Number Properties and Arithmetic Practice Questions — GMAT — Page 5

Question 81Question →

If $x$ is a negative integer and $y$ is a positive integer, what is the value of $\frac{|x - y| - |y - x|}{x + y + 100}$ ?

Show answer & explanation

Answer: 0

Answer

0

Given

x < 0

and

y > 0

, the quantity

x - y

is strictly negative, which means

|x - y| = -(x - y) = y - x

. The quantity

y - x

is strictly positive, so

|y - x| = y - x

. Thus, the numerator simplifies to

(y - x) - (y - x) = 0

. Dividing zero by any non-zero denominator yields

0

.

Step-by-Step Solution

1

Determine the signs of the terms inside the absolute value functions

Since

x < 0

and

y > 0

,

x - y < 0

and

y - x > 0

.

Subtracting a positive number from a negative number yields a negative value, while subtracting a negative number from a positive number yields a positive value.

2

Simplify the absolute value expressions

|x - y| = y - x

and

|y - x| = y - x

.

The absolute value of a negative number is its negation, and the absolute value of a positive number is the number itself.

3

Compute the difference in the numerator

|x - y| - |y - x| = (y - x) - (y - x) = 0

.

Subtracting an algebraic expression from an identical expression results in zero.

4

Evaluate the entire fraction

\frac{0}{x + y + 100} = 0

.

Zero divided by any non-zero real number is zero.

Key Concept

Positive and Negative Number Properties with Absolute Value

Question 82Question →

At the beginning of a fiscal year, a municipal transit authority allocated its capital expenditure budget among three projects: Bus Rapid Transit, Rail Modernization, and Station Upgrades. Exactly $0.30$ of the total budget was allocated to Bus Rapid Transit. Of the remaining budget, exactly $\frac{3}{7}$ was allocated to Rail Modernization, and the rest was allocated to Station Upgrades. By the end of the year, expenditures on Bus Rapid Transit exceeded its initial allocation by $25\%$ , expenditures on Rail Modernization were $15\%$ below its initial allocation, and expenditures on Station Upgrades exceeded its initial allocation by $10\%$ . By what percent did the transit authority's total expenditures across all three projects exceed its initial total budget?

Show answer & explanation

Answer: 7

Answer

The total expenditures across all three projects exceeded the initial total budget by

7\%

.

To solve this problem, represent the total initial budget as

B

. The Bus Rapid Transit allocation is

0.30B

, leaving

0.70B

. Rail Modernization receives

\frac{3}{7}

of

0.70B

, which equals

0.30B

. The remaining portion for Station Upgrades is

0.70B - 0.30B = 0.40B

. End-of-year expenditures are calculated by multiplying each allocation by its respective growth multiplier: Bus Rapid Transit is

0.30B \times 1.25 = 0.375B

, Rail Modernization is

0.30B \times 0.85 = 0.255B

, and Station Upgrades is

0.40B \times 1.10 = 0.44B

. Summing these expenditures gives

0.375B + 0.255B + 0.44B = 1.07B

. Comparing

1.07B

to the initial

1.00B

reveals an overall increase of

0.07B

, or

7\%

.

Step-by-Step Solution

1

Express the initial allocations for each project as fractions of the total budget B

Bus Rapid Transit =

0.30B

, Rail Modernization =

0.30B

, Station Upgrades =

0.40B

Bus Rapid Transit is explicitly

0.30B

. The remaining

0.70B

is split such that Rail Modernization receives

\frac{3}{7} \times 0.70B = 0.30B

, leaving

0.70B - 0.30B = 0.40B

for Station Upgrades.

2

Apply the individual percentage changes to determine end-of-year expenditures

Bus Rapid Transit =

0.375B

, Rail Modernization =

0.255B

, Station Upgrades =

0.44B

A

25\%

increase corresponds to a multiplier of

1.25

, a

15\%

decrease corresponds to a multiplier of

0.85

, and a

10\%

increase corresponds to a multiplier of

1.10

.

3

Sum the project expenditures and calculate the net percent change relative to B

Total expenditure =

1.07B

, corresponding to a

7\%

net increase

Adding

0.375B + 0.255B + 0.44B

yields

1.07B

. Subtracting the original budget

1.00B

gives

0.07B

, which is

7\%

of

B

.

