All practice questions

If the six-digit number $45231x$ is completely divisible by $9$ , what is the value of the digit $x$ ?

Show answer & explanation

Answer: 3

Answer

3

According to the divisibility rule for 9, a number is divisible by 9 if the sum of its digits is a multiple of 9. For the number

45231x

, the sum of the digits is

4 + 5 + 2 + 3 + 1 + x = 15 + x

. The smallest multiple of 9 that is greater than or equal to 15 is 18. Setting

15 + x = 18

gives

x = 3

.

Step-by-Step Solution

1

Find the sum of the known digits in the given number.

4 + 5 + 2 + 3 + 1 = 15

The divisibility rule for 9 requires analyzing the sum of all digits.

2

Formulate the condition for divisibility by 9.

15 + x

must be a multiple of 9.

Including the unknown unit digit

x

, the total digit sum is

15 + x

.

3

Solve for the single-digit integer

x

where

0 \le x \le 9

.

x = 3

The smallest multiple of 9 greater than or equal to 15 is 18, giving

15 + x = 18

, so

x = 3

.

Key Concept

Divisibility Rule for 9

Alternative Method

Dividing 452,310 by 9 yields 50,256 with a remainder of 6. To make the number divisible by 9, the remaining amount needed is

9 - 6 = 3

, so the unit digit

x

must be 3.

Estimated Time:45s

Question 62Question →

An inventory analyst is auditing two specific electronic components in a warehouse. He calculates that the product of their exact unit quantities is $32,175$ , and the Highest Common Factor (HCF) of these two quantities is $15$ . If there are more than $100$ units of each component currently in stock, what is the total combined quantity of both components?

Show answer & explanation

Answer: 360

Answer

The correct total combined quantity is 360 units.

By representing the numbers as

15x

and

15y

, we determine that

15x \times 15y = 32,175

, which simplifies to

xy = 143

. Factoring

143

into co-prime pairs gives

(1, 143)

and

(11, 13)

. Multiplying these by the HCF (

15

) gives potential quantities of

(15, 2145)

and

(165, 195)

. Only the pair

(165, 195)

satisfies the condition that both quantities must be greater than

100

. Their sum is

165 + 195 = 360

.

Step-by-Step Solution

1

Express the two unknown quantities mathematically using their HCF.

Let the quantities be

15x

and

15y

, where

x

and

y

are co-prime integers.

Since the HCF of the two numbers is

15

, both numbers must be multiples of

15

. Their remaining factors (

x

and

y

) cannot share any common factors other than

1

.

2

Formulate an equation using the given product of the quantities.

15x \times 15y = 32,175

The problem states the product of the two component quantities is equal to

32,175

.

3

Solve the equation for the product of the co-prime variables

x

and

y

.

225xy = 32,175 \Rightarrow xy = \frac{32,175}{225} = 143

Isolating

xy

simplifies the problem to finding two co-prime factors that multiply to

143

.

4

Identify all co-prime factor pairs of

143

.

The co-prime pairs are

(1, 143)

and

(11, 13)

.

We must list all integer pairs that multiply to

143

and verify they share no common divisors.

5

Calculate the possible original quantities and apply the boundary constraints.

Pair 1 gives

(15, 2145)

. Pair 2 gives

(165, 195)

. Because the problem states both quantities are

>100

, we must choose

(165, 195)

.

The constraint strictly eliminates the first pair, as

15

is not greater than

100

.

6

Sum the valid quantities.

165 + 195 = 360

The question asks for the total combined quantity of both components.

Key Concept

The relationship between the Highest Common Factor (HCF) and the product of two numbers, utilizing co-prime factor pairs.

Question 63Question →

Consider all positive two-digit integers where the sum of their digits is exactly $10$ . How many of these integers are prime numbers?

Show answer & explanation

Answer: 3

Answer

3

By listing all two-digit numbers with a digit sum of 10 (19, 28, 37, 46, 55, 64, 73, 82, 91) and testing them for primality, we find that only 19, 37, and 73 are prime. The number 91 is composite (

7 \times 13

), resulting in exactly 3 prime numbers meeting the condition.

Step-by-Step Solution

1

Identify all two-digit positive integers whose digits sum to 10.

The numbers are 19, 28, 37, 46, 55, 64, 73, 82, and 91.

Systematically listing the pairs of digits (from 1 to 9) that add up to 10 ensures no valid numbers are missed.

2

Evaluate each number to determine if it is prime or composite.

Even numbers (28, 46, 64, 82) and multiples of 5 (55) are composite. The remaining odd numbers to check are 19, 37, 73, and 91.

Applying basic divisibility rules for 2 and 5 quickly eliminates obvious composite numbers.

3

Test the remaining odd numbers for primality.

19, 37, and 73 are prime numbers. 91 is a composite number because

91 = 7 \times 13

.

A prime number has exactly two distinct positive divisors: 1 and itself. 91 is a common trap as its divisibility by 7 is not always immediately obvious.

4

Count the total number of prime numbers identified.

There are exactly 3 prime numbers in the set: 19, 37, and 73.

This directly answers the specific question asked in the stem.

