23

Show answer & explanation

Answer: 9

Answer

The total number of terms in the sequence,

k

, is 9.

The arithmetic mean of 3 consecutive odd integers is their middle term, so the three largest terms are 27, 29, and 31. The largest term is 31. Using the sum formula for an arithmetic progression,

S = k \cdot \frac{a + 31}{2} = 207

. Expressing the first term as

a = 31 - 2(k - 1) = 33 - 2k

yields

k(32 - k) = 207

, which simplifies to

k^2 - 32k + 207 = 0

. The roots are

k = 9

and

k = 23

. Because all integers in the sequence must be positive,

a = 33 - 2k > 0

, requiring

k \le 16

. Therefore,

k = 9

.

Step-by-Step Solution

1

Determine the largest integer in the sequence.

The largest integer is

31

.

For any 3 consecutive odd integers, the arithmetic mean is equal to the middle integer. Since the mean is

29

, the three largest integers are

27, 29, 31

, so the maximum term is

31

.

2

Express the smallest term

a

in terms of

k

.

a = 33 - 2k

.

The

k

-th term of a consecutive odd integer sequence starting at

a

is given by

31 = a + 2(k - 1) \implies a = 33 - 2k

.

3

Apply the positivity constraint.

k \le 16

.

Since all terms are positive integers, the smallest term must satisfy

a \ge 1 \implies 33 - 2k \ge 1 \implies k \le 16

.

4

Set up and solve the sum equation for

k

.

k = 9

.

The sum of an arithmetic sequence is

S = k \times \frac{a + L}{2}

. Substituting

S = 207

,

L = 31

, and

a = 33 - 2k

gives

207 = k \times \frac{(33 - 2k) + 31}{2} = k(32 - k) \implies k^2 - 32k + 207 = 0

. Factoring yields

(k - 9)(k - 23) = 0

. Since

k \le 16

,

k = 9

.

Key Concept

The sum of a sequence of consecutive evenly-spaced numbers equals the number of terms multiplied by the average of the first and last terms.

Estimated Time:1m 45s

Question 220Question →

Let $N = 504$ . If $M$ is the smallest positive integer such that $N \times M$ is a perfect cube, what is the value of $M$ ?

Show answer & explanation

Answer: 147

Answer

The smallest positive integer

M

such that

504 \times M

is a perfect cube is 147.

For an integer to be a perfect cube, the exponent of each prime factor in its prime factorization must be a multiple of 3. Prime factorizing 504 yields

2^3 \times 3^2 \times 7^1

. The exponent of 2 is 3 (already a multiple of 3). The exponent of 3 is 2, which requires 1 additional factor of 3 to reach 3. The exponent of 7 is 1, which requires 2 additional factors of 7 (

7^2

) to reach 3. Therefore, the minimum value for

M

is

3^1 \times 7^2 = 3 \times 49 = 147

.

Step-by-Step Solution

1

Express 504 as a product of its prime factors.

504 = 2^3 \times 3^2 \times 7^1

Decomposing

N

into prime factors allows analysis of the exponents required for perfect power conditions.

2

Apply the prime exponent rule for perfect cubes.

Every prime factor in

N \times M

must have an exponent that is a multiple of 3.

A number

K

is a perfect cube if and only if

K = p_1^{3a} \times p_2^{3b} \times \dots

3

Calculate the missing prime factors needed to complete the cube.

M = 3^{3-2} \times 7^{3-1} = 3^1 \times 7^2

To minimize

M

, we raise each prime to the smallest non-negative power that rounds the existing exponent up to the nearest multiple of 3.

4

Evaluate the value of

M

.

M = 3 \times 49 = 147

Direct arithmetic computation.

Key Concept

Prime Factorization and Exponent Requirements for Perfect Powers

Estimated Time:1m 30s

Number Properties and Arithmetic

Answer

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