Basic Numeracy

295 soru

Soru 61Soru

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×S)(P + Q) - (R \times S), where the variables are defined based on the mathematical classification of numbers:

* PP: The sum of all single-digit prime numbers.
* QQ: The product of the smallest positive composite number and the smallest odd prime number.
* RR: The total count of whole numbers strictly less than 1010 that are classified as neither prime nor composite.
* SS: The smallest positive integer nn for which the expression 12n\sqrt{12n} results in a rational number.

What is the final numerical access code?

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Cevap: 23

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Classification properties of primes, composites, whole numbers, and rational numbers.
Soru 62Soru

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

X=36+144÷6×4X = 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 XX. What number must the user input?

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Cevap: 12

Cevap

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÷6=24144 \div 6 = 24, and 24×4=9624 \times 4 = 96. Adding 36 yields X=132X = 132. The prime factorization of 132 is 22×31×1112^2 \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)=3×2×2=12(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12.

Adım Adım Çözüm

1
Evaluate the division and multiplication parts of the expression from left to right according to the BODMAS rule.
144÷6=24144 \div 6 = 24, followed by 24×4=9624 \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 XX.
X=36+96=132X = 36 + 96 = 132.
Addition is performed after higher-precedence operations are resolved.
3
Determine the prime factorization of the computed value 132.
132=4×33=22×31×111132 = 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)×(1+1)×(1+1)=3×2×2=12(2 + 1) \times (1 + 1) \times (1 + 1) = 3 \times 2 \times 2 = 12.
The formula (a+1)(b+1)...(a+1)(b+1)... systematically counts every possible product combination of the prime factors.

Anahtar Kavram

Factors, Multiples, and Prime Factorization
Tahmini Süre:1m 30s
Soru 63Soru

An integer PP is given by the mathematical expression P=436203365820332892033P = 4 \cdot 36^{2033} - 6 \cdot 58^{2033} - 2 \cdot 89^{2033}. Determine the positive remainder when PP is divided by 1717.

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Cevap: 9

Cevap

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 (a161(mod17)a^{16} \equiv 1 \pmod{17}), the exponent 2033 is reduced to 2033(mod16)=12033 \pmod{16} = 1. Substituting these simplified values back into the expression yields 4(2)6(7)2(4)=424(2) - 6(7) - 2(4) = -42. Because a remainder must be positive, adding the next highest multiple of 17 (which is 17×3=5117 \times 3 = 51) to -42 gives the final valid remainder of 9.

Adım Adım Çözüm

1
Find the remainder of each base when divided by 17.
362(mod17)36 \equiv 2 \pmod{17}, 587(mod17)58 \equiv 7 \pmod{17}, and 894(mod17)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, a161(mod17)a^{16} \equiv 1 \pmod{17}. The exponent is 2033=16×127+12033 = 16 \times 127 + 1, meaning a2033a1(mod17)a^{2033} \equiv a^1 \pmod{17}.
Fermat's Little Theorem allows us to reduce massive exponents by dividing them by (p1)(p-1) and keeping only the remainder.
3
Substitute the reduced bases and exponents into the original expression.
P4(21)6(71)2(41)(mod17)P \equiv 4(2^1) - 6(7^1) - 2(4^1) \pmod{17}, which evaluates to 8428=428 - 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.
4242+51=9(mod17)-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.

Anahtar Kavram

Divisibility Rules and Remainder Theorem
Soru 64Soru

In a centralized database architecture, data records are distributed across 1717 server nodes numbered from 00 to 1616. A record with a numerical key KK is assigned to a node using the hash function H(K)=K(mod17)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=220075K = 2^{200} - 75, which server node will process this batch?

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Cevap: 11

Cevap

The batch of records will be processed by server node 11.
By applying modular arithmetic rules, the expression (220075)(mod17)(2^{200} - 75) \pmod{17} simplifies individually to (17)=6(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.

Adım Adım Çözüm

1
Identify the mathematical goal of the problem.
Calculate (220075)(mod17)(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 22002^{200} divided by 17.
24=161(mod17)2^4 = 16 \equiv -1 \pmod{17}. Raising this to the 50th power gives (24)50(1)50=1(mod17)(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=17×4+775 = 17 \times 4 + 7, so 757(mod17)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.
17=6(mod17)1 - 7 = -6 \pmod{17}.
Substitute the individual reduced remainders back into the original expression K=220075K = 2^{200} - 75.
5
Convert the negative remainder into a valid positive node number.
6+17=11-6 + 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.