Key Concept

Identification of Prime and Composite Numbers

Question 64Question →

An environmental agency uses three autonomous drones to patrol a protected reserve. The drones fly on continuous looping routes. Drone X completes one full route in $\frac{42}{15}$ hours, Drone Y completes a route in $\frac{35}{20}$ hours, and Drone Z completes a route in $\frac{63}{30}$ hours.

If all three drones depart simultaneously from the base station, after how many hours will they all meet at the base station again for the first time?

Show answer & explanation

Answer: 42

Answer

42

The drones will meet again at a time that is a common multiple of all their individual cycle times. The very first time this happens is represented by the Least Common Multiple (LCM). When calculating the LCM of fractions, it is mathematically required to reduce them to their simplest terms first:

\frac{14}{5}

,

\frac{7}{4}

, and

\frac{21}{10}

. Using the correct formula,

\frac{\text{LCM}(14, 7, 21)}{\text{HCF}(5, 4, 10)}

, we get

\frac{42}{1} = 42

hours.

Step-by-Step Solution

1

Identify the mathematical operation required.

The convergence time is the Least Common Multiple (LCM) of the three cycle times.

The drones will meet again at a time that is a common multiple of all their individual route completion times.

2

Simplify the given fractions to their lowest terms.

\frac{42}{15} \rightarrow \frac{14}{5}

;

\frac{35}{20} \rightarrow \frac{7}{4}

;

\frac{63}{30} \rightarrow \frac{21}{10}

The formula for the LCM of fractions requires all fractions to be in their simplest form to yield the correct result.

3

Calculate the LCM of the simplified numerators.

\text{LCM}(14, 7, 21) = 42

The numerator of the resulting fraction must be divisible by all original numerators.

4

Calculate the HCF of the simplified denominators.

\text{HCF}(5, 4, 10) = 1

The denominator of the resulting fraction must evenly divide all original denominators.

5

Compute the final fraction.

\frac{42}{1} = 42

hours

Applying the formula

\frac{\text{LCM of Numerators}}{\text{HCF of Denominators}}

gives the exact time of the next simultaneous meeting.

Key Concept

Calculating the Least Common Multiple (LCM) of fractions, emphasizing the critical prerequisite of simplifying the fractions first.

Estimated Time:2m 30s

Question 65Question →

A large agricultural cooperative is dividing a massive tract of land for different crops. They allocate $0.\overline{428571}$ of the total land to cultivate sunflowers, $0.1\overline{6}$ of the total land to cultivate maize, and $0.0\overline{5}$ of the total land to cultivate organic vegetables. The remaining land, which measures exactly $110$ hectares, is preserved as a wildlife reserve. What is the total area of the tract of land, in hectares?

Show answer & explanation

Answer: 315

Answer

315

The total area is found by properly converting all recurring decimals into exact fractions (

3/7

,

1/6

, and

1/18

), summing them to find the total allocated land (

41/63

), determining the remaining land fraction (

22/63

), and setting it equal to the given

110

hectares. Solving for the whole yields exactly

315

hectares.

Step-by-Step Solution

1

Convert the pure recurring decimal

0.\overline{428571}

into a simplified fraction.

The fraction is

\frac{3}{7}

.

Recognizing that

\frac{1}{7} = 0.\overline{142857}

, we can multiply by

3

to get

0.\overline{428571}

. Alternatively, using the algebraic method:

\frac{428571}{999999} = \frac{3}{7}

.

2

Convert the mixed recurring decimals

0.1\overline{6}

and

0.0\overline{5}

into fractions.

0.1\overline{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}

and

0.0\overline{5} = \frac{5-0}{90} = \frac{5}{90} = \frac{1}{18}

.

To operate with mixed recurring decimals, subtract the non-repeating part from the entire number, and place it over a denominator consisting of 9s (for repeating digits) followed by 0s (for non-repeating digits after the decimal point).

3

Calculate the total fraction of land allocated to the three crops.

\frac{3}{7} + \frac{1}{6} + \frac{1}{18} = \frac{3}{7} + \frac{3+1}{18} = \frac{3}{7} + \frac{4}{18} = \frac{3}{7} + \frac{2}{9} = \frac{27+14}{63} = \frac{41}{63}

.

Finding a common denominator (

63

) allows us to sum the individual crop fractions to determine the total proportion of cultivated land.

4

Determine the fraction representing the wildlife reserve and calculate the total land area.

Reserve fraction =

1 - \frac{41}{63} = \frac{22}{63}

. Total Area =

110 \times \frac{63}{22} = 315

hectares.

The unallocated fraction represents the reserve area. Setting this fraction of the total area (

T

) equal to

110

hectares (

\frac{22}{63} \times T = 110

) gives the final answer.

Key Concept

Fractions and Decimals

Estimated Time:2m 30s

Question 66Question →

A city's public transport network features three distinct tram lines that operate on continuous circular routes departing from a central station. Tram Line 1 completes its route every $\frac{45}{4}$ minutes. Tram Line 2 completes its route every $\frac{25}{2}$ minutes, and Tram Line 3 takes $\frac{75}{8}$ minutes per loop. If all three trams depart from the central station simultaneously, how many minutes will it take for them to depart together again for the first time?

Show answer & explanation

Answer: 112.5

Answer

It will take 112.5 minutes for all three trams to depart together again.

