Tüm alıştırma soruları

188 soru

Soru 61Soru

If the six-digit number 45231x45231x is completely divisible by 99, what is the value of the digit xx?

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Cevap: 3

Cevap

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 45231x45231x, the sum of the digits is 4+5+2+3+1+x=15+x4 + 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=1815 + x = 18 gives x=3x = 3.

Adım Adım Çözüm

1
Find the sum of the known digits in the given number.
4+5+2+3+1=154 + 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+x15 + x must be a multiple of 9.
Including the unknown unit digit xx, the total digit sum is 15+x15 + x.
3
Solve for the single-digit integer xx where 0x90 \le x \le 9.
x=3x = 3
The smallest multiple of 9 greater than or equal to 15 is 18, giving 15+x=1815 + x = 18, so x=3x = 3.

Anahtar Kavram

Divisibility Rule for 9

Alternatif Yöntem

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 96=39 - 6 = 3, so the unit digit xx must be 3.
Tahmini Süre:45s
Soru 62Soru

An inventory analyst is auditing two specific electronic components in a warehouse. He calculates that the product of their exact unit quantities is 32,17532,175, and the Highest Common Factor (HCF) of these two quantities is 1515. If there are more than 100100 units of each component currently in stock, what is the total combined quantity of both components?

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Cevap: 360

Cevap

The correct total combined quantity is 360 units.
By representing the numbers as 15x15x and 15y15y, we determine that 15x×15y=32,17515x \times 15y = 32,175, which simplifies to xy=143xy = 143. Factoring 143143 into co-prime pairs gives (1,143)(1, 143) and (11,13)(11, 13). Multiplying these by the HCF (1515) gives potential quantities of (15,2145)(15, 2145) and (165,195)(165, 195). Only the pair (165,195)(165, 195) satisfies the condition that both quantities must be greater than 100100. Their sum is 165+195=360165 + 195 = 360.

Adım Adım Çözüm

1
Express the two unknown quantities mathematically using their HCF.
Let the quantities be 15x15x and 15y15y, where xx and yy are co-prime integers.
Since the HCF of the two numbers is 1515, both numbers must be multiples of 1515. Their remaining factors (xx and yy) cannot share any common factors other than 11.
2
Formulate an equation using the given product of the quantities.
15x×15y=32,17515x \times 15y = 32,175
The problem states the product of the two component quantities is equal to 32,17532,175.
3
Solve the equation for the product of the co-prime variables xx and yy.
225xy=32,175xy=32,175225=143225xy = 32,175 \Rightarrow xy = \frac{32,175}{225} = 143
Isolating xyxy simplifies the problem to finding two co-prime factors that multiply to 143143.
4
Identify all co-prime factor pairs of 143143.
The co-prime pairs are (1,143)(1, 143) and (11,13)(11, 13).
We must list all integer pairs that multiply to 143143 and verify they share no common divisors.
5
Calculate the possible original quantities and apply the boundary constraints.
Pair 1 gives (15,2145)(15, 2145). Pair 2 gives (165,195)(165, 195). Because the problem states both quantities are >100>100, we must choose (165,195)(165, 195).
The constraint strictly eliminates the first pair, as 1515 is not greater than 100100.
6
Sum the valid quantities.
165+195=360165 + 195 = 360
The question asks for the total combined quantity of both components.

Anahtar Kavram

The relationship between the Highest Common Factor (HCF) and the product of two numbers, utilizing co-prime factor pairs.
Soru 63Soru

Consider all positive two-digit integers where the sum of their digits is exactly 1010. How many of these integers are prime numbers?

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Cevap: 3

Cevap

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 (7×137 \times 13), resulting in exactly 3 prime numbers meeting the condition.

Adım Adım Çözüm

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=7×1391 = 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.

Anahtar Kavram

Identification of Prime and Composite Numbers
Soru 64Soru

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 4215\frac{42}{15} hours, Drone Y completes a route in 3520\frac{35}{20} hours, and Drone Z completes a route in 6330\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?

