Consecutive Integers and Number Sets Practice Questions — GMAT

Question 1Question →

Set $S$ consists of all consecutive integers from $-m$ to $n$ , inclusive, where $m$ and $n$ are positive integers with $n > m$ . If set $S$ contains exactly $25$ integers and the sum of all integers in set $S$ is $75$ , what is the value of $m$ ?

Show answer & explanation

Answer: 9

Answer

The value of

m

is

9

.

To find

m

, use the two fundamental properties of consecutive integer sets: term count and set sum. The number of terms from

-m

to

n

inclusive is

n - (-m) + 1 = n + m + 1 = 25

, which yields

n + m = 24

. The sum of an evenly spaced set is the product of the number of terms and the arithmetic mean of the smallest and largest terms:

25 \times \frac{-m + n}{2} = 75

, which simplifies to

n - m = 6

. Subtracting

n - m = 6

from

n + m = 24

gives

2m = 18

, so

m = 9

.

Step-by-Step Solution

1

Set up the equation for the number of terms in the set.

n + m = 24

The number of integers from

-m

to

n

inclusive is

n - (-m) + 1 = n + m + 1 = 25

.

2

Set up the equation for the sum of the integers in the set.

n - m = 6

The sum of an arithmetic progression is given by

\text{number of terms} \times \text{mean} = 25 \times \frac{-m + n}{2} = 75

, leading to

\frac{n - m}{2} = 3

.

3

Solve for

m

using the two linear equations.

m = 9

Subtracting

n - m = 6

from

n + m = 24

yields

2m = 18

, giving

m = 9

.

Key Concept

Properties of consecutive integer sets: inclusive term counting (

n - \text{start} + 1

) and set sum calculation (

\text{count} \times \text{mean}

).

Estimated Time:1m 30s

Question 2Question →

A set $S$ consists of $k$ consecutive integers. The sum of the first $m$ integers in set $S$ is $34$ , and the sum of the last $m$ integers in set $S$ is $74$ . If the sum of all $k$ integers in set $S$ is $189$ , what is the value of $k$ ?

Show answer & explanation

Answer: 14

Answer

The total number of integers in set S is 14.

By applying the property that the arithmetic mean of an evenly spaced set is the average of the mean of its first

m

elements and the mean of its last

m

elements, we establish that the mean of the set is

\frac{54}{m}

. Since the total sum is

189

,

k \cdot \frac{54}{m} = 189

, which yields

k = 3.5m

. Substituting this ratio into the difference between the two subset sums

m(k-m) = 40

gives

2.5m^2 = 40

, so

m = 4

and

k = 14

.

Step-by-Step Solution

1

Set up algebraic expressions for the sums of the first m terms and last m terms.

Let the set be

S = \{a, a+1, \dots, a+k-1\}

. The first

m

terms sum to

S_1 = m a + \frac{m(m-1)}{2} = 34

. The last

m

terms sum to

S_2 = m(a+k-m) + \frac{m(m-1)}{2} = 74

.

Consecutive integer sums can be represented by the starting term and the number of terms.

2

Subtract the sum of the first m terms from the sum of the last m terms.

S_2 - S_1 = m(a+k-m) - ma = m(k-m) = 74 - 34 = 40

.

Subtracting eliminates the initial term

a

and quadratic term

\frac{m(m-1)}{2}

, giving a clean relationship between

m

and

k

.

3

Determine the arithmetic mean of set S using subset averages.

The average of the first

m

terms is

\frac{34}{m}

and the average of the last

m

terms is

\frac{74}{m}

. The average of the entire set is the midpoint of these two averages:

\text{Mean} = \frac{1}{2}\left(\frac{34}{m} + \frac{74}{m}\right) = \frac{54}{m}

.

In any evenly spaced set, the overall median/mean is equal to the average of the lower-bound subset mean and upper-bound subset mean.

4

Relate the total sum to the set size k and overall mean.

S_{total} = k \times \text{Mean} \implies 189 = k \left(\frac{54}{m}\right) \implies \frac{k}{m} = \frac{189}{54} = 3.5 \implies k = 3.5m

.

The sum of a set of consecutive integers is always equal to the number of terms times the mean of the set.

5

Solve for m and k using the system of equations.