The correct answer is found by taking the Least Common Multiple of the fractional times. By finding the LCM of the numerators (225) and dividing it by the Highest Common Factor of the denominators (2), we get 225/2, which evaluates to exactly 112.5 minutes.

Step-by-Step Solution

1

Determine the mathematical operation required to find when the events will synchronize.

Identify the need to calculate the Least Common Multiple (LCM) of the fractions

\frac{45}{4}

,

\frac{25}{2}

, and

\frac{75}{8}

.

The trams will meet again at a time that is a common multiple of their individual loop durations. The 'first time' indicates the least common multiple is needed.

2

Apply the rule for calculating the LCM of fractional values.

Use the formula:

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

.

To synchronize fractional frequencies, the numerators must reach a common multiple while strictly maintaining the largest common baseline unit defined by the denominators.

3

Calculate the LCM of the numerators: 45, 25, and 75.

The LCM of 45, 25, and 75 is 225.

Prime factorization:

45 = 3^2 \times 5

;

25 = 5^2

;

75 = 3 \times 5^2

. Taking the highest powers gives

3^2 \times 5^2 = 9 \times 25 = 225

.

4

Calculate the HCF of the denominators: 4, 2, and 8.

The HCF of 4, 2, and 8 is 2.

2 is the largest integer that can divide 4, 2, and 8 without leaving a remainder.

5

Compute the final synchronized time.

Divide the LCM of numerators by the HCF of denominators:

\frac{225}{2} = 112.5

.

Applying the values to the fraction LCM formula yields the exact time in minutes.

Key Concept

Calculating the Least Common Multiple (LCM) for fractions to solve simultaneous event problems.

Question 67Question →

An event organizer is arranging chairs for a large conference. When the chairs are arranged in rows of $18$ , $24$ , or $36$ , there are always exactly $5$ chairs left over. However, when the chairs are arranged in rows of $13$ , all chairs are perfectly accommodated with none left over. What is the minimum possible total number of chairs the organizer has?

Show answer & explanation

Answer: 221

Answer

The minimum possible total number of chairs is 221.

The correct answer is derived by first establishing that any number leaving a remainder of

5

when divided by

18

,

24

, and

36

must be of the form

72k + 5

, where

72

is the LCM of the divisors. By systematically checking values of

k

, we find that

k=3

is the smallest integer that makes the expression

(72k + 5)

perfectly divisible by

13

, resulting in

72(3) + 5 = 221

.

Step-by-Step Solution

1

Set up the conditions for the total number of chairs mathematically.

Let the total number of chairs be

N

. We are given

N \equiv 5 \pmod{18}

,

N \equiv 5 \pmod{24}

,

N \equiv 5 \pmod{36}

, and

N \equiv 0 \pmod{13}

.

Translating the word problem into modular arithmetic helps systematically apply the concepts of LCM and divisibility.

2

Find the Least Common Multiple (LCM) of the first set of divisors.

The divisors are

18

,

24

, and

36

. Their prime factorizations are

18 = 2 \times 3^2

,

24 = 2^3 \times 3

, and

36 = 2^2 \times 3^2

. The LCM is

2^3 \times 3^2 = 8 \times 9 = 72

.

Any number that leaves the same remainder when divided by multiple divisors must be a multiple of their LCM plus that remainder.

3

Express

N

using the LCM and the common remainder.

Since

N

leaves a remainder of

5

when divided by

18

,

24

, or

36

, we can write

N = 72k + 5

, where

k

is a non-negative integer (

k = 0, 1, 2, \dots

).

This general formula captures all possible numbers of chairs that satisfy the first condition.

4

Apply the final divisibility condition to find

k

.

We require

N

to be perfectly divisible by

13

, meaning

72k + 5 \equiv 0 \pmod{13}

.

This guarantees the solution satisfies the second condition where arranging chairs in rows of 13 leaves no remainder.

5

Simplify the congruence modulo 13 and solve for

k

.

Divide

72

by

13

to find the remainder:

72 = 13 \times 5 + 7

. So,

72k \equiv 7k \pmod{13}

. The equation becomes

7k + 5 \equiv 0 \pmod{13}

. Testing values for

k

: if

k=1

,

7(1)+5=12

(not divisible); if

k=2

,

7(2)+5=19

(not divisible); if

k=3

,

7(3)+5=26

(divisible by

13

, since

26 = 13 \times 2

). The smallest valid

k

is

3

.

Finding the smallest non-negative integer

k

ensures we find the minimum possible number of chairs.

6

Calculate the final value of

N

.

N = 72(3) + 5 = 216 + 5 = 221

.

Substituting

k=3

back into our general formula gives the final answer.

Key Concept

Solving simultaneous remainder and divisibility conditions using the Least Common Multiple (LCM).

Question 68Question →

A mechanical clock is synchronized to the correct standard time at exactly 00:00 (midnight) on February 15th of a leap year. This particular clock is known to gain exactly $4$ minutes every $24$ hours of true time.

What is the acute angle (in degrees) between the hour and minute hands of this faulty clock at exactly 12:00 noon (true time) on March 2nd of the same leap year?

Show answer & explanation

Answer: 3

Answer

The acute angle between the hands is 3 degrees.

