Basic Numeracy Practice Questions — State PSC Exam — Page 4

Question 61Question →

A secure digital vault requires a 2-digit numerical access code. The code is determined by calculating the value of the expression $(P + Q) - (R \times S)$ , where the variables are defined based on the mathematical classification of numbers:

* $P$ : The sum of all single-digit prime numbers.
* $Q$ : The product of the smallest positive composite number and the smallest odd prime number.
* $R$ : The total count of whole numbers strictly less than $10$ that are classified as neither prime nor composite.
* $S$ : The smallest positive integer $n$ for which the expression $\sqrt{12n}$ results in a rational number.

What is the final numerical access code?

Show answer & explanation

Answer: 23

Answer

23

The correct calculation evaluates each subset definition perfectly: P = 17 (sum of primes 2, 3, 5, 7), Q = 12 (4 * 3), R = 2 (0 and 1 are neither prime nor composite), and S = 3 (making 12 * 3 = 36 a perfect square). Plugging these into the equation (17 + 12) - (2 * 3) yields exactly 23.

Step-by-Step Solution

1

Identify single-digit primes and sum them to find P.

P = 17

The single-digit primes are 2, 3, 5, and 7. Summing them yields 2 + 3 + 5 + 7 = 17.

2

Identify the smallest composite and smallest odd prime to find Q.

Q = 12

The smallest positive composite number is 4, and the smallest odd prime number is 3. Multiplying them yields 4 * 3 = 12.

3

Count whole numbers less than 10 that are neither prime nor composite to find R.

R = 2

The set of whole numbers strictly less than 10 is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Only 0 and 1 fit the classification of being neither prime nor composite.

4

Find the smallest positive integer n making \sqrt{12n} rational.

S = 3

For the square root to be rational, 12n must be a perfect square. Since 12 factors to 2^2 * 3, the smallest integer n to pair the remaining 3 is 3.

5

Evaluate the final expression.

23

(P + Q) - (R * S) = (17 + 12) - (2 * 3) = 29 - 6 = 23.

Key Concept

Classification properties of primes, composites, whole numbers, and rational numbers.

Question 62Question →

A numerical access code $X$ is generated based on the evaluation of the following mathematical expression:

X = 36 + 144 \div 6 \times 4

To successfully bypass the security protocol, a user must input the exact total number of positive factors (divisors) of the value $X$ . What number must the user input?

3

8

12

16

Show answer & explanation

Answer: 12

Answer

The user must input 12, as the evaluated number 132 possesses exactly 12 positive factors.

To find the correct access code, the expression must first be properly evaluated using standard order of operations (BODMAS). By performing division and then multiplication left-to-right, we get

144 \div 6 = 24

, and

24 \times 4 = 96

. Adding 36 yields

X = 132

. The prime factorization of 132 is

2^2 \times 3^1 \times 11^1

. Using the standard formula for finding total divisors, we add 1 to each exponent and multiply them together:

(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12

.

Step-by-Step Solution

1

Evaluate the division and multiplication parts of the expression from left to right according to the BODMAS rule.

144 \div 6 = 24

, followed by

24 \times 4 = 96

.

Division and multiplication hold equal precedence and must be executed left-to-right before any addition.

2

Complete the addition to find the final integer value of

X

.

X = 36 + 96 = 132

.

Addition is performed after higher-precedence operations are resolved.

3

Determine the prime factorization of the computed value 132.

132 = 4 \times 33 = 2^2 \times 3^1 \times 11^1

.

Finding the prime bases and their respective powers is required to apply the divisor count formula.

4

Calculate the total number of positive factors by adding 1 to each exponent and multiplying the results.

(2 + 1) \times (1 + 1) \times (1 + 1) = 3 \times 2 \times 2 = 12

.

The formula

(a+1)(b+1)...

systematically counts every possible product combination of the prime factors.

Key Concept

Factors, Multiples, and Prime Factorization

Estimated Time:1m 30s

Question 63Question →

An integer $P$ is given by the mathematical expression $P = 4 \cdot 36^{2033} - 6 \cdot 58^{2033} - 2 \cdot 89^{2033}$ . Determine the positive remainder when $P$ is divided by $17$ .

Show answer & explanation

Answer: 9

Answer

9

The correct answer is derived by first reducing the bases modulo 17, giving 2, 7, and 4 respectively. Then, utilizing Fermat's Little Theorem (

a^{16} \equiv 1 \pmod{17}

), the exponent 2033 is reduced to

2033 \pmod{16} = 1

. Substituting these simplified values back into the expression yields

4(2) - 6(7) - 2(4) = -42

. Because a remainder must be positive, adding the next highest multiple of 17 (which is

17 \times 3 = 51

) to -42 gives the final valid remainder of 9.

Step-by-Step Solution

1

Find the remainder of each base when divided by 17.

36 \equiv 2 \pmod{17}

,

58 \equiv 7 \pmod{17}

, and

89 \equiv 4 \pmod{17}

.

Reducing bases before dealing with large exponents simplifies the modular arithmetic calculation.

2

Apply Fermat's Little Theorem to simplify the exponent.

Since 17 is a prime number,

a^{16} \equiv 1 \pmod{17}

. The exponent is

2033 = 16 \times 127 + 1

, meaning

a^{2033} \equiv a^1 \pmod{17}

.

Fermat's Little Theorem allows us to reduce massive exponents by dividing them by

(p-1)

and keeping only the remainder.

3

Substitute the reduced bases and exponents into the original expression.