Substitute

k = 3.5m

into

m(k-m) = 40 \implies m(2.5m) = 40 \implies 2.5m^2 = 40 \implies m^2 = 16 \implies m = 4

. Thus,

k = 3.5 \times 4 = 14

.

Since

m > 0

, taking the positive square root gives

m = 4

, which leads directly to

k = 14

.

Key Concept

Average and Sum Equivalences in Evenly Spaced Sets

Question 3Question →

A set $S$ consists of consecutive integers. The arithmetic mean of all the positive integers in set $S$ is $18.5$ , and the arithmetic mean of all the negative integers in set $S$ is $-12$ . How many integers are in set $S$ ?

Show answer & explanation

Answer: 60

Answer

The total number of integers in set

S

is

60

.

The positive integers in set

S

are

1, 2, \dots, 36

, which have an arithmetic mean of

\frac{1 + 36}{2} = 18.5

. The negative integers in set

S

are

-23, -22, \dots, -1

, which have an arithmetic mean of

\frac{-23 + (-1)}{2} = -12

. Since set

S

consists of consecutive integers spanning from

-23

to

36

, it includes

23

negative integers,

36

positive integers, and the integer

0

. The total number of elements is

23 + 1 + 36 = 60

.

Step-by-Step Solution

1

Find the largest positive integer in set

S

.

The largest positive integer is

36

, meaning there are

36

positive integers in

S

.

The positive integers in

S

must form a consecutive sequence starting at

1

up to some maximum integer

m

. The average of consecutive integers from

1

to

m

is

\frac{1 + m}{2}

. Setting

\frac{1 + m}{2} = 18.5

yields

1 + m = 37

, so

m = 36

.

2

Find the smallest negative integer in set

S

.

The smallest negative integer is

-23

, meaning there are

23

negative integers in

S

.

The negative integers in

S

must form a consecutive sequence ending at

-1

down to some minimum integer

-k

. The average of consecutive integers from

-k

to

-1

is

\frac{-k + (-1)}{2}

. Setting

\frac{-k - 1}{2} = -12

yields

-k - 1 = -24

, so

k = 23

.

3

Calculate the total number of elements in set

S

.

Set

S

contains

60

integers.

Because set

S

contains consecutive integers ranging from negative to positive values, it must also contain

0

. The total count is

23\text{ (negative integers)} + 1\text{ (the integer zero)} + 36\text{ (positive integers)} = 60

.

Key Concept

Evenly spaced set averages and classification of zero in consecutive integer sets.

Estimated Time:2m 0s

Question 4Question →

Set $S$ consists of $n$ consecutive even integers. The arithmetic mean of all the integers in set $S$ is $45$ . If the difference between the largest integer and the smallest integer in set $S$ is $34$ , what is the value of the largest integer in set $S$ ?

Show answer & explanation

Answer: 62

Answer

The largest integer in set

S

is

62

.

In any set of evenly spaced numbers, such as consecutive even integers, the arithmetic mean is equal to the average of the smallest term and the largest term. Since the arithmetic mean is 45, the sum of the smallest term and the largest term must be

45 \times 2 = 90

. Combining this with the given fact that the difference between the largest term and the smallest term is 34 creates a system of equations: Smallest + Largest = 90 and Largest - Smallest = 34. Adding these two equations cancels out the smallest term, resulting in 2 * Largest = 124, which gives 62 for the largest integer.

Step-by-Step Solution

1

Express the arithmetic mean of the set in terms of the smallest element (

F

) and largest element (

L

).

F + L = 90

For any evenly spaced set of numbers, the arithmetic mean is equal to the average of the first and last elements:

\frac{F + L}{2} = 45

.

2

Set up the equation for the difference between the largest and smallest elements.

L - F = 34

The question specifies that the largest integer exceeds the smallest integer by 34.

3

Solve the system of two linear equations for

L

.

L = 62

Adding

(F + L) + (L - F) = 90 + 34

eliminates

F

, leaving

2L = 124

, which yields

L = 62

.

Key Concept

Equivalence of arithmetic mean to the average of the first and last terms in an evenly spaced set

Question 5Question →

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 6Question →

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 7Question →

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 8Question →

Set $S$ consists of $n$ consecutive integers. The sum of all elements in set $S$ is equal to $0$ . If the product of the smallest element in set $S$ and the total number of elements $n$ is $-300$ , how many positive integers are in set $S$ ?