Anahtar Kavram

Modular arithmetic with large exponents and handling negative remainders appropriately.
Tahmini Süre:1m 30s
Soru 65Soru

Three automated data backup protocols in a cloud server execute network synchronization pulses at precise intervals. Protocol Alpha pulses every 185\frac{18}{5} seconds, Protocol Beta every 2425\frac{24}{25} seconds, and Protocol Gamma every 3635\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 1212 minutes?

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Cevap: 5050

Cevap

The data protocols will pulse together 5050 times in the 1212-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 (18,24,3618, 24, 36) by the HCF of their denominators (5,25,355, 25, 35). The LCM of 18,24,18, 24, and 3636 is 7272. The HCF of 5,25,5, 25, and 3535 is 55. This gives a synchronized interval of 72/5=14.472/5 = 14.4 seconds. In 1212 minutes (720720 seconds), the number of synchronized intervals is 720÷14.4=50720 \div 14.4 = 50. Since the initial pulse is excluded as per the instructions, the correct answer remains 5050.

Adım Adım Çözüm

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: 185\frac{18}{5}, 2425\frac{24}{25}, and 3635\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.
LCM of fractions=LCM of numeratorsHCF of denominators\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 (18,24,3618, 24, 36) and the HCF of the denominators (5,25,355, 25, 35).
LCM(18,24,36)=72\text{LCM}(18, 24, 36) = 72 and HCF(5,25,35)=5\text{HCF}(5, 25, 35) = 5. Thus, the synchronized interval is 725\frac{72}{5} seconds, or 14.414.4 seconds.
Prime factorization gives 18=2×3218 = 2 \times 3^2, 24=23×324 = 2^3 \times 3, 36=22×3236 = 2^2 \times 3^2 (LCM = 23×32=722^3 \times 3^2 = 72). For denominators, 55 is the largest common divisor.
4
Convert the given total time window into seconds.
12 minutes=12×60=720 seconds12 \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÷14.4=50720 \div 14.4 = 50. Since the initial pulse is excluded, the final count remains 5050.
Dividing the total time window by the synchronized interval yields the number of subsequent events.

Anahtar Kavram

Least Common Multiple (LCM) of fractions and its application to periodic simultaneous events.
Soru 66Soru

Consider the positive integer N=3600N = 3600. What is the total number of positive divisors of NN that are multiples of either 44 or 99?

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Cevap: 33

Cevap

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 (24×32×522^4 \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+159=3327 + 15 - 9 = 33.

Adım Adım Çözüm

1
Prime factorize 3600 to identify the structure of its divisors.
3600=36×100=24×32×523600 = 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 222^2. The valid powers for 2 are {22,23,24}\{2^2, 2^3, 2^4\} (3 choices). For 3, {30,31,32}\{3^0, 3^1, 3^2\} (3 choices). For 5, {50,51,52}\{5^0, 5^1, 5^2\} (3 choices). Total = 3×3×3=273 \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 323^2. The valid powers for 2 are 5 choices (from 0 to 4). For 3, {32}\{3^2\} (1 choice). For 5, 3 choices. Total = 5×1×3=155 \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 222^2 and 323^2. Choices: 3 (for 2) ×\times 1 (for 3) ×\times 3 (for 5) = 99 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+159=3327 + 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.

Anahtar Kavram

Applying set theory (inclusion-exclusion principle) to the combinations of prime factors of a number to find constrained subsets of divisors.
Soru 67Soru

Calculate the remainder when the integer value of 11×177311 \times 17^{73} is divided by 1919.

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Cevap: 16

Cevap

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.

Adım Adım Çözüm

1
Simplify the base of the exponent modulo 19.
172(mod19)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.
a181(mod19)a^{18} \equiv 1 \pmod{19} for any integer aa 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 7373 using the identified cyclicity.
73=18×4+173 = 18 \times 4 + 1, which means (2)73(2)1(mod19)(-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×(2)=2211 \times (-2) = -22
The original mathematical expression contains a coefficient of 11.
5
Convert the negative intermediate result to a valid positive remainder.
2222+2×19=22+38=16(mod19)-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.

Anahtar Kavram

Modular Arithmetic and Fermat's Little Theorem
Soru 68Soru

A logistics manager is arranging shipping containers into equal stacks. When the containers are stacked in groups of 1818, 2424, or 3232, there are always exactly 1111 containers left over. However, when they are stacked in groups of 3535, there are no containers left over. What is the minimum total number of containers the manager could have?