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Cevap: 42

Cevap

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: 145\frac{14}{5}, 74\frac{7}{4}, and 2110\frac{21}{10}. Using the correct formula, LCM(14,7,21)HCF(5,4,10)\frac{\text{LCM}(14, 7, 21)}{\text{HCF}(5, 4, 10)}, we get 421=42\frac{42}{1} = 42 hours.

Adım Adım Çözüm

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.
4215145\frac{42}{15} \rightarrow \frac{14}{5}; 352074\frac{35}{20} \rightarrow \frac{7}{4}; 63302110\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.
LCM(14,7,21)=42\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.
HCF(5,4,10)=1\text{HCF}(5, 4, 10) = 1
The denominator of the resulting fraction must evenly divide all original denominators.
5
Compute the final fraction.
421=42\frac{42}{1} = 42 hours
Applying the formula LCM of NumeratorsHCF of Denominators\frac{\text{LCM of Numerators}}{\text{HCF of Denominators}} gives the exact time of the next simultaneous meeting.

Anahtar Kavram

Calculating the Least Common Multiple (LCM) of fractions, emphasizing the critical prerequisite of simplifying the fractions first.
Tahmini Süre:2m 30s
Soru 65Soru

A large agricultural cooperative is dividing a massive tract of land for different crops. They allocate 0.4285710.\overline{428571} of the total land to cultivate sunflowers, 0.160.1\overline{6} of the total land to cultivate maize, and 0.050.0\overline{5} of the total land to cultivate organic vegetables. The remaining land, which measures exactly 110110 hectares, is preserved as a wildlife reserve. What is the total area of the tract of land, in hectares?

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Cevap: 315

Cevap

315
The total area is found by properly converting all recurring decimals into exact fractions (3/73/7, 1/61/6, and 1/181/18), summing them to find the total allocated land (41/6341/63), determining the remaining land fraction (22/6322/63), and setting it equal to the given 110110 hectares. Solving for the whole yields exactly 315315 hectares.

Adım Adım Çözüm

1
Convert the pure recurring decimal 0.4285710.\overline{428571} into a simplified fraction.
The fraction is 37\frac{3}{7}.
Recognizing that 17=0.142857\frac{1}{7} = 0.\overline{142857}, we can multiply by 33 to get 0.4285710.\overline{428571}. Alternatively, using the algebraic method: 428571999999=37\frac{428571}{999999} = \frac{3}{7}.
2
Convert the mixed recurring decimals 0.160.1\overline{6} and 0.050.0\overline{5} into fractions.
0.16=16190=1590=160.1\overline{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6} and 0.05=5090=590=1180.0\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.
37+16+118=37+3+118=37+418=37+29=27+1463=4163\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 (6363) 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 = 14163=22631 - \frac{41}{63} = \frac{22}{63}. Total Area = 110×6322=315110 \times \frac{63}{22} = 315 hectares.
The unallocated fraction represents the reserve area. Setting this fraction of the total area (TT) equal to 110110 hectares (2263×T=110\frac{22}{63} \times T = 110) gives the final answer.

Anahtar Kavram

Fractions and Decimals
Tahmini Süre:2m 30s
Soru 66Soru

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 454\frac{45}{4} minutes. Tram Line 2 completes its route every 252\frac{25}{2} minutes, and Tram Line 3 takes 758\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?

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Cevap: 112.5

Cevap

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.

Adım Adım Çözüm

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 454\frac{45}{4}, 252\frac{25}{2}, and 758\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: LCM of fractions=LCM of numeratorsHCF of denominators\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=32×545 = 3^2 \times 5; 25=5225 = 5^2; 75=3×5275 = 3 \times 5^2. Taking the highest powers gives 32×52=9×25=2253^2 \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: 2252=112.5\frac{225}{2} = 112.5.
Applying the values to the fraction LCM formula yields the exact time in minutes.