P \equiv 4(2^1) - 6(7^1) - 2(4^1) \pmod{17}

, which evaluates to

8 - 42 - 8 = -42

.

This step calculates the combined overall remainder before adjusting for the strict definition of a positive modulo.

4

Convert the negative result to a valid positive remainder.

-42 \equiv -42 + 51 = 9 \pmod{17}

.

A remainder must be a non-negative integer strictly less than the divisor (17). Adding the next highest multiple of 17 (which is 51) provides the correct positive remainder.

Key Concept

Divisibility Rules and Remainder Theorem

Question 64Question →

In a centralized database architecture, data records are distributed across $17$ server nodes numbered from $0$ to $16$ . A record with a numerical key $K$ is assigned to a node using the hash function $H(K) = K \pmod{17}$ , which always yields a positive remainder. If a specific batch of records is assigned a master key given by the expression $K = 2^{200} - 75$ , which server node will process this batch?

Show answer & explanation

Answer: 11

Answer

The batch of records will be processed by server node 11.

By applying modular arithmetic rules, the expression

(2^{200} - 75) \pmod{17}

simplifies individually to

(1 - 7) = -6

. Because node numbers in this system must be positive, adding the divisor 17 to -6 provides the correct positive remainder, which is 11.

Step-by-Step Solution

1

Identify the mathematical goal of the problem.

Calculate

(2^{200} - 75) \pmod{17}

and ensure the final result is a positive integer between 0 and 16.

The server nodes are assigned strictly based on the positive remainder when the key is divided by 17.

2

Find the remainder of the exponential term

2^{200}

divided by 17.

2^4 = 16 \equiv -1 \pmod{17}

. Raising this to the 50th power gives

(2^4)^{50} \equiv (-1)^{50} = 1 \pmod{17}

.

Finding a power of the base that is 1 or -1 modulo the divisor simplifies extremely large exponent calculations.

3

Find the remainder of the constant term 75 divided by 17.

75 = 17 \times 4 + 7

, so

75 \equiv 7 \pmod{17}

.

Each term in a modular arithmetic expression can and should be reduced individually.

4

Combine the reduced terms to find the overall remainder.

1 - 7 = -6 \pmod{17}

.

Substitute the individual reduced remainders back into the original expression

K = 2^{200} - 75

.

5

Convert the negative remainder into a valid positive node number.

-6 + 17 = 11

.

A remainder of -6 indicates the value is 6 units short of a full multiple of 17. The equivalent positive remainder is found by adding the divisor.

Key Concept

Modular arithmetic with large exponents and handling negative remainders appropriately.

Estimated Time:1m 30s

Question 65Question →

Three automated data backup protocols in a cloud server execute network synchronization pulses at precise intervals. Protocol Alpha pulses every $\frac{18}{5}$ seconds, Protocol Beta every $\frac{24}{25}$ seconds, and Protocol Gamma every $\frac{36}{35}$ seconds. They all initiate their first pulse simultaneously at 08:00:00 AM. Assuming the initial pulse at 08:00:00 AM is excluded from the count, which of the following represents the total number of times all three protocols will pulse together in exactly $12$ minutes?

21000

10

50

51

Show answer & explanation

Answer: $50$

Answer

The data protocols will pulse together

50

times in the

12

-minute period.

To find when all three protocols pulse together, we must determine the Least Common Multiple (LCM) of their cycle times. The LCM of fractions is calculated by dividing the LCM of their numerators (

18, 24, 36

) by the HCF of their denominators (

5, 25, 35

). The LCM of

18, 24,

and

36

is

72

. The HCF of

5, 25,

and

35

is

5

. This gives a synchronized interval of

72/5 = 14.4

seconds. In

12

minutes (

720

seconds), the number of synchronized intervals is

720 \div 14.4 = 50

. Since the initial pulse is excluded as per the instructions, the correct answer remains

50

.

Step-by-Step Solution

1

Identify the mathematical concept required to find when all three protocols pulse simultaneously.

The time of simultaneous pulsing is the Least Common Multiple (LCM) of their individual cycle times:

\frac{18}{5}

,

\frac{24}{25}

, and

\frac{36}{35}

.

Simultaneous occurrence of periodic events always aligns at multiples common to all individual intervals.

2

Apply the formula for finding the LCM of fractions.

\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

.

This is the standard algebraic rule to find the least common multiple for fractional values.

3

Calculate the LCM of the numerators (

18, 24, 36

) and the HCF of the denominators (

5, 25, 35

).

\text{LCM}(18, 24, 36) = 72

and

\text{HCF}(5, 25, 35) = 5

. Thus, the synchronized interval is

\frac{72}{5}

seconds, or

14.4

seconds.

Prime factorization gives

18 = 2 \times 3^2

,

24 = 2^3 \times 3

,

36 = 2^2 \times 3^2

(LCM =

2^3 \times 3^2 = 72

). For denominators,

5

is the largest common divisor.

4

Convert the given total time window into seconds.

12 \text{ minutes} = 12 \times 60 = 720 \text{ seconds}

.

Units must match the cycle interval (seconds) before division.

5

Calculate the total number of simultaneous pulses.

720 \div 14.4 = 50

. Since the initial pulse is excluded, the final count remains

50

.

Dividing the total time window by the synchronized interval yields the number of subsequent events.

Key Concept

Least Common Multiple (LCM) of fractions and its application to periodic simultaneous events.

Question 66Question →

Consider the positive integer $N = 3600$ . What is the total number of positive divisors of $N$ that are multiples of either $4$ or $9$ ?