11

13

12

24

25

Show answer & explanation

Answer: 12

Answer

12

For a set of consecutive integers to sum to zero, the elements must be symmetric about zero. Therefore, the set has an odd number of terms

n = 2k + 1

ranging from

-k

to

k

. Setting up the product of the smallest term

(-k)

and the term count

(2k + 1)

gives

(-k)(2k + 1) = -300

, which simplifies to

2k^2 + k - 300 = 0

. Factoring this quadratic yields

(2k + 25)(k - 12) = 0

, so

k = 12

. The positive integers in the set are

1

through

12

, totaling 12 elements.

Step-by-Step Solution

1

Determine the symmetry and structure of set

S

using its sum.

The median of set

S

is

0

, and the set contains an odd number of terms

n = 2k + 1

centered at

0

, expressed as

\{-k, -(k-1), \dots, 0, \dots, k-1, k\}

.

For a set of consecutive integers to sum to

0

, the terms must be symmetric around

0

. Since

0

is an element of the set,

n

must be odd.

2

Set up the quadratic equation using the given product.

The smallest element is

-k

and the total number of elements is

n = 2k + 1

, yielding

(-k)(2k + 1) = -300

, which simplifies to

2k^2 + k - 300 = 0

.

The question specifies that the product of the smallest element and the number of elements is

-300

.

3

Solve the quadratic equation for

k

.

Factoring

(2k + 25)(k - 12) = 0

gives

k = 12

as the only positive integer solution.

Since

k

represents a count of elements strictly above zero,

k

must be a positive integer.

4

Count the number of positive integers in set

S

.

The positive integers are

1, 2, \dots, 12

, which gives a total of

12

positive integers.

Zero is neither positive nor negative, so only the integers from

1

to

k

(

12

) are counted.

Key Concept

Symmetry and median property of consecutive integers centered at zero.

Question 9Question →

Set $A$ consists of $k$ consecutive odd integers, and Set $B$ consists of $k$ consecutive even integers, where $k > 1$ . The smallest integer in Set $B$ is $3$ greater than the median of Set $A$ . If all integers in Set $A$ are positive, the sum of all integers in Set $A$ is $145$ , and the median of Set $B$ is $36$ , what is the smallest integer in Set $A$ ?

21

23

25

27

29

Show answer & explanation

Answer: 25

Answer

The smallest integer in Set A is 25.

For an arithmetic sequence of

k

consecutive odd integers, the mean and median are equal to

\frac{\text{Sum}}{k} = \frac{145}{k}

. The smallest integer in Set

B

is

b_1 = \frac{145}{k} + 3

. Since Set

B

consists of

k

consecutive even integers, its median is

b_1 + k - 1

. Equating this to

36

yields

\frac{145}{k} + k + 2 = 36

, leading to

k^2 - 34k + 145 = 0

, whose roots are

k=5

and

k=29

. If

k=29

, the median of Set

A

is

5

, which implies negative integers exist in Set

A

. Since all integers in Set

A

are positive,

k=5

. With

k=5

, the median of Set

A

is

29

, and the smallest integer is

29 - 2(2) = 25

.

Step-by-Step Solution

1

Express the median of Set A in terms of k.

M_A = \frac{145}{k}

For any evenly spaced set with an odd number of terms or symmetry, the arithmetic mean equals the median. The mean is the total sum divided by the number of terms

k

.

2

Express the median of Set B in terms of k using the given relationship for the smallest element of Set B.

M_B = \left(\frac{145}{k} + 3\right) + (k - 1) = \frac{145}{k} + k + 2

The smallest element in Set

B

is

b_1 = M_A + 3 = \frac{145}{k} + 3

. Since Set

B

contains

k

consecutive even integers (spacing

d=2

), its median is

b_1 + \frac{2(k-1)}{2} = b_1 + k - 1

.

3

Set the median of Set B to 36 and solve the quadratic equation for k.

k = 5

or

k = 29

Setting

\frac{145}{k} + k + 2 = 36

gives

\frac{145}{k} + k = 34

, which rearranges to

k^2 - 34k + 145 = 0

. Factoring gives

(k-5)(k-29) = 0

.