Anahtar Kavram

Calculating the Least Common Multiple (LCM) for fractions to solve simultaneous event problems.
Soru 67Soru

An event organizer is arranging chairs for a large conference. When the chairs are arranged in rows of 1818, 2424, or 3636, there are always exactly 55 chairs left over. However, when the chairs are arranged in rows of 1313, all chairs are perfectly accommodated with none left over. What is the minimum possible total number of chairs the organizer has?

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Cevap: 221

Cevap

The minimum possible total number of chairs is 221.
The correct answer is derived by first establishing that any number leaving a remainder of 55 when divided by 1818, 2424, and 3636 must be of the form 72k+572k + 5, where 7272 is the LCM of the divisors. By systematically checking values of kk, we find that k=3k=3 is the smallest integer that makes the expression (72k+5)(72k + 5) perfectly divisible by 1313, resulting in 72(3)+5=22172(3) + 5 = 221.

Adım Adım Çözüm

1
Set up the conditions for the total number of chairs mathematically.
Let the total number of chairs be NN. We are given N5(mod18)N \equiv 5 \pmod{18}, N5(mod24)N \equiv 5 \pmod{24}, N5(mod36)N \equiv 5 \pmod{36}, and N0(mod13)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 1818, 2424, and 3636. Their prime factorizations are 18=2×3218 = 2 \times 3^2, 24=23×324 = 2^3 \times 3, and 36=22×3236 = 2^2 \times 3^2. The LCM is 23×32=8×9=722^3 \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 NN using the LCM and the common remainder.
Since NN leaves a remainder of 55 when divided by 1818, 2424, or 3636, we can write N=72k+5N = 72k + 5, where kk is a non-negative integer (k=0,1,2,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 kk.
We require NN to be perfectly divisible by 1313, meaning 72k+50(mod13)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 kk.
Divide 7272 by 1313 to find the remainder: 72=13×5+772 = 13 \times 5 + 7. So, 72k7k(mod13)72k \equiv 7k \pmod{13}. The equation becomes 7k+50(mod13)7k + 5 \equiv 0 \pmod{13}. Testing values for kk: if k=1k=1, 7(1)+5=127(1)+5=12 (not divisible); if k=2k=2, 7(2)+5=197(2)+5=19 (not divisible); if k=3k=3, 7(3)+5=267(3)+5=26 (divisible by 1313, since 26=13×226 = 13 \times 2). The smallest valid kk is 33.
Finding the smallest non-negative integer kk ensures we find the minimum possible number of chairs.
6
Calculate the final value of NN.
N=72(3)+5=216+5=221N = 72(3) + 5 = 216 + 5 = 221.
Substituting k=3k=3 back into our general formula gives the final answer.

Anahtar Kavram

Solving simultaneous remainder and divisibility conditions using the Least Common Multiple (LCM).
Soru 68Soru

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 44 minutes every 2424 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?

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Cevap: 3

Cevap

The acute angle between the hands is 3 degrees.
By determining the exact time elapsed across the leap year boundary (16.516.5 days), the total time gained by the clock is 16.5×4=6616.5 \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 3636^\circ (6×66 \times 6^\circ) and the hour hand has moved past the 1 o'clock mark (3030^\circ) by an additional 33^\circ ((6/60)×30(6/60) \times 30^\circ). The difference between 3636^\circ and 3333^\circ is 33^\circ.

Adım Adım Çözüm

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.

Anahtar Kavram

Calculating elapsed time across leap year month boundaries combined with continuous clock drift and geometric clock angle formulas.
Tahmini Süre:2m 30s
Soru 69Soru

A city hall maintains an antique mechanical tower clock that has a constant drift, falling behind standard time by exactly 2.52.5 minutes every 2424 hours. The maintenance crew calibrates the clock to the precise standard time on January 14, 19001900, 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, 19001900.

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Cevap: 150

Cevap

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.