By determining the exact time elapsed across the leap year boundary (

16.5

days), the total time gained by the clock is

16.5 \times 4 = 66

minutes. Adding this to the true time of 12:00 yields a faulty clock time of 13:06 (or 1:06). At 1:06, the minute hand is at

36^\circ

(

6 \times 6^\circ

) and the hour hand has moved past the 1 o'clock mark (

30^\circ

) by an additional

3^\circ

(

(6/60) \times 30^\circ

). The difference between

36^\circ

and

33^\circ

is

3^\circ

.

Step-by-Step Solution

1

Calculate the total true time elapsed in days.

The total elapsed time is 16.5 days.

From February 15th 00:00 to March 1st 00:00 in a leap year (29 days in February) is exactly 15 days. From March 1st to March 2nd 00:00 is 1 day. From 00:00 to 12:00 noon is 0.5 days. Total: 15 + 1 + 0.5 = 16.5 days.

2

Calculate the total time gained by the faulty clock.

The clock gains a total of 66 minutes.

The clock gains 4 minutes per 24 hours (1 day). For 16.5 days, the total gain is 16.5 * 4 = 66 minutes.

3

Determine the time shown on the faulty clock.

The faulty clock shows 13:06 (or 1:06 PM).

The true time is 12:00 noon. Adding the gained 66 minutes (1 hour and 6 minutes) gives 13:06.

4

Calculate the precise position of the minute and hour hands at 1:06.

Minute hand is at 36 degrees; Hour hand is at 33 degrees.

The minute hand moves 6 degrees per minute: 6 * 6 = 36 degrees from the 12 o'clock position. The hour hand moves 30 degrees per hour plus 0.5 degrees per minute: (1 * 30) + (6 * 0.5) = 30 + 3 = 33 degrees.

5

Find the acute angle between the two hands.

The acute angle is 3 degrees.

The difference between the two positions is |36 - 33| = 3 degrees.

Key Concept

Calculating elapsed time across leap year month boundaries combined with continuous clock drift and geometric clock angle formulas.

Estimated Time:2m 30s

Question 69Question →

A city hall maintains an antique mechanical tower clock that has a constant drift, falling behind standard time by exactly $2.5$ minutes every $24$ hours. The maintenance crew calibrates the clock to the precise standard time on January 14, $1900$ , at exactly 12:00 PM (Noon). Calculate the total amount of time, in minutes, that the clock will have drifted behind standard time by exactly 12:00 PM (Noon) on March 15, $1900$ .

Show answer & explanation

Answer: 150

Answer

The clock will have accumulated a total drift of 150 minutes.

The correct answer is derived by accurately counting the exact number of days between January 14, 1900, and March 15, 1900. Since 1900 is not a leap year (it is a century year not divisible by 400), February has exactly 28 days. Adding the 17 days in January, 28 days in February, and 15 days in March gives a total of 60 days. Multiplying the 60 days by the daily drift rate of 2.5 minutes yields exactly 150 minutes.

Step-by-Step Solution

1

Calculate the days elapsed in January.

31 - 14 = 17

days.

To find the time accumulated during the first month.

2

Determine the number of days in February

1900

.

February has

28

days.

The year

1900

is a century year not divisible by

400

, making it a standard non-leap year.

3

Sum the total days from January 14 to March 15.

17 + 28 + 15 = 60

days.

To find the total

24

-hour periods over which the drift occurred.

4

Calculate the total drift in minutes.

60 \times 2.5 = 150

minutes.

The clock loses exactly

2.5

minutes for every full

24

-hour period elapsed.

Key Concept

Century Leap Year Rule and Time Accumulation

Estimated Time:1m 30s

Question 70Question →

A botanist leaves her base camp to collect soil samples in a dense forest. She first walks $35\text{ m}$ straight East. She then turns $135^{\circ}$ to her left and walks $20\sqrt{2}\text{ m}$ . Finally, she turns $135^{\circ}$ to her left again and walks $12\text{ m}$ . What is the shortest straight-line distance, in meters, between her current position and the base camp?

Show answer & explanation

Answer: 17

Answer

The shortest straight-line distance between the botanist's current position and the base camp is 17 meters.

By breaking each movement into its horizontal and vertical components on a Cartesian plane, the final position is found to be 15 meters East and 8 meters North of the origin. The shortest direct distance is the hypotenuse of the triangle formed by these coordinates: √(15² + 8²) = 17.

Step-by-Step Solution

1

Plot the first movement on a coordinate plane.

Position 1 is at

(35, 0)

.

Setting the starting base camp at the origin

(0, 0)

, moving East means traveling along the positive x-axis.

2

Calculate the vector and position for the second movement.

Position 2 is at

(15, 20)

.

Facing East (0°), a left turn of 135° results in facing North-West. Moving

20\sqrt{2}\text{ m}

diagonally at a 45° angle to the axes translates to

\Delta x = -20\text{ m}

(West) and

\Delta y = +20\text{ m}

(North). New coordinate:

(35 - 20, 0 + 20)

.

3

Calculate the vector and position for the third movement.

Position 3 is at

(15, 8)

.

Facing North-West, another left turn of 135° places her facing directly South. Walking

12\text{ m}

South subtracts 12 from the y-coordinate. New coordinate:

(15, 20 - 12)

.