Key Concept

Weighted Percentage Changes and Sequential Fraction-Decimal Operations

Question 83Question →

If $x$ is a negative real number and $y$ is a positive real number, what is the value of $\frac{x}{|x|} + \frac{|y|}{y}$ ?

Show answer & explanation

Answer: 0

Answer

The value of the expression is 0.

For any negative number

x

, the ratio

\frac{x}{|x|}

evaluates to

-1

because

|x| = -x

. For any positive number

y

, the ratio

\frac{|y|}{y}

evaluates to

1

because

|y| = y

. Summing

-1

and

1

results in

0

.

Step-by-Step Solution

1

Evaluate the first term for a negative variable

-1

By definition of absolute value, if

x < 0

, then

|x| = -x

, making

\frac{x}{|x|} = \frac{x}{-x} = -1

.

2

Evaluate the second term for a positive variable

1

If

y > 0

, then

|y| = y

, making

\frac{|y|}{y} = \frac{y}{y} = 1

.

3

Sum the simplified values

0

Combining

-1

and

1

yields

-1 + 1 = 0

.

Key Concept

Properties of Absolute Value and Signs of Numbers

Question 84Question →

If $k$ is a positive integer such that $\sqrt{8^{10} + 2^{23} + 4^7} = 2^k + 128$ , what is the value of $k$ ?

Show answer & explanation

Answer: 15

Answer

The value of

k

is 15.

Converting all terms under the square root to base 2 produces

\sqrt{2^{30} + 2^{23} + 2^{14}}

. Recognizing that

(2^{15} + 2^7)^2 = (2^{15})^2 + 2(2^{15})(2^7) + (2^7)^2 = 2^{30} + 2^{23} + 2^{14}

, taking the square root yields

2^{15} + 2^7 = 2^{15} + 128

. Matching this with

2^k + 128

yields

k = 15

.

Step-by-Step Solution

1

Convert terms under the square root to base 2.

The radical expression becomes

\sqrt{2^{30} + 2^{23} + 2^{14}}

.

Expressing terms with the same base allows exponent rules and algebraic identity recognition.

2

Identify the expression under the radical as a perfect square of the form

(a + b)^2

.

Setting

a = 2^{15}

and

b = 2^7

gives

2ab = 2 \cdot 2^{15} \cdot 2^7 = 2^{23}

, so

2^{30} + 2^{23} + 2^{14} = (2^{15} + 2^7)^2

.

The middle term

2^{23}

satisfies

2 \cdot \sqrt{2^{30}} \cdot \sqrt{2^{14}}

.

3

Evaluate the square root and solve for

k

.

\sqrt{(2^{15} + 2^7)^2} = 2^{15} + 128

. Setting

2^{15} + 128 = 2^k + 128

yields

k = 15

.

Comparing terms directly after evaluating

2^7 = 128

isolates

2^k = 2^{15}

.

Key Concept

Application of exponent rules combined with perfect square algebraic identities under radicals

Question 85Question →

If $n$ is a positive integer such that $140n$ is a perfect square and $105n$ is a perfect cube, what is the minimum possible number of positive divisors of $n$ ?

12

27

108

756

15

Show answer & explanation

Answer: 108

Answer

108

To minimize the number of divisors of

n

, express

n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d

. The condition that

140n = 2^{a+2} \cdot 3^b \cdot 5^{c+1} \cdot 7^{d+1}

is a perfect square implies

a

and

b

must be even, while

c

and

d

must be odd. The condition that

105n = 2^a \cdot 3^{b+1} \cdot 5^{c+1} \cdot 7^{d+1}

is a perfect cube requires

a

to be a multiple of 3, and

b, c, d

to leave a remainder of 2 when divided by 3. The minimum non-negative integers satisfying both requirements are

a=0

(since 0 is even and a multiple of 3),

b=2

(even and leaves remainder 2 mod 3),

c=5

(odd and leaves remainder 2 mod 3), and

d=5

(odd and leaves remainder 2 mod 3). The minimum number of positive divisors is

(0+1)(2+1)(5+1)(5+1) = 108

.

Step-by-Step Solution

1

Find the prime factorizations of 140 and 105

140 = 2^2 \cdot 5 \cdot 7

and

105 = 3 \cdot 5 \cdot 7

Decomposing into prime factors allows analyzing exponent constraints for perfect powers.