12

42

33

9

Show answer & explanation

Answer: 33

Answer

The total number of positive divisors of 3600 that are multiples of either 4 or 9 is 33.

The correct answer is derived by finding the prime factorization of 3600 (

2^4 \times 3^2 \times 5^2

), determining the separate counts of divisors that are multiples of 4 (27 factors) and multiples of 9 (15 factors), and then using the inclusion-exclusion principle to subtract the intersection (divisors that are multiples of 36, which is 9 factors). This yields

27 + 15 - 9 = 33

.

Step-by-Step Solution

1

Prime factorize 3600 to identify the structure of its divisors.

3600 = 36 \times 100 = 2^4 \times 3^2 \times 5^2

.

A divisor's properties depend entirely on the combinations of its prime factors.

2

Calculate the total number of divisors that are multiples of 4.

Multiples of 4 must contain at least

2^2

. The valid powers for 2 are

\{2^2, 2^3, 2^4\}

(3 choices). For 3,

\{3^0, 3^1, 3^2\}

(3 choices). For 5,

\{5^0, 5^1, 5^2\}

(3 choices). Total =

3 \times 3 \times 3 = 27

divisors.

To be a multiple of 4, the divisor's prime factorization must contain a power of 2 that is at least 2.

3

Calculate the total number of divisors that are multiples of 9.

Multiples of 9 must contain at least

3^2

. The valid powers for 2 are 5 choices (from 0 to 4). For 3,

\{3^2\}

(1 choice). For 5, 3 choices. Total =

5 \times 1 \times 3 = 15

divisors.

To be a multiple of 9, the divisor must contain a power of 3 that is at least 2.

4

Calculate the number of divisors that are multiples of both 4 and 9 (i.e., multiples of 36).

These divisors must contain at least

2^2

and

3^2

. Choices: 3 (for 2)

\times

1 (for 3)

\times

3 (for 5) =

9

divisors.

Since 4 and 9 are co-prime, multiples of both must be multiples of their Least Common Multiple (LCM), which is 36.

5

Apply the Principle of Inclusion-Exclusion to find divisors that are multiples of either 4 or 9.

27 + 15 - 9 = 33

divisors.

Adding the multiples of 4 and 9 counts their intersection (multiples of 36) twice, so it must be subtracted once to find the correct union.

Key Concept

Applying set theory (inclusion-exclusion principle) to the combinations of prime factors of a number to find constrained subsets of divisors.

Question 67Question →

Calculate the remainder when the integer value of $11 \times 17^{73}$ is divided by $19$ .

Show answer & explanation

Answer: 16

Answer

16

The correct remainder is found by applying Fermat's Little Theorem to reduce the large exponent, yielding an intermediate calculation of -22. Converting this negative value to a proper positive remainder modulo 19 gives exactly 16.

Step-by-Step Solution

1

Simplify the base of the exponent modulo 19.

17 \equiv -2 \pmod{19}

Working with a smaller absolute value simplifies subsequent exponentiation steps.

2

Apply Fermat's Little Theorem to identify the cyclicity.

a^{18} \equiv 1 \pmod{19}

for any integer

a

not divisible by 19.

Since 19 is a prime number, the remainders of powers will repeat in cycles of 18.

3

Reduce the large exponent

73

using the identified cyclicity.

73 = 18 \times 4 + 1

, which means

(-2)^{73} \equiv (-2)^1 \pmod{19}

Because the powers cycle every 18, only the remainder of the exponent divided by 18 affects the final result.

4

Multiply the reduced exponential term by the leading coefficient.

11 \times (-2) = -22

The original mathematical expression contains a coefficient of 11.

5

Convert the negative intermediate result to a valid positive remainder.

-22 \equiv -22 + 2 \times 19 = -22 + 38 = 16 \pmod{19}

By standard definition, a remainder must be a positive integer strictly less than the divisor.

Key Concept

Modular Arithmetic and Fermat's Little Theorem

Question 68Question →

A logistics manager is arranging shipping containers into equal stacks. When the containers are stacked in groups of $18$ , $24$ , or $32$ , there are always exactly $11$ containers left over. However, when they are stacked in groups of $35$ , there are no containers left over. What is the minimum total number of containers the manager could have?

Show answer & explanation

Answer: 875

Answer

875

The correct answer is derived by recognizing that the total number of containers must take the form of

LCM(18, 24, 32) \times k + 11

. The LCM is

288

, so the number is

288k + 11

. We then find the smallest integer

k

such that

(288k + 11)

is perfectly divisible by

35

. By evaluating the expression for sequential values of

k

(or using modular arithmetic,

8k \equiv 24 \pmod{35}

), we find that

k = 3

is the smallest valid multiplier. Substituting this back yields

288(3) + 11 = 875

.

Step-by-Step Solution

1

Calculate the Least Common Multiple (LCM) of the initial stack group sizes (

18

,

24

, and

32

).

The prime factorizations are

18 = 2 \times 3^2

,

24 = 2^3 \times 3

, and

32 = 2^5

. The LCM is

2^5 \times 3^2 = 32 \times 9 = 288

.

The base cycle for the stacks without remainders requires finding the smallest number perfectly divisible by all three group sizes.

2

Express the total number of containers algebraically.

Let the total number of containers be

N

. We can write

N = 288k + 11

, where

k

is a positive integer.

Since there is always a remainder of

11

containers when divided by these numbers, the total must be

11

more than a multiple of their LCM.

3

Apply the secondary divisibility condition.