4

Calculate the shortest straight-line distance to the origin.

The final distance is

17\text{ m}

.

The straight-line distance from

(0, 0)

to

(15, 8)

forms a right-angled triangle. Using Pythagoras:

d = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17

.

Key Concept

Calculating continuous displacement using angular turns, vector breakdown, and the Pythagorean theorem.

Estimated Time:1m 30s

Question 71Question →

When the mathematical expression $44^{63} - 17^2$ is divided by $45$ , what is the final positive remainder?

Show answer & explanation

Answer: 25

Answer

25

Applying modular arithmetic rules, the base

44

is congruent to

-1 \pmod{45}

. Raising

-1

to an odd power (

63

) keeps the value as

-1

. For the second term,

17^2

equals

289

, which leaves a remainder of

19

when divided by

45

. Subtracting the second remainder from the first gives

-1 - 19 = -20

. Because a standard remainder must be positive, adding the divisor (

45

) to

-20

yields the final correct answer of

25

.

Step-by-Step Solution

1

Apply modular arithmetic to the first term,

44^{63}

, relative to the divisor

45

.

44 \equiv -1 \pmod{45}

, so

44^{63} \equiv (-1)^{63} = -1 \pmod{45}

Using negative remainders for bases close to the divisor significantly simplifies large power calculations.

2

Evaluate the second term,

17^2

, and find its remainder when divided by

45

.

17^2 = 289

. Dividing

289

by

45

yields a quotient of

6

(

270

) with a remainder of

19

.

The constant term must be reduced modulo 45 to properly combine it with the first term.

3

Combine the simplified terms according to the original expression structure.

-1 - 19 = -20 \pmod{45}

The remainder of a difference is equivalent to the difference of the individual remainders.

4

Convert the resulting negative remainder into an equivalent positive remainder.

-20 + 45 = 25

Standard remainders must be non-negative. Adding the divisor to a negative remainder provides the mathematically correct positive value.

Key Concept

Modular Arithmetic and Negative Remainders

Question 72Question →

A solid wooden block has dimensions $12 \text{ cm} \times 15 \text{ cm} \times 18 \text{ cm}$ . Three of its mutually adjacent faces (which meet at a single corner) are painted Red, and the remaining three faces are painted Blue. The block is then cut into smaller, identical cubes of size $1 \text{ cm} \times 1 \text{ cm} \times 1 \text{ cm}$ . How many of these smaller cubes have AT LEAST one Red face AND at least one Blue face?

Show answer & explanation

Answer: 84

Answer

84

The unit cubes with both colors lie exactly on the 6 boundary edges separating the Red and Blue faces. By summing the lengths of these 6 edges (

12+12+15+15+18+18 = 90

) and subtracting the 6 overlapping corner cubes to correct for double-counting, we arrive at exactly 84 cubes.

Step-by-Step Solution

1

Identify the configuration of the painted faces on the block.

The 3 Red faces share one corner, and the 3 Blue faces share the diagonally opposite corner.

To accurately locate where the Red and Blue paints meet on the surface.

2

Determine which unit cubes satisfy the condition of having both Red and Blue faces.

Only the cubes lying exactly on the boundary edges between a Red face and a Blue face will have both colors.

Cubes fully within a Red face have no Blue paint, internal cubes have no paint, and purely corner cubes not on the boundary have only one color.

3

Count the number of boundary edges.

There are exactly 6 boundary edges forming a continuous zig-zag hexagonal ring around the block.

Each of the 3 Red faces shares an edge with exactly 2 Blue faces.

4

Determine the lengths of these 6 boundary edges.

The ring wraps around all three dimensions symmetrically, covering two edges of 12 cm, two of 15 cm, and two of 18 cm.

A cuboid has 4 edges of each dimension; the boundary ring traverses exactly half of them.

5

Calculate the total number of unit cubes on these edges using the inclusion-exclusion principle.

2 \times (12 + 15 + 18) - 6 = 90 - 6 = 84

cubes.

Summing the edge lengths gives 90 cubes, but the 6 corners connecting these edges in the ring are counted twice, so we must subtract 6 to prevent double-counting.

Key Concept

Spatial reasoning and painted cube boundary edge calculation.

Question 73Question →

Calculate the true positive remainder obtained upon dividing the numerical expression $67^{95} - 3 \times 53^{42}$ by $17$ .

Show answer & explanation

Answer: 4

Answer

The correct positive remainder is 4.

By evaluating the expression under modulo

17

, we first simplify the base numbers:

67 \equiv -1 \pmod{17}

and

53 \equiv 2 \pmod{17}

. This reduces the overarching expression to

(-1)^{95} - 3 \times 2^{42}

. The first term trivially becomes

-1

. For the second term, we can utilize the fact that

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

. Therefore,

2^{42}

can be broken down into

(2^4)^{10} \times 2^2 \equiv (-1)^{10} \times 4 \equiv 4 \pmod{17}

. Substituting these simplified values back into the expression yields

-1 - 3(4) = -13

. Because standard remainders must be positive, we adjust the negative result by adding the divisor

17

to

-13

, which gives the true positive remainder of

4

.

Step-by-Step Solution

1

Reduce the base numbers

67

and

53

to smaller equivalent values modulo

17

.