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Cevap: 875

Cevap

875
The correct answer is derived by recognizing that the total number of containers must take the form of LCM(18,24,32)×k+11LCM(18, 24, 32) \times k + 11. The LCM is 288288, so the number is 288k+11288k + 11. We then find the smallest integer kk such that (288k+11)(288k + 11) is perfectly divisible by 3535. By evaluating the expression for sequential values of kk (or using modular arithmetic, 8k24(mod35)8k \equiv 24 \pmod{35}), we find that k=3k = 3 is the smallest valid multiplier. Substituting this back yields 288(3)+11=875288(3) + 11 = 875.

Adım Adım Çözüm

1
Calculate the Least Common Multiple (LCM) of the initial stack group sizes (1818, 2424, and 3232).
The prime factorizations are 18=2×3218 = 2 \times 3^2, 24=23×324 = 2^3 \times 3, and 32=2532 = 2^5. The LCM is 25×32=32×9=2882^5 \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 NN. We can write N=288k+11N = 288k + 11, where kk is a positive integer.
Since there is always a remainder of 1111 containers when divided by these numbers, the total must be 1111 more than a multiple of their LCM.
3
Apply the secondary divisibility condition.
The problem states that NN is exactly divisible by 3535. Therefore, 288k+110(mod35)288k + 11 \equiv 0 \pmod{35}.
When the containers are grouped by 3535, there is no remainder.
4
Solve the congruence to find the smallest integer kk.
First, simplify 288(mod35)288 \pmod{35}: since 35×8=28035 \times 8 = 280, we have 2888(mod35)288 \equiv 8 \pmod{35}. The equation becomes 8k+110(mod35)8k + 11 \equiv 0 \pmod{35}, or 8k1124(mod35)8k \equiv -11 \equiv 24 \pmod{35}. Dividing both sides by 88 gives k=3k = 3.
Finding the smallest valid positive integer kk will yield the minimum total number of containers.
5
Calculate the final number of containers using k=3k = 3.
N=288(3)+11=864+11=875N = 288(3) + 11 = 864 + 11 = 875.
Substitute the multiplier back into the original algebraic expression.

Anahtar Kavram

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

Alternatif Yöntem

Instead of using modular arithmetic to solve 288k+110(mod35)288k + 11 \equiv 0 \pmod{35}, you can manually test integer values for kk in the formula N=288k+11N = 288k + 11. For k=1k=1, N=288(1)+11=299N = 288(1) + 11 = 299 (299÷35299 \div 35 leaves remainder 1919). For k=2k=2, N=288(2)+11=587N = 288(2) + 11 = 587 (587÷35587 \div 35 leaves remainder 2727). For k=3k=3, N=288(3)+11=875N = 288(3) + 11 = 875 (875÷35=25875 \div 35 = 25 with no remainder). Therefore, 875875 is the smallest valid number.
Tahmini Süre:2m 30s
Soru 69Soru

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 45\frac{4}{5}, 815\frac{8}{15}, and 1225\frac{12}{25} of a second, respectively. Let LL be the least common multiple (LCM) of these three time intervals, representing the exact time when all three dispensers fire simultaneously. Let HH 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 LH\frac{L}{H}?

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Cevap: 9090

Cevap

The value of the ratio L/HL/H is 90.
To solve the problem, we first find the LCM and HCF of the fractions 45\frac{4}{5}, 815\frac{8}{15}, and 1225\frac{12}{25}. The formula for the LCM of fractions is LCM of numeratorsHCF of denominators\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=245L = \frac{24}{5}. The formula for the HCF of fractions is HCF of numeratorsLCM of denominators\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=475H = \frac{4}{75}. Finally, dividing LL by HH gives 245÷475\frac{24}{5} \div \frac{4}{75}, which equals 245×754=6×15=90\frac{24}{5} \times \frac{75}{4} = 6 \times 15 = 90.

Adım Adım Çözüm

1
Calculate LL, the LCM of the three fractions.
L=245L = \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 HH, the HCF of the three fractions.
H=475H = \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 LH\frac{L}{H}.
24/54/75=245×754=6×15=90\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.

Anahtar Kavram

Calculating the LCM and HCF of fractional values
Soru 70Soru

An industrial cutting machine is programmed to divide a rectangular metallic sheet measuring 245\frac{24}{5} meters in length and 163\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?

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Cevap: 64225 m2\frac{64}{225} \text{ m}^2

Cevap

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 245\frac{24}{5} and 163\frac{16}{3} is HCF(24,16)LCM(5,3)=815\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 (815)2=64225 m2(\frac{8}{15})^2 = \frac{64}{225} \text{ m}^2.

Adım Adım Çözüm

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.

Anahtar Kavram

HCF of Fractions and Geometric Application
Soru 71Soru

Four navigation buoys in a harbor flash at regular intervals of 3133 \frac{1}{3} minutes, 4164 \frac{1}{6} minutes, 5595 \frac{5}{9} minutes, and 7127 \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 1212 hours?