The problem states that

N

is exactly divisible by

35

. Therefore,

288k + 11 \equiv 0 \pmod{35}

.

When the containers are grouped by

35

, there is no remainder.

4

Solve the congruence to find the smallest integer

k

.

First, simplify

288 \pmod{35}

: since

35 \times 8 = 280

, we have

288 \equiv 8 \pmod{35}

. The equation becomes

8k + 11 \equiv 0 \pmod{35}

, or

8k \equiv -11 \equiv 24 \pmod{35}

. Dividing both sides by

8

gives

k = 3

.

Finding the smallest valid positive integer

k

will yield the minimum total number of containers.

5

Calculate the final number of containers using

k = 3

.

N = 288(3) + 11 = 864 + 11 = 875

.

Substitute the multiplier back into the original algebraic expression.

Key Concept

Solving for an unknown quantity using the Least Common Multiple (LCM) combined with constant remainder logic and a secondary divisibility condition.

Alternative Method

Instead of using modular arithmetic to solve

288k + 11 \equiv 0 \pmod{35}

, you can manually test integer values for

k

in the formula

N = 288k + 11

. For

k=1

,

N = 288(1) + 11 = 299

(

299 \div 35

leaves remainder

19

). For

k=2

,

N = 288(2) + 11 = 587

(

587 \div 35

leaves remainder

27

). For

k=3

,

N = 288(3) + 11 = 875

(

875 \div 35 = 25

with no remainder). Therefore,

875

is the smallest valid number.

Estimated Time:2m 30s

Question 69Question →

A robotics engineering team is programming three distinct automated paint dispensers on an assembly line. They are calibrated to release a burst of paint every $\frac{4}{5}$ , $\frac{8}{15}$ , and $\frac{12}{25}$ of a second, respectively. Let $L$ be the least common multiple (LCM) of these three time intervals, representing the exact time when all three dispensers fire simultaneously. Let $H$ be the highest common factor (HCF) of the intervals, representing the largest fundamental time unit that can perfectly measure all three intervals. What is the value of the ratio $\frac{L}{H}$ ?

90

\frac{1}{90}

\frac{2}{5}

\frac{32}{125}

Show answer & explanation

Answer: $90$

Answer

The value of the ratio

L/H

is 90.

To solve the problem, we first find the LCM and HCF of the fractions

\frac{4}{5}

,

\frac{8}{15}

, and

\frac{12}{25}

. The formula for the LCM of fractions is

\frac{\text{LCM of numerators}}{\text{HCF of denominators}}

. The numerators are 4, 8, and 12, which have an LCM of 24. The denominators are 5, 15, and 25, which have an HCF of 5. Thus,

L = \frac{24}{5}

. The formula for the HCF of fractions is

\frac{\text{HCF of numerators}}{\text{LCM of denominators}}

. The HCF of 4, 8, and 12 is 4. The LCM of 5, 15, and 25 is 75. Thus,

H = \frac{4}{75}

. Finally, dividing

L

by

H

gives

\frac{24}{5} \div \frac{4}{75}

, which equals

\frac{24}{5} \times \frac{75}{4} = 6 \times 15 = 90

.

Step-by-Step Solution

1

Calculate

L

, the LCM of the three fractions.

L = \frac{24}{5}

The LCM of fractions is the LCM of their numerators divided by the HCF of their denominators. LCM(4, 8, 12) = 24 and HCF(5, 15, 25) = 5.

2

Calculate

H

, the HCF of the three fractions.

H = \frac{4}{75}

The HCF of fractions is the HCF of their numerators divided by the LCM of their denominators. HCF(4, 8, 12) = 4 and LCM(5, 15, 25) = 75.

3

Calculate the ratio

\frac{L}{H}

.

\frac{24/5}{4/75} = \frac{24}{5} \times \frac{75}{4} = 6 \times 15 = 90

To divide by a fraction, multiply by its reciprocal. Simplifying the resulting expression yields 90.

Key Concept

Calculating the LCM and HCF of fractional values

Question 70Question →

An industrial cutting machine is programmed to divide a rectangular metallic sheet measuring $\frac{24}{5}$ meters in length and $\frac{16}{3}$ meters in width into identical square plates. If the square plates must be of the maximum possible size without leaving any wasted material, what is the area of each resulting square plate?

\frac{64}{225} \text{ m}^2

2304 \text{ m}^2

64 \text{ m}^2

\frac{256}{25} \text{ m}^2

Show answer & explanation

Answer: $\frac{64}{225} \text{ m}^2$

Answer

The area of each resulting square plate is \frac{64}{225} \text{ m}^2.

To cut the rectangle into identical squares of maximum size without waste, the side length of the square must be the Highest Common Factor (HCF) of the rectangle's dimensions. Using the fraction formula, the HCF of

\frac{24}{5}

and

\frac{16}{3}

is

\frac{\text{HCF}(24, 16)}{\text{LCM}(5, 3)} = \frac{8}{15}

meters. The area of the square is the side length squared, which is

(\frac{8}{15})^2 = \frac{64}{225} \text{ m}^2

.

Step-by-Step Solution

1

Determine the required mathematical operation for finding the maximum square size.

The side length of the square must be the Highest Common Factor (HCF) of the length (24/5) and width (16/3).

To divide a rectangle into identical squares without waste, the square's side must perfectly divide both dimensions. The 'maximum size' implies finding the greatest common divisor.

2

Apply the rule for finding the HCF of fractions.