67 = 17 \times 4 - 1 \implies 67 \equiv -1 \pmod{17}

. And

53 = 17 \times 3 + 2 \implies 53 \equiv 2 \pmod{17}

.

To drastically simplify large exponentiations by substituting smaller, manageable equivalent bases.

2

Evaluate the remainder of the first term,

67^{95} \pmod{17}

.

(-1)^{95} = -1 \pmod{17}

.

An odd exponent applied to a base of

-1

preserves the negative sign.

3

Simplify the second term's exponentiation,

2^{42} \pmod{17}

, by identifying a nearby power of

2

that relates to

17

.

Observe that

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

.

Finding a power that equals

1

or

-1

modulo

17

creates a highly efficient shortcut for reducing massive exponents.

4

Break down

2^{42}

using the established property of

2^4

.

2^{42} = (2^4)^{10} \times 2^2 \equiv (-1)^{10} \times 4 \equiv 1 \times 4 = 4 \pmod{17}

.

To substitute the

-1

equivalence and systematically compute the modular value of the second term.

5

Combine both simplified terms into the original arithmetic expression.

The expression evaluates to

-1 - 3 \times 4 = -1 - 12 = -13 \pmod{17}

.

To find the overall aggregate modular value of the complete mathematical expression.

6

Convert the negative result into the equivalent true positive remainder.

-13 + 17 = 4

.

By definition, a valid remainder must be a non-negative integer strictly less than the divisor.

Key Concept

Modular Arithmetic, Exponent Rules, and Negative Remainders

Estimated Time:3m 0s

Question 74Question →

Consider the exponential equation:

4^x - 3 \cdot 2^{x+2} + 32 = 0

Determine the sum of all real values of $x$ that satisfy this equation.

Show answer & explanation

Answer: 5

Answer

5

By applying the laws of indices, the original expression transforms into a quadratic equation in terms of

2^x

. Solving

y^2 - 12y + 32 = 0

yields

y=4

and

y=8

, which correspond precisely to

x=2

and

x=3

. Their sum is 5.

Step-by-Step Solution

1

Express all terms with a common base of 2.

4^x

becomes

(2^x)^2

and

2^{x+2}

becomes

4 \cdot 2^x

.

Creating a common base allows the equation to be transformed into a standard polynomial form.

2

Rewrite the equation using the new terms.

(2^x)^2 - 12(2^x) + 32 = 0

Simplifying the coefficients makes it easier to spot the quadratic structure.

3

Perform a substitution to solve the quadratic equation.

Letting

y = 2^x

gives

y^2 - 12y + 32 = 0

. Factoring yields

(y - 4)(y - 8) = 0

, so

y = 4

or

y = 8

.

Substitution converts a complex exponential equation into a simple quadratic one.

4

Solve for the original variable

x

.

2^x = 4 \implies x = 2

, and

2^x = 8 \implies x = 3

.

The question asks for the values of

x

, not the intermediate substitution variable

y

.

5

Calculate the sum of all valid

x

values.

2 + 3 = 5

This addresses the specific final requirement of the question stem.

Key Concept

Solving exponential equations reducible to quadratics using index laws.

Estimated Time:1m 30s

Question 75Question →

A synchronized scheduling system operates on a repeating 19-millisecond cycle. A specific event is triggered at a timestamp $T$ in milliseconds, given by the formula $T = 37^{45} - 17 \times 40^{23}$ . To find the exact position within the current cycle when the event occurs, the system calculates the positive remainder when $T$ is divided by $19$ . At what millisecond mark within the cycle does the event trigger?

Show answer & explanation

Answer: 6

Answer

6

By applying modular arithmetic rules, Fermat's Little Theorem reduces the large exponents. The first term evaluates to

-1 \pmod{19}

and the second term evaluates to

12 \pmod{19}

. Their difference is

-13 \pmod{19}

, which corresponds to a positive remainder of

6

.

Step-by-Step Solution

1

Simplify the first term

37^{45}

modulo 19.

37^{45} \equiv -1 \pmod{19}

Since

37 = 19 \times 2 - 1

, it follows that

37 \equiv -1 \pmod{19}

. An odd power of

-1

is

-1

.

2

Simplify the base of the second term,

40^{23}

, modulo 19.

40^{23} \equiv 2^{23} \pmod{19}

Because

40 = 19 \times 2 + 2

, we can replace the base

40

with its remainder

2

.

3

Use Fermat's Little Theorem to reduce the exponent in

2^{23} \pmod{19}

.

2^{23} \equiv 13 \pmod{19}

Fermat's Little Theorem states

a^{p-1} \equiv 1 \pmod{p}

for a prime

p

. Here,

2^{18} \equiv 1 \pmod{19}

. Therefore,

2^{23} = 2^{18} \times 2^5 \equiv 1 \times 32 \equiv 13 \pmod{19}

.

4

Multiply by 17 and find the remainder of the second term.

17 \times 40^{23} \equiv 12 \pmod{19}

We can write

17 \equiv -2 \pmod{19}

. Then,

(-2) \times 13 = -26

. Adding a multiple of 19 (which is 38) gives

-26 + 38 = 12

.

5

Subtract the second term from the first and convert to a positive remainder.