Adım Adım Çözüm

1
Calculate the days elapsed in January.
3114=1731 - 14 = 17 days.
To find the time accumulated during the first month.
2
Determine the number of days in February 19001900.
February has 2828 days.
The year 19001900 is a century year not divisible by 400400, making it a standard non-leap year.
3
Sum the total days from January 14 to March 15.
17+28+15=6017 + 28 + 15 = 60 days.
To find the total 2424-hour periods over which the drift occurred.
4
Calculate the total drift in minutes.
60×2.5=15060 \times 2.5 = 150 minutes.
The clock loses exactly 2.52.5 minutes for every full 2424-hour period elapsed.

Anahtar Kavram

Century Leap Year Rule and Time Accumulation
Tahmini Süre:1m 30s
Soru 70Soru

A botanist leaves her base camp to collect soil samples in a dense forest. She first walks 35 m35\text{ m} straight East. She then turns 135135^{\circ} to her left and walks 202 m20\sqrt{2}\text{ m}. Finally, she turns 135135^{\circ} to her left again and walks 12 m12\text{ m}. What is the shortest straight-line distance, in meters, between her current position and the base camp?

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Cevap: 17

Cevap

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.

Adım Adım Çözüm

1
Plot the first movement on a coordinate plane.
Position 1 is at (35,0)(35, 0).
Setting the starting base camp at the origin (0,0)(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)(15, 20).
Facing East (0°), a left turn of 135° results in facing North-West. Moving 202 m20\sqrt{2}\text{ m} diagonally at a 45° angle to the axes translates to Δx=20 m\Delta x = -20\text{ m} (West) and Δy=+20 m\Delta y = +20\text{ m} (North). New coordinate: (3520,0+20)(35 - 20, 0 + 20).
3
Calculate the vector and position for the third movement.
Position 3 is at (15,8)(15, 8).
Facing North-West, another left turn of 135° places her facing directly South. Walking 12 m12\text{ m} South subtracts 12 from the y-coordinate. New coordinate: (15,2012)(15, 20 - 12).
4
Calculate the shortest straight-line distance to the origin.
The final distance is 17 m17\text{ m}.
The straight-line distance from (0,0)(0, 0) to (15,8)(15, 8) forms a right-angled triangle. Using Pythagoras: d=152+82=225+64=289=17d = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17.

Anahtar Kavram

Calculating continuous displacement using angular turns, vector breakdown, and the Pythagorean theorem.
Tahmini Süre:1m 30s
Soru 71Soru

When the mathematical expression 446317244^{63} - 17^2 is divided by 4545, what is the final positive remainder?

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Cevap: 25

Cevap

25
Applying modular arithmetic rules, the base 4444 is congruent to 1(mod45)-1 \pmod{45}. Raising 1-1 to an odd power (6363) keeps the value as 1-1. For the second term, 17217^2 equals 289289, which leaves a remainder of 1919 when divided by 4545. Subtracting the second remainder from the first gives 119=20-1 - 19 = -20. Because a standard remainder must be positive, adding the divisor (4545) to 20-20 yields the final correct answer of 2525.

Adım Adım Çözüm

1
Apply modular arithmetic to the first term, 446344^{63}, relative to the divisor 4545.
441(mod45)44 \equiv -1 \pmod{45}, so 4463(1)63=1(mod45)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, 17217^2, and find its remainder when divided by 4545.
172=28917^2 = 289. Dividing 289289 by 4545 yields a quotient of 66 (270270) with a remainder of 1919.
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.
119=20(mod45)-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=25-20 + 45 = 25
Standard remainders must be non-negative. Adding the divisor to a negative remainder provides the mathematically correct positive value.

Anahtar Kavram

Modular Arithmetic and Negative Remainders
Soru 72Soru

A solid wooden block has dimensions 12 cm×15 cm×18 cm12 \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 cm×1 cm×1 cm1 \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?

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Cevap: 84

Cevap

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=9012+12+15+15+18+18 = 90) and subtracting the 6 overlapping corner cubes to correct for double-counting, we arrive at exactly 84 cubes.

Adım Adım Çözüm

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×(12+15+18)6=906=842 \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.

Anahtar Kavram

Spatial reasoning and painted cube boundary edge calculation.
Soru 73Soru

Calculate the true positive remainder obtained upon dividing the numerical expression 67953×534267^{95} - 3 \times 53^{42} by 1717.