HCF(a/b, c/d) = HCF(a, c) / LCM(b, d). Thus, HCF(24/5, 16/3) = HCF(24, 16) / LCM(5, 3).

The standard formula for the HCF of fractions requires taking the HCF of the numerators divided by the LCM of the denominators.

3

Calculate the specific HCF and LCM values.

HCF(24, 16) = 8 and LCM(5, 3) = 15. The side length of the square is \frac{8}{15} meters.

8 is the largest integer dividing both 24 and 16. 15 is the smallest integer divisible by both 5 and 3.

4

Calculate the area of the square.

Area = (\frac{8}{15})^2 = \frac{64}{225} \text{ m}^2.

The question specifically asks for the area of the plate, which is the square of its side length.

Key Concept

HCF of Fractions and Geometric Application

Question 71Question →

Four navigation buoys in a harbor flash at regular intervals of $3 \frac{1}{3}$ minutes, $4 \frac{1}{6}$ minutes, $5 \frac{5}{9}$ minutes, and $7 \frac{1}{2}$ minutes, respectively. If all four buoys flash simultaneously at exactly 12:00 noon, how many additional times will they all flash together again within the next $12$ hours?

Show answer & explanation

Answer: 4

Answer

4

The correct answer is derived by finding the Least Common Multiple of the fractional intervals to determine the duration between simultaneous flashes (150 minutes). Then, by converting the 12-hour period into 720 minutes, dividing 720 by 150 yields 4.8, indicating the buoys will flash together 4 additional times within that exact timeframe.

Step-by-Step Solution

1

Convert the mixed numbers into improper fractions.

The intervals are

\frac{10}{3}

,

\frac{25}{6}

,

\frac{50}{9}

, and

\frac{15}{2}

minutes.

Mathematical operations involving LCM and HCF of fractions require them to be in proper or improper fractional forms, rather than mixed numbers.

2

Determine the mathematical formula required for simultaneous events.

We must calculate the Least Common Multiple (LCM) of the intervals using the formula:

\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

.

Simultaneous cyclic events synchronize at the lowest common multiple of their individual intervals.

3

Calculate the LCM of the numerators.

The numerators are

10, 25, 50

, and

15

. Their LCM is

150

.

150

is the smallest integer divisible by all four numerators (

150 = 10 \times 15 = 25 \times 6 = 50 \times 3 = 15 \times 10

).

4

Calculate the HCF of the denominators.

The denominators are

3, 6, 9

, and

2

. Their HCF is

1

.

The numbers

2

and

3

are coprime (they share no common factors greater than

1

), forcing the overall HCF of the set to be

1

.

5

Calculate the exact time interval between simultaneous flashes.

The LCM of the intervals is

\frac{150}{1} = 150

minutes.

Applying the results from steps 3 and 4 directly into the fractional LCM formula.

6

Determine how many times this 150-minute interval fits into the 12-hour window.

12 \text{ hours} = 720 \text{ minutes}

.

\lfloor \frac{720}{150} \rfloor = \lfloor 4.8 \rfloor = 4

.

Dividing the total available time by the simultaneous interval length gives the exact count of additional flashes.

Key Concept

LCM of fractions in simultaneous cyclic events

Question 72Question →

Consider the composite number $X = p^3 \cdot q^2$ , where $p$ and $q$ are distinct prime numbers. Which of the following statements regarding the factors and properties of $X$ are correct?

Select all that apply

The total number of positive factors of

X

is exactly

12

.

If

p=2

and

q=3

, the sum of all positive factors of

X

is

195

.

The total number of positive factors can be derived by the expression

3 + 1 \times 2 + 1

, which yields exactly

6

factors.

The total number of positive factors of

X

that are multiples of either

p

or

q

is

17

, calculated by directly adding the

9

factors divisible by

p

to the

8

factors divisible by

q

.

Show answer & explanation

Answer: The total number of positive factors of $X$ is exactly $12$ .; If $p=2$ and $q=3$ , the sum of all positive factors of $X$ is $195$ .

Answer

The correct statements are that the total number of positive factors is exactly 12, and if p=2 and q=3, the sum of all positive factors is 195.

The correct options properly apply the theorems of prime factorization. A composite number with prime factorization

p^a \cdot q^b

always has

(a+1)(b+1)

total factors, which confirms the count of

12

. Furthermore, the geometric series expansion for the sum of factors correctly evaluates to

195

when substituting

p=2

and

q=3

.

Step-by-Step Solution

1

Determine the total number of positive factors of X.

The prime powers are 3 and 2. Using the formula (a+1)(b+1), we get (3+1)(2+1) = 4 * 3 = 12.

This establishes the baseline count of factors necessary for evaluating the first statement and the arithmetic logic of the third statement.

2

Calculate the sum of factors for the specific case where p=2 and q=3.

The sum is (1 + 2 + 4 + 8) * (1 + 3 + 9) = 15 * 13 = 195.

This verifies the explicit calculation claim made in the second statement.

3

Analyze the arithmetic expression proposed for factor counting: 3 + 1 * 2 + 1.

The expression lacks brackets. By standard order of operations, 3 + (1*2) + 1 = 6, which is mathematically flawed for factor counting.

Identifying the missing parentheses reveals a fundamental BODMAS order of operations violation.

4

Apply the inclusion-exclusion principle to count factors that are multiples of p or q.

Factors divisible by p: 9. Factors divisible by q: 8. Intersection (divisible by pq): 6. Union = 9 + 8 - 6 = 11.

This verifies whether directly adding the two sets is mathematically sound, proving that the fourth statement double-counts the overlap.