4

Determine the valid value of k and find the smallest integer in Set A.

Smallest integer in Set A is 25.

If

k = 29

,

M_A = \frac{145}{29} = 5

, and the smallest integer in Set

A

would be

5 - 2(14) = -23

, violating the condition that all integers in Set

A

are positive. Thus

k = 5

, making

M_A = 29

. The 5 consecutive odd integers are

25, 27, 29, 31, 33

, so the smallest integer is

25

.

Key Concept

Properties of consecutive integer sets, median-mean equivalence in arithmetic sequences, and term indexing.

Question 10Question →

Set $S$ consists of $n$ consecutive integers. The arithmetic mean of the $5$ smallest integers in Set $S$ is $-12$ , and the arithmetic mean of the $5$ largest integers in Set $S$ is $24$ . What is the value of $n$ ?

36

37

40

41

42

Show answer & explanation

Answer: 41

Answer

41

For an evenly spaced set of 5 consecutive integers, the arithmetic mean equals the middle term. Therefore, the 3rd smallest term of Set

S

is

-12

, making the 1st term

x = -14

. Similarly, the 3rd term from the end of Set

S

is

24

. Representing the 3rd term from the end as

x + n - 3

and substituting

x = -14

yields

-14 + n - 3 = 24

, which simplifies to

n = 41

.

Step-by-Step Solution

1

Express the 5 smallest integers and find the first term of the set.

Let the set

S

be represented as

\{x, x+1, x+2, \dots, x+n-1\}

. The 5 smallest integers are

x, x+1, x+2, x+3, x+4

. Their arithmetic mean is the middle term,

x+2

. Setting

x+2 = -12

yields

x = -14

.

In any set of consecutive integers with an odd number of elements, the arithmetic mean equals the median (middle term).

2

Express the 5 largest integers and set up an equation for

n

.

The 5 largest integers in Set

S

are

x+n-5, x+n-4, x+n-3, x+n-2, x+n-1

. Their arithmetic mean is the middle term,

x+n-3

. Setting

x+n-3 = 24

and substituting

x = -14

gives

-14 + n - 3 = 24

.

The 5 largest elements also form an evenly spaced set whose mean is the middle of those 5 terms.

3

Solve for

n

.

n - 17 = 24 \implies n = 41

.

Simplifying the linear equation gives the total number of consecutive integers in Set

S

.

Key Concept

Arithmetic Mean and Median Equivalence in Consecutive Integer Subsets

Question 11Question →

Set $S$ consists of $n$ consecutive integers. The product of the smallest and largest integers in $S$ is $-144$ . If the sum of all elements in $S$ is $91$ , what is the value of $n$ ?

Show answer & explanation

Answer: 26

Answer

The number of consecutive integers

n

in Set

S

is 26.

By combining the consecutive set sum formula

n(a + b) = 182

with the term span relation

b - a = n - 1

and the product constraint

ab = -144

, testing the factors of

182

uniquely determines

n = 26

, corresponding to the set of consecutive integers from

-9

to

16

.

Step-by-Step Solution

1

Relate the number of elements

n

to the smallest element

a

and largest element

b

.

b - a = n - 1

In any set of

n

consecutive integers, the distance between the maximum and minimum elements is

n - 1

.

2

Use the sum formula for an evenly spaced set.

n(a + b) = 182

The sum is

\text{Sum} = n \times \frac{a + b}{2} = 91

, which simplifies to

n(a + b) = 182

.

3

Factor

182

to find integer solutions satisfying

a \cdot b = -144

.

n = 26

,

a = -9

,

b = 16

For

n = 26

,

a + b = 7

and

b - a = 25

, giving

a = -9

and

b = 16

. Their product is

(-9)(16) = -144

.

Key Concept

Properties of consecutive integer sets, average/median sum formula, and term boundary indexing

Question 12Question →

Set $S$ consists of consecutive integers. The sum of all positive integers in set $S$ is $105$ , and the sum of all integers in set $S$ is $-31$ . How many negative integers are contained in set $S$ ?

Show answer & explanation

Answer: 16

Answer

There are 16 negative integers in set

S

.