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Cevap: 4

Cevap

The correct positive remainder is 4.
By evaluating the expression under modulo 1717, we first simplify the base numbers: 671(mod17)67 \equiv -1 \pmod{17} and 532(mod17)53 \equiv 2 \pmod{17}. This reduces the overarching expression to (1)953×242(-1)^{95} - 3 \times 2^{42}. The first term trivially becomes 1-1. For the second term, we can utilize the fact that 24=161(mod17)2^4 = 16 \equiv -1 \pmod{17}. Therefore, 2422^{42} can be broken down into (24)10×22(1)10×44(mod17)(2^4)^{10} \times 2^2 \equiv (-1)^{10} \times 4 \equiv 4 \pmod{17}. Substituting these simplified values back into the expression yields 13(4)=13-1 - 3(4) = -13. Because standard remainders must be positive, we adjust the negative result by adding the divisor 1717 to 13-13, which gives the true positive remainder of 44.

Adım Adım Çözüm

1
Reduce the base numbers 6767 and 5353 to smaller equivalent values modulo 1717.
67=17×41    671(mod17)67 = 17 \times 4 - 1 \implies 67 \equiv -1 \pmod{17}. And 53=17×3+2    532(mod17)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, 6795(mod17)67^{95} \pmod{17}.
(1)95=1(mod17)(-1)^{95} = -1 \pmod{17}.
An odd exponent applied to a base of 1-1 preserves the negative sign.
3
Simplify the second term's exponentiation, 242(mod17)2^{42} \pmod{17}, by identifying a nearby power of 22 that relates to 1717.
Observe that 24=161(mod17)2^4 = 16 \equiv -1 \pmod{17}.
Finding a power that equals 11 or 1-1 modulo 1717 creates a highly efficient shortcut for reducing massive exponents.
4
Break down 2422^{42} using the established property of 242^4.
242=(24)10×22(1)10×41×4=4(mod17)2^{42} = (2^4)^{10} \times 2^2 \equiv (-1)^{10} \times 4 \equiv 1 \times 4 = 4 \pmod{17}.
To substitute the 1-1 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 13×4=112=13(mod17)-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=4-13 + 17 = 4.
By definition, a valid remainder must be a non-negative integer strictly less than the divisor.

Anahtar Kavram

Modular Arithmetic, Exponent Rules, and Negative Remainders
Tahmini Süre:3m 0s
Soru 74Soru

Consider the exponential equation:

4x32x+2+32=04^x - 3 \cdot 2^{x+2} + 32 = 0

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

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Cevap: 5

Cevap

5
By applying the laws of indices, the original expression transforms into a quadratic equation in terms of 2x2^x. Solving y212y+32=0y^2 - 12y + 32 = 0 yields y=4y=4 and y=8y=8, which correspond precisely to x=2x=2 and x=3x=3. Their sum is 5.

Adım Adım Çözüm

1
Express all terms with a common base of 2.
4x4^x becomes (2x)2(2^x)^2 and 2x+22^{x+2} becomes 42x4 \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.
(2x)212(2x)+32=0(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=2xy = 2^x gives y212y+32=0y^2 - 12y + 32 = 0. Factoring yields (y4)(y8)=0(y - 4)(y - 8) = 0, so y=4y = 4 or y=8y = 8.
Substitution converts a complex exponential equation into a simple quadratic one.
4
Solve for the original variable xx.
2x=4    x=22^x = 4 \implies x = 2, and 2x=8    x=32^x = 8 \implies x = 3.
The question asks for the values of xx, not the intermediate substitution variable yy.
5
Calculate the sum of all valid xx values.
2+3=52 + 3 = 5
This addresses the specific final requirement of the question stem.

Anahtar Kavram

Solving exponential equations reducible to quadratics using index laws.
Tahmini Süre:1m 30s
Soru 75Soru

A synchronized scheduling system operates on a repeating 19-millisecond cycle. A specific event is triggered at a timestamp TT in milliseconds, given by the formula T=374517×4023T = 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 TT is divided by 1919. At what millisecond mark within the cycle does the event trigger?