Key Concept

Prime factorization properties, including total factor counting, sum of factors, and applying set theory (inclusion-exclusion) to factor subsets.

Estimated Time:1m 30s

Question 73Question →

Evaluate the remainder when the numeric expression $47^{101} + 111^{101} - 25$ is divided by $16$ .

Show answer & explanation

Answer: 5

Answer

5

By applying the properties of modular arithmetic, we determine that

47 \equiv -1 \pmod{16}

and

111 \equiv -1 \pmod{16}

. Raising these to the 101st power yields

-1

and

-1

. Subtracting 25 gives a total of

-27

. To find the positive remainder, we add the next highest multiple of 16 (which is 32) to -27, yielding a final valid positive remainder of 5.

Step-by-Step Solution

1

Find the remainder of the bases when divided by 16.

47 \equiv -1 \pmod{16}

and

111 \equiv -1 \pmod{16}

.

Using negative remainders (since

47 = 16 \times 3 - 1

and

111 = 16 \times 7 - 1

) significantly simplifies the calculation of large powers.

2

Substitute the simplified bases into the original expression.

(47)^{101} + (111)^{101} - 25 \equiv (-1)^{101} + (-1)^{101} - 25 \pmod{16}

.

The properties of modular arithmetic allow replacing bases with their congruent values modulo the divisor.

3

Evaluate the exponents and the constant term.

(-1) + (-1) - 25 = -27

.

An odd power of a negative number remains negative, so

(-1)^{101} = -1

.

4

Find the equivalent positive remainder for -27 modulo 16.

-27 \equiv 5 \pmod{16}

.

Adding multiples of 16 (in this case,

16 \times 2 = 32

) to

-27

yields a positive remainder in the valid range of

0

to

15

. Calculating

-27 + 32 = 5

gives the final remainder.

Key Concept

Modular arithmetic with large powers using negative remainders

Alternative Method

Instead of converting the bases to -1, one could convert them to positive 15. The expression becomes

15^{101} + 15^{101} - 25

. However, evaluating this requires recognizing that

15 \equiv -1 \pmod{16}

to easily compute the large exponents, which ultimately merges back into the primary solution path.

Estimated Time:1m 30s

Question 74Question →

Consider the number $N = 2^4 \times 3^3 \times 5^2$ . How many positive factors of $N$ are also multiples of $120$ ?

12

16

24

44

Show answer & explanation

Answer: 12

Answer

12

For a factor of

N

to be a multiple of

120

, it must contain at least the prime factorization of

120

, which is

2^3 \times 3^1 \times 5^1

. We can factor this out from

N

:

N = 120 \times (2^1 \times 3^2 \times 5^1)

. The number of such factors is simply the total number of divisors of the remaining portion

(2^1 \times 3^2 \times 5^1)

, which is calculated by adding

1

to each exponent and multiplying them:

(1+1) \times (2+1) \times (1+1) = 12

.

Step-by-Step Solution

1

Determine the prime factorization of 120.

120 = 2^3 \times 3^1 \times 5^1

To find factors of N that are multiples of 120, we must know the minimum prime powers required.

2

Determine the prime factorization of the quotient N / 120.

\frac{2^4 \times 3^3 \times 5^2}{2^3 \times 3^1 \times 5^1} = 2^{(4-3)} \times 3^{(3-1)} \times 5^{(2-1)} = 2^1 \times 3^2 \times 5^1

Any factor of N that is a multiple of 120 can be expressed as

120 \times k

, where

k

must be a factor of the remaining prime powers of N.

3

Calculate the total number of positive factors for the quotient

k

.

(1 + 1) \times (2 + 1) \times (1 + 1) = 2 \times 3 \times 2 = 12

The number of choices for

k

corresponds exactly to the number of factors of N that are multiples of 120.

Key Concept

Identifying restricted factors and multiples using prime factorization.

Question 75Question →

A digital encryption protocol derives a validation key from the mathematical properties of the integer $12600$ . The key is calculated by multiplying the total quantity of its even positive divisors by the total quantity of its odd positive divisors. What is the exact value of this validation key?

Show answer & explanation

Answer: 972

Answer

972

By finding the prime factorization of

12600

as

2^3 \times 3^2 \times 5^2 \times 7^1

, we can separate the divisors into even and odd categories. Odd divisors utilize only the odd prime factors, yielding a total of

(2+1)(2+1)(1+1) = 18

odd divisors. Even divisors require at least one power of

2

, giving

3 \times (2+1)(2+1)(1+1) = 54

even divisors. Following the protocol's rule, their product is

18 \times 54 = 972

.

Step-by-Step Solution

1

Find the prime factorization of

12600

.

12600 = 2^3 \times 3^2 \times 5^2 \times 7^1

Prime factorization is necessary to determine the number and types of divisors an integer has.

2

Calculate the quantity of odd positive divisors.

Number of odd divisors =

18

Odd divisors are formed by using only the odd prime factors (

3

,

5

, and

7

). By adding

1

to each of their exponents and multiplying them, we get

(2+1) \times (2+1) \times (1+1) = 3 \times 3 \times 2 = 18

.

3

Calculate the quantity of even positive divisors.

Number of even divisors =

54

Even divisors must include at least one factor of

2

. The number of choices for the power of

2

is equal to its exponent (

3

choices:

2^1, 2^2, 2^3

). Multiplying this by the choices for odd prime factors gives

3 \times (2+1) \times (2+1) \times (1+1) = 3 \times 3 \times 3 \times 2 = 54

.