The final remainder is

6

.

-1 - 12 = -13

. To get the positive remainder, add the modulus 19:

-13 + 19 = 6

.

Key Concept

Applying modular arithmetic rules, properties of negative remainders, and Fermat's Little Theorem to simplify large exponential expressions.

Estimated Time:1m 30s

Question 76Question →

Consider the set of the first 100 positive integers (from 1 to 100 inclusive).

An integer $N$ from this set satisfies all of the following three conditions simultaneously:
1. $N$ is a composite number.
2. $N$ is neither divisible by 2 nor divisible by 3.
3. The square root of $N$ is an irrational number.

What is the total number of possible values for $N$ ?

Show answer & explanation

Answer: 7

Answer

There are exactly 7 values for N that satisfy all three conditions.

The complete set of numbers not divisible by 2 or 3 from 1 to 100 contains 33 integers. Excluding the number 1 (neither prime nor composite) and the 23 prime numbers leaves exactly 9 composite numbers. From these 9 composites, 25 and 49 must be excluded because their square roots are 5 and 7 (rational numbers). This leaves 7 valid integers.

Step-by-Step Solution

1

Determine the total number of integers from 1 to 100 that are neither divisible by 2 nor divisible by 3.

There are 33 such numbers.

Using the inclusion-exclusion principle: there are 50 multiples of 2, 33 multiples of 3, and 16 multiples of 6. Multiples of 2 or 3 = 50 + 33 - 16 = 67. The remaining numbers are 100 - 67 = 33.

2

Filter the 33 remaining integers to find those that are composite.

There are 9 composite numbers: 25, 35, 49, 55, 65, 77, 85, 91, and 95.

Of the 33 numbers, 1 is neither prime nor composite. There are 25 primes up to 100, and excluding 2 and 3 leaves 23 primes. Thus, the composites are 33 - 1 - 23 = 9. These are the products of primes 5 and greater.

3

Eliminate numbers from the composite list whose square roots are rational.

Remove 25 and 49.

Condition 3 requires the square root of N to be irrational. An integer has a rational square root if and only if it is a perfect square. In our list, 25 and 49 are perfect squares.

4

Count the final remaining valid numbers.

7 numbers remain.

Subtracting the 2 perfect squares from the 9 composite numbers leaves 7 numbers that satisfy all three conditions.

Key Concept

Classification of numbers combining prime/composite definitions, divisibility principles, and properties of rational and irrational roots.

Question 77Question →

A logistics manager is packing identical relief kits into crates. When she attempts to pack them in equal batches of $16$ , $24$ , $30$ , or $36$ kits per crate, there are always exactly $8$ kits left over. However, if she packs them in batches of exactly $19$ kits per crate, there are zero kits left over. What is the least possible total number of relief kits she could be packing?

Show answer & explanation

Answer: 2888

Answer

The least possible total number of relief kits is 2888.

The correct answer is derived by finding the general form of a number that leaves a remainder of 8 when divided by 16, 24, 30, and 36. This form is

720k + 8

. By applying the final condition that the total number must be exactly divisible by 19, we find the smallest valid multiplier is

k=4

. Substituting this back gives

720(4) + 8 = 2888

.

Step-by-Step Solution

1

Establish the relationship for the total number of kits based on the remainders.

The number of kits,

N

, leaves a remainder of

8

when divided by

16

,

24

,

30

, and

36

. Thus,

N = \text{LCM}(16, 24, 30, 36) \times k + 8

.

Any number that leaves the same remainder when divided by multiple divisors can be expressed as a multiple of their least common multiple plus that remainder.

2

Calculate the least common multiple (LCM) of

16

,

24

,

30

, and

36

.

The prime factorizations are

16 = 2^4

,

24 = 2^3 \times 3

,

30 = 2 \times 3 \times 5

, and

36 = 2^2 \times 3^2

. The LCM is the product of the highest powers:

2^4 \times 3^2 \times 5 = 720

.

The LCM is required to find the base repeating cycle for the division condition.

3

Formulate the exact divisibility condition.

Substitute the LCM into the equation to get

N = 720k + 8

. The problem states

N

is exactly divisible by

19

, so

(720k + 8) \equiv 0 \pmod{19}

.

This applies the second constraint of the problem to find the specific multiplier

k

.

4

Simplify the modular arithmetic equation to solve for

k

.

Divide

720

by

19

to find the remainder:

720 = 19 \times 37 + 17

. Substitute

17

for

720

to get

(17k + 8) \equiv 0 \pmod{19}

. This can be written as

(-2k + 8) \equiv 0 \pmod{19}

.

Simplifying large numbers using modulo properties makes finding the integer

k

manageable.

5

Find the smallest positive integer

k

that satisfies the equation.

Solving

-2k + 8 = 0

yields

2k = 8

, which means

k = 4

. Checking:

17(4) + 8 = 68 + 8 = 76

, and

76 \div 19 = 4

, which is exactly divisible.

Finding the smallest valid

k

ensures we calculate the least possible total number of kits.

6

Calculate the final total number of kits.

N = 720(4) + 8 = 2880 + 8 = 2888

.

Substituting

k = 4

back into the original formula for

N

provides the final numerical answer.