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Cevap: 6

Cevap

6
By applying modular arithmetic rules, Fermat's Little Theorem reduces the large exponents. The first term evaluates to 1(mod19)-1 \pmod{19} and the second term evaluates to 12(mod19)12 \pmod{19}. Their difference is 13(mod19)-13 \pmod{19}, which corresponds to a positive remainder of 66.

Adım Adım Çözüm

1
Simplify the first term 374537^{45} modulo 19.
37451(mod19)37^{45} \equiv -1 \pmod{19}
Since 37=19×2137 = 19 \times 2 - 1, it follows that 371(mod19)37 \equiv -1 \pmod{19}. An odd power of 1-1 is 1-1.
2
Simplify the base of the second term, 402340^{23}, modulo 19.
4023223(mod19)40^{23} \equiv 2^{23} \pmod{19}
Because 40=19×2+240 = 19 \times 2 + 2, we can replace the base 4040 with its remainder 22.
3
Use Fermat's Little Theorem to reduce the exponent in 223(mod19)2^{23} \pmod{19}.
22313(mod19)2^{23} \equiv 13 \pmod{19}
Fermat's Little Theorem states ap11(modp)a^{p-1} \equiv 1 \pmod{p} for a prime pp. Here, 2181(mod19)2^{18} \equiv 1 \pmod{19}. Therefore, 223=218×251×3213(mod19)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.
17×402312(mod19)17 \times 40^{23} \equiv 12 \pmod{19}
We can write 172(mod19)17 \equiv -2 \pmod{19}. Then, (2)×13=26(-2) \times 13 = -26. Adding a multiple of 19 (which is 38) gives 26+38=12-26 + 38 = 12.
5
Subtract the second term from the first and convert to a positive remainder.
The final remainder is 66.
112=13-1 - 12 = -13. To get the positive remainder, add the modulus 19: 13+19=6-13 + 19 = 6.

Anahtar Kavram

Applying modular arithmetic rules, properties of negative remainders, and Fermat's Little Theorem to simplify large exponential expressions.
Tahmini Süre:1m 30s
Soru 76Soru

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

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

What is the total number of possible values for NN?

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Cevap: 7

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Classification of numbers combining prime/composite definitions, divisibility principles, and properties of rational and irrational roots.
Soru 77Soru

A logistics manager is packing identical relief kits into crates. When she attempts to pack them in equal batches of 1616, 2424, 3030, or 3636 kits per crate, there are always exactly 88 kits left over. However, if she packs them in batches of exactly 1919 kits per crate, there are zero kits left over. What is the least possible total number of relief kits she could be packing?

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Cevap: 2888

Cevap

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+8720k + 8. By applying the final condition that the total number must be exactly divisible by 19, we find the smallest valid multiplier is k=4k=4. Substituting this back gives 720(4)+8=2888720(4) + 8 = 2888.

Adım Adım Çözüm

1
Establish the relationship for the total number of kits based on the remainders.
The number of kits, NN, leaves a remainder of 88 when divided by 1616, 2424, 3030, and 3636. Thus, N=LCM(16,24,30,36)×k+8N = \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 1616, 2424, 3030, and 3636.
The prime factorizations are 16=2416 = 2^4, 24=23×324 = 2^3 \times 3, 30=2×3×530 = 2 \times 3 \times 5, and 36=22×3236 = 2^2 \times 3^2. The LCM is the product of the highest powers: 24×32×5=7202^4 \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+8N = 720k + 8. The problem states NN is exactly divisible by 1919, so (720k+8)0(mod19)(720k + 8) \equiv 0 \pmod{19}.
This applies the second constraint of the problem to find the specific multiplier kk.
4
Simplify the modular arithmetic equation to solve for kk.
Divide 720720 by 1919 to find the remainder: 720=19×37+17720 = 19 \times 37 + 17. Substitute 1717 for 720720 to get (17k+8)0(mod19)(17k + 8) \equiv 0 \pmod{19}. This can be written as (2k+8)0(mod19)(-2k + 8) \equiv 0 \pmod{19}.
Simplifying large numbers using modulo properties makes finding the integer kk manageable.
5
Find the smallest positive integer kk that satisfies the equation.
Solving 2k+8=0-2k + 8 = 0 yields 2k=82k = 8, which means k=4k = 4. Checking: 17(4)+8=68+8=7617(4) + 8 = 68 + 8 = 76, and 76÷19=476 \div 19 = 4, which is exactly divisible.
Finding the smallest valid kk ensures we calculate the least possible total number of kits.
6
Calculate the final total number of kits.
N=720(4)+8=2880+8=2888N = 720(4) + 8 = 2880 + 8 = 2888.
Substituting k=4k = 4 back into the original formula for NN provides the final numerical answer.