4

Multiply the number of even divisors by the number of odd divisors to find the validation key.

54 \times 18 = 972

The encryption protocol specifically defines the key as the product of these two calculated quantities.

Key Concept

Prime Factorization and Number of Divisors (Even and Odd)

Question 76Question →

Three different cryptographic algorithms in a secure server refresh their encryption keys at regular intervals. Algorithm X refreshes every $\frac{8}{3}$ milliseconds, Algorithm Y every $\frac{10}{9}$ milliseconds, and Algorithm Z every $\frac{14}{5}$ milliseconds. They all start a refresh cycle simultaneously. Let $M$ be the minimum number of milliseconds until they all start a refresh cycle together again. Let $H$ be the Highest Common Factor (HCF) of their respective refresh intervals in milliseconds. What is the value of $M \div H$ ?

\frac{1}{6300}

140

78400

6300

Show answer & explanation

Answer: $6300$

Answer

The value of

M \div H

is

6300

.

The value

M

requires finding the LCM of the fractions, which represents the synchronization time.

\text{LCM}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5}) = \frac{\text{LCM}(8, 10, 14)}{\text{HCF}(3, 9, 5)} = \frac{280}{1} = 280

. The value

H

requires finding the HCF of the fractions.

\text{HCF}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5}) = \frac{\text{HCF}(8, 10, 14)}{\text{LCM}(3, 9, 5)} = \frac{2}{45}

. Finally, dividing

M

by

H

gives

280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300

.

Step-by-Step Solution

1

Determine the synchronization time

M

by finding the LCM of the three fractional intervals.

M = \text{LCM}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5})

Simultaneous repeating events synchronize at the Lowest Common Multiple of their individual cycle times.

2

Calculate the LCM using the fraction rule:

\frac{\text{LCM of numerators}}{\text{HCF of denominators}}

.

Numerators

(8, 10, 14)

have an LCM of

280

. Denominators

(3, 9, 5)

have an HCF of

1

. Thus,

M = \frac{280}{1} = 280

.

This is the mathematical formula required for finding the LCM of rational numbers.

3

Determine the Highest Common Factor

H

of the three fractional intervals.

H = \text{HCF}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5})

The question explicitly requests the HCF of the three given refresh times.

4

Calculate the HCF using the fraction rule:

\frac{\text{HCF of numerators}}{\text{LCM of denominators}}

.

Numerators

(8, 10, 14)

have an HCF of

2

. Denominators

(3, 9, 5)

have an LCM of

45

. Thus,

H = \frac{2}{45}

.

This is the mathematical formula required for finding the HCF of rational numbers.

5

Calculate the final required ratio

M \div H

.

280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300

.

This satisfies the final computational step requested by the problem stem.

Key Concept

Calculating the Lowest Common Multiple (LCM) and Highest Common Factor (HCF) for fractions using their specific formulas.

Question 77Question →

A computer algorithm processes a data block and assigns it a routing ID based on modular arithmetic. This ID corresponds to the strictly positive remainder left after dividing the value of $56^{103} - 45$ by $13$ .

Find the value of this routing ID.

Show answer & explanation

Answer: 11

Answer

11

By reducing each component of the expression modulo 13,

56^{103}

reduces to

4^{103}

, which simplifies to

4

because

4^3 \equiv -1 \pmod{13}

. Subtracting

45 \pmod{13}

, which is

6

, yields

-2

. Adding the divisor 13 to

-2

gives the strictly positive remainder of 11.

Step-by-Step Solution

1

Reduce the base 56 modulo 13.

56 \equiv 4 \pmod{13}

Modular arithmetic allows simplifying the base before exponentiation.

2

Find a pattern to easily evaluate

4^{103} \pmod{13}

.

4^3 = 64 \equiv -1 \pmod{13}

Finding a small power that is congruent to 1 or -1 simplifies large exponents significantly.

3

Evaluate

4^{103} \pmod{13}

using the established pattern.

4^{103} = (4^3)^{34} \times 4^1 \equiv (-1)^{34} \times 4 = 4 \pmod{13}

Exponent rules allow breaking down 103 into

3 \times 34 + 1

.

4

Reduce the subtracted constant 45 modulo 13.

45 = 13 \times 3 + 6 \equiv 6 \pmod{13}

All terms in the expression must be evaluated under the same modulus.

5

Subtract the reduced values.

4 - 6 = -2

Substitute the individual modulo results back into the original expression structure.

6

Convert the negative remainder into a positive remainder.

-2 + 13 = 11

The question asks for a strictly positive remainder, which is standard in modular division contexts.

Key Concept

Modular Arithmetic and Negative Remainders

Question 78Question →

A cybersecurity protocol requires transmitting a data file of exactly $5400$ kilobytes. The system must divide this file into equally sized data packets without leaving any remaining data. For the encryption algorithm to function correctly, the size of each packet (in kilobytes) must be an integer that is a multiple of $12$ . Based on these conditions, how many different packet sizes are possible?

18

48

30

6

Show answer & explanation

Answer: 18

Answer

18

To find the number of factors of

5400

that are multiples of

12

, we first divide

5400

by

12

, which gives

450

. The number of possible packet sizes is exactly equal to the number of divisors of

450

. The prime factorization of

450

is

2^1 \times 3^2 \times 5^2

. Using the divisor-counting formula (adding

1

to each exponent and multiplying), we find there are

(1+1) \times (2+1) \times (2+1) = 2 \times 3 \times 3 = 18

valid packet sizes.