The correct response is 16. By setting up the sum of consecutive positive integers starting at 1, we determine that the set contains positive integers up to 14, which sum to 105. Subtracting 105 from the total set sum of -31 reveals that the negative integers must sum to -136. The consecutive negative integers -1, -2, ..., -p sum to -136 when p = 16, since 16 × 17 / 2 = 136.

Step-by-Step Solution

1

Find the maximum positive integer

k

in set

S

The largest positive integer in set

S

is

14

Because set

S

consists of consecutive integers, the positive integers are

1, 2, \dots, k

. The sum formula

\frac{k(k+1)}{2} = 105

leads to

k(k+1) = 210

. Factoring

210

into two consecutive integers gives

14 \times 15

, so

k = 14

.

2

Calculate the sum of all negative integers in set

S

The sum of all negative integers is

-136

The total sum of set

S

is the sum of its negative integers plus

0

plus the sum of its positive integers:

-31 = S_{\text{neg}} + 0 + 105 \implies S_{\text{neg}} = -136

.

3

Determine the count

p

of negative integers

The number of negative integers is

16

The negative integers are

-1, -2, \dots, -p

. Their sum is

-\frac{p(p+1)}{2} = -136 \implies p(p+1) = 272

. Solving

p(p+1) = 272

gives

p = 16

because

16 \times 17 = 272

.

Key Concept

Consecutive integer set properties and partitioning sets into positive and negative components using arithmetic series formulas.

Question 13Question →

Set $S$ consists of a sequence of consecutive integers, ordered from least to greatest. If Set $S$ contains exactly $45$ integers and the sum of all the integers in Set $S$ is $405$ , what is the value of the smallest integer in Set $S$ ?

Show answer & explanation

Answer: -13

Answer

The smallest integer in Set

S

is

-13

.

The average (arithmetic mean) of the set is calculated by dividing the total sum (

405

) by the number of terms (

45

), which equals

9

. In an evenly spaced set with an odd number of elements, the mean is equal to the median (middle term). Because there are

45

terms, exactly

22

terms lie below the median. Subtracting

22

from

9

yields

-13

as the smallest integer in the set.

Step-by-Step Solution

1

Calculate the arithmetic mean and median of the set

Arithmetic mean = Median = 9

For an evenly spaced set, the arithmetic mean is equal to the median. Dividing the sum (

405

) by the number of terms (

45

) gives

9

.

2

Determine the number of terms preceding the median

22 terms precede the median

With

45

terms in total, the median is the

23\text{rd}

term, which leaves

\frac{45 - 1}{2} = 22

terms smaller than the median.

3

Calculate the smallest integer

Smallest integer =

-13

Subtracting

22

from the median gives

9 - 22 = -13

.

Key Concept

In any set of consecutive integers with

n

terms, the arithmetic mean equals the median. If

n

is odd, the smallest integer is given by

\text{Median} - \frac{n-1}{2}

.

Question 14Question →

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

21

23

24

25

26

Show answer & explanation

Answer: 25

Answer

The largest integer in the set is 25.

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

115

by

5

gives a median of

23

. The five consecutive integers are

21, 22, 23, 24, 25

, so the largest integer is

25

.

Step-by-Step Solution

1

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

Dividing the sum

115

by

5

yields a median of

23

.

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

2

Determine the five consecutive integers centered around the median.

The integers are

21, 22, 23, 24, 25

.

Consecutive integers differ by

1

, so the two terms above the median

23

are

24

and

25

.

3

Identify the largest integer.

The largest integer is

25

.

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

Key Concept

Average and Median Equivalence in Consecutive Integer Sets

Estimated Time:45s

Question 15Question →

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

Show answer & explanation

Answer: 15

Answer

The median integer of the set is 15.

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

7

integers is

105

, the mean is

105 \div 7 = 15

. Therefore, the median of the set is

15

.

Step-by-Step Solution

1

Apply the property of evenly spaced sets.

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

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

2

Calculate the arithmetic mean.

Mean =

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

.

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

3

Determine the median.

Median =

15

.

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

Key Concept

Average and Median Equivalence in Evenly Spaced Sets

Estimated Time:45s

Question 16Question →

If the sum of three consecutive even integers is $42$ , what is the smallest of these integers?