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Cevap: 4

Cevap

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.

Adım Adım Çözüm

1
Convert the mixed numbers into improper fractions.
The intervals are 103\frac{10}{3}, 256\frac{25}{6}, 509\frac{50}{9}, and 152\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: LCM of fractions=LCM of numeratorsHCF of denominators\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 10,25,5010, 25, 50, and 1515. Their LCM is 150150.
150150 is the smallest integer divisible by all four numerators (150=10×15=25×6=50×3=15×10150 = 10 \times 15 = 25 \times 6 = 50 \times 3 = 15 \times 10).
4
Calculate the HCF of the denominators.
The denominators are 3,6,93, 6, 9, and 22. Their HCF is 11.
The numbers 22 and 33 are coprime (they share no common factors greater than 11), forcing the overall HCF of the set to be 11.
5
Calculate the exact time interval between simultaneous flashes.
The LCM of the intervals is 1501=150\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 hours=720 minutes12 \text{ hours} = 720 \text{ minutes}. 720150=4.8=4\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.

Anahtar Kavram

LCM of fractions in simultaneous cyclic events
Soru 72Soru

Consider the composite number X=p3q2X = p^3 \cdot q^2, where pp and qq are distinct prime numbers. Which of the following statements regarding the factors and properties of XX are correct?

Geçerli olan tümünü seçin

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Cevap: The total number of positive factors of XX is exactly 1212.; If p=2p=2 and q=3q=3, the sum of all positive factors of XX is 195195.

Cevap

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 paqbp^a \cdot q^b always has (a+1)(b+1)(a+1)(b+1) total factors, which confirms the count of 1212. Furthermore, the geometric series expansion for the sum of factors correctly evaluates to 195195 when substituting p=2p=2 and q=3q=3.

Adım Adım Çözüm

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.

Anahtar Kavram

Prime factorization properties, including total factor counting, sum of factors, and applying set theory (inclusion-exclusion) to factor subsets.
Tahmini Süre:1m 30s
Soru 73Soru

Evaluate the remainder when the numeric expression 47101+1111012547^{101} + 111^{101} - 25 is divided by 1616.

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Cevap: 5

Cevap

5
By applying the properties of modular arithmetic, we determine that 471(mod16)47 \equiv -1 \pmod{16} and 1111(mod16)111 \equiv -1 \pmod{16}. Raising these to the 101st power yields 1-1 and 1-1. Subtracting 25 gives a total of 27-27. 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.

Adım Adım Çözüm

1
Find the remainder of the bases when divided by 16.
471(mod16)47 \equiv -1 \pmod{16} and 1111(mod16)111 \equiv -1 \pmod{16}.
Using negative remainders (since 47=16×3147 = 16 \times 3 - 1 and 111=16×71111 = 16 \times 7 - 1) significantly simplifies the calculation of large powers.
2
Substitute the simplified bases into the original expression.
(47)101+(111)10125(1)101+(1)10125(mod16)(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(-1) + (-1) - 25 = -27.
An odd power of a negative number remains negative, so (1)101=1(-1)^{101} = -1.
4
Find the equivalent positive remainder for -27 modulo 16.
275(mod16)-27 \equiv 5 \pmod{16}.
Adding multiples of 16 (in this case, 16×2=3216 \times 2 = 32) to 27-27 yields a positive remainder in the valid range of 00 to 1515. Calculating 27+32=5-27 + 32 = 5 gives the final remainder.

Anahtar Kavram

Modular arithmetic with large powers using negative remainders

Alternatif Yöntem

Instead of converting the bases to -1, one could convert them to positive 15. The expression becomes 15101+151012515^{101} + 15^{101} - 25. However, evaluating this requires recognizing that 151(mod16)15 \equiv -1 \pmod{16} to easily compute the large exponents, which ultimately merges back into the primary solution path.
Tahmini Süre:1m 30s
Soru 74Soru

Consider the number N=24×33×52N = 2^4 \times 3^3 \times 5^2. How many positive factors of NN are also multiples of 120120?

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Cevap: 12

Cevap

12
For a factor of NN to be a multiple of 120120, it must contain at least the prime factorization of 120120, which is 23×31×512^3 \times 3^1 \times 5^1. We can factor this out from NN: N=120×(21×32×51)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 (21×32×51)(2^1 \times 3^2 \times 5^1), which is calculated by adding 11 to each exponent and multiplying them: (1+1)×(2+1)×(1+1)=12(1+1) \times (2+1) \times (1+1) = 12.