Step-by-Step Solution

1

Find the prime factorization of the total file size (

5400

).

5400 = 54 \times 100 = 2^3 \times 3^3 \times 5^2

.

Prime factorization is required to find the total number of divisors.

2

Determine the mathematical condition for the packet size

S

.

S

must be a divisor of

5400

and a multiple of

12

(

2^2 \times 3^1

). This means

S = 12 \times k

.

Translates the real-world constraints into a mathematical relationship.

3

Find the maximum possible value and prime factorization of the multiplier

k

.

k

must be a divisor of

\frac{5400}{12} = 450

. The prime factorization of

450

is

2^1 \times 3^2 \times 5^2

.

The number of valid packet sizes

S

is directly equal to the number of divisors of

k

.

4

Calculate the total number of divisors of

450

.

(1+1) \times (2+1) \times (2+1) = 2 \times 3 \times 3 = 18

.

Applies the standard formula for counting factors (adding 1 to each prime exponent and multiplying them).

Key Concept

Finding the number of factors of a number that are multiples of another given number.

Question 79Question →

During a materials testing process, the density of four composite alloys (A, B, C, and D) is recorded in grams per cubic centimeter ( $\text{g/cm}^3$ ):

- Alloy A: $2.\overline{15}$
- Alloy B: $\frac{15}{7}$
- Alloy C: $2.1\overline{5}$
- Alloy D: $\frac{28}{13}$

Arrange the alloys in ascending order based on their density, from the lowest to the highest.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order is Alloy B, Alloy A, Alloy D, Alloy C.

By converting all values to their decimal or pure fractional equivalents, we can observe that Alloy B (

2.142...

) is the smallest. The remaining values all begin with

2.15

, but expanding the subsequent decimal places reveals that Alloy A is

2.1515...

, Alloy D is

2.1538...

, and Alloy C is

2.1555...

. This yields the exact ascending sequence: Alloy B < Alloy A < Alloy D < Alloy C.

Step-by-Step Solution

1

Separate the whole number from the fractional part for all given values to simplify the comparison.

All values have a whole number part of 2. We only need to compare their fractional parts: A (

0.\overline{15}

), B (

\frac{1}{7}

), C (

0.1\overline{5}

), and D (

\frac{2}{13}

).

Since all numbers start with 2, isolating the fractional or decimal part reduces computational load.

2

Convert all fractional and decimal parts into a common format (decimal approximation) to establish initial bounds.

A =

0.1515...

, B

\approx 0.1428...

, C =

0.1555...

, D

\approx 0.1538...

Decimal expansion allows for rapid estimation and sequencing without finding a large common denominator.

3

Verify the exact order by cross-multiplying the fractional equivalents of the closest values to eliminate any rounding uncertainties.

Fractions to compare: A (

\frac{15}{99} = \frac{5}{33}

), B (

\frac{1}{7}

), C (

\frac{14}{90} = \frac{7}{45}

), D (

\frac{2}{13}

). Comparing B and A:

1 \times 33 = 33

,

7 \times 5 = 35

, so B < A. Comparing A and D:

5 \times 13 = 65

,

33 \times 2 = 66

, so A < D. Comparing D and C:

2 \times 45 = 90

,

13 \times 7 = 91

, so D < C.

Cross-multiplication of fractions provides absolute mathematical proof of the inequalities:

\frac{1}{7} < \frac{5}{33} < \frac{2}{13} < \frac{7}{45}

.

Key Concept

Comparing mixed fractions and recurring decimals via decimal expansion and cross-multiplication

Question 80Question →

A wholesale merchant has three batches of specialty syrups measuring $\frac{63}{5}$ liters, $\frac{72}{25}$ liters, and $\frac{81}{10}$ liters in volume. To distribute them in uniform sample sizes, the merchant wants to repackage all the syrup into identical smaller bottles filled exactly to the brim, ensuring no syrup is left over from any of the original batches. What is the maximum possible capacity (in liters) of each sample bottle?

Show answer & explanation

Answer: 0.18

Answer

The maximum possible capacity of each sample bottle is 0.18 liters.

To find the maximum capacity of uniform bottles that can hold the syrup from any batch without leftover, we must find the Highest Common Factor (HCF) of the respective volumes. For fractions, the rule is

\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}

. The HCF of the numerators (

63, 72, 81

) is

9

, and the LCM of the denominators (

5, 25, 10

) is

50

. The resulting maximum capacity is

\frac{9}{50}

, which equals

0.18

liters.

Step-by-Step Solution

1

Determine the mathematical operation required.

Calculate the HCF of

\frac{63}{5}

,

\frac{72}{25}

, and

\frac{81}{10}

.

Finding the largest equal container size that leaves no remainder requires computing the Highest Common Factor.

2

Apply the rule for finding the HCF of fractions.

\text{HCF of fractions} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}

This standard formula correctly identifies the greatest common divisor for a set of rational numbers.

3

Find the HCF of the numerators.

\text{HCF}(63, 72, 81) = 9

The largest integer that perfectly divides 63, 72, and 81 is 9.

4

Find the LCM of the denominators.

\text{LCM}(5, 25, 10) = 50

The smallest integer that is a multiple of 5, 25, and 10 is 50.

5

Compute the final fraction and convert it to a decimal.

\frac{9}{50} = 0.18

Dividing the HCF of the numerators by the LCM of the denominators yields the exact capacity in liters.

Key Concept

HCF and LCM of fractions

Basic Numeracy

Answer

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