2

Express

n

in terms of prime factors

n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d

and determine constraints for

140n

to be a perfect square

140n = 2^{a+2} \cdot 3^b \cdot 5^{c+1} \cdot 7^{d+1}

requires

a

to be even,

b

to be even,

c

to be odd, and

d

to be odd.

All prime factor exponents in a perfect square must be even numbers.

3

Determine constraints for

105n

to be a perfect cube

105n = 2^a \cdot 3^{b+1} \cdot 5^{c+1} \cdot 7^{d+1}

requires

a \equiv 0 \pmod 3

,

b \equiv 2 \pmod 3

,

c \equiv 2 \pmod 3

, and

d \equiv 2 \pmod 3

.

All prime factor exponents in a perfect cube must be multiples of 3.

4

Find the smallest non-negative integers satisfying both sets of constraints for each exponent

a = 0

,

b = 2

,

c = 5

, and

d = 5

For

a

: smallest non-negative even multiple of 3 is 0. For

b

: smallest non-negative even number congruent to

2 \pmod 3

is 2. For

c

and

d

: smallest odd numbers congruent to

2 \pmod 3

are 5.

5

Calculate the total number of positive divisors of

n = 3^2 \cdot 5^5 \cdot 7^5

(2 + 1)(5 + 1)(5 + 1) = 3 \times 6 \times 6 = 108

The divisor counting formula multiplies

(e_i + 1)

for each prime exponent

e_i

.

Key Concept

Prime factor exponent constraints for perfect powers and the divisor counting formula

Question 86Question →

For any positive integer $n$ , let $S_n$ denote the units digit of the sum $2^n + 3^n + 4^n + 7^n$ . What is the remainder when the sum $T = \sum_{n=1}^{102} S_n$ is divided by $9$ ?

Show answer & explanation

Answer: 1

Answer

The remainder when the sum T is divided by 9 is 1.

Evaluating the units digits of each exponent term reveals that

S_n

follows a repeating 4-term cycle

(6, 8, 2, 4)

with a sum of 20 per cycle. For 102 terms, there are 25 full cycles plus the first two terms (

S_1=6

and

S_2=8

). The total sum is

T = 25 \times 20 + 6 + 8 = 514

. Dividing 514 by 9 yields

514 = 9 \times 57 + 1

, so the remainder is 1.

Step-by-Step Solution

1

Find the cyclicity of the units digit of each term

2^n, 3^n, 4^n,

and

7^n

.

Units digits repeat in patterns of length 4: for

2^n

(2, 4, 8, 6), for

3^n

(3, 9, 7, 1), for

4^n

(4, 6, 4, 6), and for

7^n

(7, 9, 3, 1).

Units digits of positive integer powers cycle with periodicities that divide 4.

2

Compute

S_n

for

n = 1, 2, 3, 4

and find the sum of one 4-term period.

S_1 = \text{units}(2+3+4+7=16) = 6

,

S_2 = \text{units}(4+9+6+9=28) = 8

,

S_3 = \text{units}(8+7+4+3=22) = 2

,

S_4 = \text{units}(6+1+6+1=14) = 4

. Sum of one period =

6+8+2+4 = 20

.

The sum of the units digits of individual terms determines the units digit of the total expression.

3

Calculate the total sum

T = \sum_{n=1}^{102} S_n

.

Since

102 = 25 \times 4 + 2

, the sequence consists of 25 complete cycles of 4 terms plus the first 2 terms (

S_1=6

and

S_2=8

). Thus,

T = 25 \times 20 + 6 + 8 = 500 + 14 = 514

.

Dividing the total number of terms by the period length gives the number of full cycles and remaining initial terms.

4

Compute the remainder when

T = 514

is divided by 9.

The sum of the digits of 514 is

5 + 1 + 4 = 10

, and

10 \equiv 1 \pmod 9

. Alternatively,

514 = 9 \times 57 + 1

. Thus, the remainder is 1.

A positive integer and the sum of its digits leave the same remainder when divided by 9.

Key Concept

Units digit cyclicity of exponential terms and modular arithmetic on sequence sums

Estimated Time:2m 0s

Question 87Question →

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 88Question →

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 89Question →

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 90Question →

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 91Question →

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 92Question →

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 93Question →

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 94Question →

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 95Question →

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 96Question →

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 97Question →

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 98Question →

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 99Question →

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 100Question →

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

Number Properties and Arithmetic

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Answer

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Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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