Key Concept

Finding a specific numerical value based on multiple simultaneous divisibility and remainder conditions using Least Common Multiple (LCM) and modular arithmetic.

Question 78Question →

Identify all two-digit prime numbers where both the tens digit and the units digit are strictly prime numbers. What is the sum of the largest and the smallest numbers that meet this criterion?

Show answer & explanation

Answer: 96

Answer

96

The correct answer requires finding the intersection of two distinct sets: two-digit numbers formed entirely by prime digits, and two-digit numbers that are mathematically prime. This resulting set is {23, 37, 53, 73}. Adding the minimum value (23) and maximum value (73) yields 96.

Step-by-Step Solution

1

Identify valid single digits.

The single-digit primes available for use are 2, 3, 5, and 7.

The problem states that both individual digits of the target number must be prime.

2

Determine valid units digits.

The units digit can only be 3 or 7.

If the units digit is 2, the number is even. If the units digit is 5, the number is a multiple of 5. Both cases result in a composite two-digit number.

3

List all potential combinations.

The possible combinations are 23, 33, 53, 73, 27, 37, 57, and 77.

These are generated by pairing any prime tens digit {2, 3, 5, 7} with the valid prime units digits {3, 7}.

4

Eliminate composite numbers from the list.

The valid primes are 23, 37, 53, and 73. The numbers 27, 33, 57, and 77 are removed.

27, 33, and 57 are divisible by 3 (sum of digits is a multiple of 3). 77 is divisible by 7.

5

Calculate the final sum.

23 + 73 = 96.

The question asks for the sum of the smallest valid number (23) and the largest valid number (73).

Key Concept

Classification of prime digits and prime numbers.

Question 79Question →

Calculate the lowest common multiple (LCM) of the fractions $\frac{3}{4}$ , $\frac{9}{10}$ , and $\frac{15}{16}$ . Express your final answer as a decimal.

Show answer & explanation

Answer: 22.5

Answer

22.5

The lowest common multiple (LCM) of a set of fractions is found by dividing the LCM of their numerators by the highest common factor (HCF) of their denominators. For the numerators

3

,

9

, and

15

, the LCM is

45

. For the denominators

4

,

10

, and

16

, the HCF is

2

. Thus, the LCM of the fractions is

\frac{45}{2}

, which equals

22.5

in decimal form.

Step-by-Step Solution

1

Recall the mathematical formula for finding the LCM of multiple fractions.

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

This formula is the standard method to determine the lowest common multiple when dealing with fractional values.

2

Identify the numerators and calculate their lowest common multiple (LCM).

The numerators are

3

,

9

, and

15

. Their LCM is

45

(since

45

is the smallest number perfectly divisible by

3

,

9

, and

15

).

The numerator of the final fraction requires the LCM of all the given numerators.

3

Identify the denominators and calculate their highest common factor (HCF).

The denominators are

4

,

10

, and

16

. Their HCF is

2

(since

2

is the largest number that divides

4

,

10

, and

16

without a remainder).

The denominator of the final fraction requires the HCF of all the given denominators.

4

Apply the calculated values to the fraction LCM formula.

\text{LCM} = \frac{45}{2}

Combining the results from the previous steps yields the LCM in fractional form.

5

Convert the resulting fraction into a decimal format.

\frac{45}{2} = 22.5

The question explicitly requires the final answer to be expressed as a decimal.

Key Concept

Calculating the LCM of fractions using the specific formula: LCM of numerators divided by the HCF of denominators.

Question 80Question →

If the expression $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$ can be expressed in the form $a - b\sqrt{6}$ where $a$ and $b$ are rational numbers, what is the exact value of $a + b$ ?

Show answer & explanation

Answer: 7

Answer

The correct value is 7.

The correct answer is derived by multiplying the numerator and denominator by the conjugate

3\sqrt{2} - 2\sqrt{3}

. This rationalizes the denominator to

6

. Expanding the numerator gives

30 - 12\sqrt{6}

. Dividing the numerator by

6

yields

5 - 2\sqrt{6}

. Setting this equal to

a - b\sqrt{6}

identifies

a = 5

and

b = 2

, giving a final sum of

7

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator,

3\sqrt{2} - 2\sqrt{3}

.

\frac{(3\sqrt{2} - 2\sqrt{3})^2}{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3})}

This process, known as rationalizing the denominator, removes the surds from the bottom of the fraction.

2

Expand the numerator using the binomial square formula

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

.

(3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 - 12\sqrt{6} + 12 = 30 - 12\sqrt{6}

Expanding the squared binomial simplifies the top part of the fraction.

3

Expand the denominator using the difference of squares formula

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

.

(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6

This guarantees that the denominator becomes a rational number.

4

Divide the terms in the numerator by the denominator.

\frac{30 - 12\sqrt{6}}{6} = 5 - 2\sqrt{6}

Simplifying the fraction allows us to match it to the given form

a - b\sqrt{6}

.

5

Equate the simplified expression to

a - b\sqrt{6}

and solve for

a + b

.

a = 5

,

b = 2

, and

a + b = 7

By direct comparison of rational and irrational parts, we determine the values of

a

and

b

to find their sum.

Key Concept

Rationalizing the denominator using conjugates and expanding binomial expressions involving surds.

Answer

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