Anahtar Kavram

Finding a specific numerical value based on multiple simultaneous divisibility and remainder conditions using Least Common Multiple (LCM) and modular arithmetic.
Soru 78Soru

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?

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Cevap: 96

Cevap

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.

Adım Adım Çözüm

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).

Anahtar Kavram

Classification of prime digits and prime numbers.
Soru 79Soru

Calculate the lowest common multiple (LCM) of the fractions 34\frac{3}{4}, 910\frac{9}{10}, and 1516\frac{15}{16}. Express your final answer as a decimal.

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Cevap: 22.5

Cevap

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 33, 99, and 1515, the LCM is 4545. For the denominators 44, 1010, and 1616, the HCF is 22. Thus, the LCM of the fractions is 452\frac{45}{2}, which equals 22.522.5 in decimal form.

Adım Adım Çözüm

1
Recall the mathematical formula for finding the LCM of multiple fractions.
LCM of fractions=LCM of numeratorsHCF of denominators\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 33, 99, and 1515. Their LCM is 4545 (since 4545 is the smallest number perfectly divisible by 33, 99, and 1515).
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 44, 1010, and 1616. Their HCF is 22 (since 22 is the largest number that divides 44, 1010, and 1616 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.
LCM=452\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.
452=22.5\frac{45}{2} = 22.5
The question explicitly requires the final answer to be expressed as a decimal.

Anahtar Kavram

Calculating the LCM of fractions using the specific formula: LCM of numerators divided by the HCF of denominators.
Soru 80Soru

If the expression 322332+23\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} can be expressed in the form ab6a - b\sqrt{6} where aa and bb are rational numbers, what is the exact value of a+ba + b?

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Cevap: 7

Cevap

The correct value is 7.
The correct answer is derived by multiplying the numerator and denominator by the conjugate 32233\sqrt{2} - 2\sqrt{3}. This rationalizes the denominator to 66. Expanding the numerator gives 3012630 - 12\sqrt{6}. Dividing the numerator by 66 yields 5265 - 2\sqrt{6}. Setting this equal to ab6a - b\sqrt{6} identifies a=5a = 5 and b=2b = 2, giving a final sum of 77.

Adım Adım Çözüm

1
Multiply the numerator and denominator by the conjugate of the denominator, 32233\sqrt{2} - 2\sqrt{3}.
(3223)2(32+23)(3223)\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 (xy)2=x22xy+y2(x-y)^2 = x^2 - 2xy + y^2.
(32)22(32)(23)+(23)2=18126+12=30126(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)(xy)=x2y2(x+y)(x-y) = x^2 - y^2.
(32)2(23)2=1812=6(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.
301266=526\frac{30 - 12\sqrt{6}}{6} = 5 - 2\sqrt{6}
Simplifying the fraction allows us to match it to the given form ab6a - b\sqrt{6}.
5
Equate the simplified expression to ab6a - b\sqrt{6} and solve for a+ba + b.
a=5a = 5, b=2b = 2, and a+b=7a + b = 7
By direct comparison of rational and irrational parts, we determine the values of aa and bb to find their sum.

Anahtar Kavram

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