10

12

13

14

16

Show answer & explanation

Answer: $12$

Answer

12

The arithmetic mean (average) of three consecutive even integers is equal to the middle integer. Dividing the total sum

42

by

3

yields

14

, which is the middle integer. Subtracting

2

gives the smallest integer,

12

.

Step-by-Step Solution

1

Define the three consecutive even integers algebraically

Let the three consecutive even integers be

n

,

n + 2

, and

n + 4

, where

n

represents the smallest integer.

Consecutive even integers differ by

2

.

2

Set up the equation for their sum

n + (n + 2) + (n + 4) = 42 \implies 3n + 6 = 42

The problem states that the sum of the three integers equals

42

.

3

Solve for

n

3n = 36 \implies n = 12

Subtract

6

from both sides and divide by

3

to find the smallest integer.

Key Concept

Properties and Sums of Consecutive Even Integers

Estimated Time:45s

Question 17Question →

Set $A$ consists of $k$ consecutive integers, where $k > 1$ . Set $B$ consists of $2k$ consecutive integers. The median of Set $A$ is equal to the least integer in Set $B$ , and the sum of all integers in Set $A$ is equal to the sum of all integers in Set $B$ . If the sum of the integers in Set $B$ is $-153$ , what is the greatest integer in Set $A$ ?

-21

-17

-13

-12

-9

Show answer & explanation

Answer: $-13$

Answer

The greatest integer in Set

A

is

-13

.

The problem establishes a relationship between two sets of consecutive integers using their medians and sums. Equating the sum formulas

S_A = k\left(a + \frac{k-1}{2}\right)

and

S_B = 2k\left(a + \frac{3k-2}{2}\right)

leads to

a = \frac{3-5k}{2}

. Substituting this into the sum expression gives

k - 2k^2 = -153

, yielding

k = 9

. Thus, the smallest element is

a = -21

, and the greatest element is

-21 + 9 - 1 = -13

.

Step-by-Step Solution

1

Express the sum of Set

A

in terms of

k

and its smallest element

a

.

The median of Set

A

is

M_A = a + \frac{k-1}{2}

, so the sum of Set

A

is

S_A = k \cdot M_A = k\left(a + \frac{k-1}{2}\right)

.

In any set of consecutive integers, the sum equals the number of terms multiplied by the average (median).

2

Express the sum of Set

B

using the given conditions.

The least integer in Set

B

is

M_A = a + \frac{k-1}{2}

. Since Set

B

has

2k

consecutive integers, its median is

M_B = \left(a + \frac{k-1}{2}\right) + \frac{2k-1}{2} = a + \frac{3k-2}{2}

. The sum of Set

B

is

S_B = 2k \cdot M_B = 2k\left(a + \frac{3k-2}{2}\right) = k(2a + 3k - 2)

.

Set

B

starts at the median of Set

A

and contains

2k

terms.

3

Equate

S_A

and

S_B

to solve for

a

in terms of

k

.

Since

S_A = S_B

and

k > 1

, we divide by

k

:

a + \frac{k-1}{2} = 2a + 3k - 2 \implies a = \frac{3 - 5k}{2}

.

The problem states that the sums of the two sets are equal.

4

Substitute

a = \frac{3 - 5k}{2}

into

S_B

and set it equal to

-153

.

S_B = k\left(2\left(\frac{3-5k}{2}\right) + 3k - 2\right) = k(3 - 5k + 3k - 2) = k(1 - 2k) = k - 2k^2

. Setting

k - 2k^2 = -153 \implies 2k^2 - k - 153 = 0

. Factoring gives

(2k + 17)(k - 9) = 0

. Since

k

must be a positive integer greater than 1,

k = 9

.

Finding

k

allows us to calculate the exact elements of Set

A

.

5

Calculate the greatest integer in Set

A

.

With

k = 9

, the smallest integer

a = \frac{3 - 5(9)}{2} = -21

. The greatest integer in Set

A

is

a + k - 1 = -21 + 9 - 1 = -13

.

The

k

-th term of a sequence of consecutive integers starting at

a

is

a + k - 1

.

Key Concept

Properties of consecutive integer sets: sum equals number of terms times median; indexing terms in evenly spaced sets.

Question 18Question →

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 19Question →

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 20Question →

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

Consecutive Integers and Number Sets

Answer

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