Adım Adım Çözüm

1
Determine the prime factorization of 120.
120=23×31×51120 = 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.
24×33×5223×31×51=2(43)×3(31)×5(21)=21×32×51\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×k120 \times k, where kk must be a factor of the remaining prime powers of N.
3
Calculate the total number of positive factors for the quotient kk.
(1+1)×(2+1)×(1+1)=2×3×2=12(1 + 1) \times (2 + 1) \times (1 + 1) = 2 \times 3 \times 2 = 12
The number of choices for kk corresponds exactly to the number of factors of N that are multiples of 120.

Anahtar Kavram

Identifying restricted factors and multiples using prime factorization.
Soru 75Soru

A digital encryption protocol derives a validation key from the mathematical properties of the integer 1260012600. 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?

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Cevap: 972

Cevap

972
By finding the prime factorization of 1260012600 as 23×32×52×712^3 \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(2+1)(2+1)(1+1) = 18 odd divisors. Even divisors require at least one power of 22, giving 3×(2+1)(2+1)(1+1)=543 \times (2+1)(2+1)(1+1) = 54 even divisors. Following the protocol's rule, their product is 18×54=97218 \times 54 = 972.

Adım Adım Çözüm

1
Find the prime factorization of 1260012600.
12600=23×32×52×7112600 = 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 = 1818
Odd divisors are formed by using only the odd prime factors (33, 55, and 77). By adding 11 to each of their exponents and multiplying them, we get (2+1)×(2+1)×(1+1)=3×3×2=18(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 = 5454
Even divisors must include at least one factor of 22. The number of choices for the power of 22 is equal to its exponent (33 choices: 21,22,232^1, 2^2, 2^3). Multiplying this by the choices for odd prime factors gives 3×(2+1)×(2+1)×(1+1)=3×3×3×2=543 \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.
54×18=97254 \times 18 = 972
The encryption protocol specifically defines the key as the product of these two calculated quantities.

Anahtar Kavram

Prime Factorization and Number of Divisors (Even and Odd)
Soru 76Soru

Three different cryptographic algorithms in a secure server refresh their encryption keys at regular intervals. Algorithm X refreshes every 83\frac{8}{3} milliseconds, Algorithm Y every 109\frac{10}{9} milliseconds, and Algorithm Z every 145\frac{14}{5} milliseconds. They all start a refresh cycle simultaneously. Let MM be the minimum number of milliseconds until they all start a refresh cycle together again. Let HH be the Highest Common Factor (HCF) of their respective refresh intervals in milliseconds. What is the value of M÷HM \div H?

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Cevap: 63006300

Cevap

The value of M÷HM \div H is 63006300.
The value MM requires finding the LCM of the fractions, which represents the synchronization time. LCM(83,109,145)=LCM(8,10,14)HCF(3,9,5)=2801=280\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 HH requires finding the HCF of the fractions. HCF(83,109,145)=HCF(8,10,14)LCM(3,9,5)=245\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 MM by HH gives 280÷245=280×452=140×45=6300280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300.

Adım Adım Çözüm

1
Determine the synchronization time MM by finding the LCM of the three fractional intervals.
M=LCM(83,109,145)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: LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
Numerators (8,10,14)(8, 10, 14) have an LCM of 280280. Denominators (3,9,5)(3, 9, 5) have an HCF of 11. Thus, M=2801=280M = \frac{280}{1} = 280.
This is the mathematical formula required for finding the LCM of rational numbers.
3
Determine the Highest Common Factor HH of the three fractional intervals.
H=HCF(83,109,145)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: HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
Numerators (8,10,14)(8, 10, 14) have an HCF of 22. Denominators (3,9,5)(3, 9, 5) have an LCM of 4545. Thus, H=245H = \frac{2}{45}.
This is the mathematical formula required for finding the HCF of rational numbers.
5
Calculate the final required ratio M÷HM \div H.
280÷245=280×452=140×45=6300280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300.
This satisfies the final computational step requested by the problem stem.

Anahtar Kavram

Calculating the Lowest Common Multiple (LCM) and Highest Common Factor (HCF) for fractions using their specific formulas.
Soru 77Soru

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 561034556^{103} - 45 by 1313.

Find the value of this routing ID.

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Cevap: 11

Cevap

11
By reducing each component of the expression modulo 13, 5610356^{103} reduces to 41034^{103}, which simplifies to 44 because 431(mod13)4^3 \equiv -1 \pmod{13}. Subtracting 45(mod13)45 \pmod{13}, which is 66, yields 2-2. Adding the divisor 13 to 2-2 gives the strictly positive remainder of 11.

Adım Adım Çözüm

1
Reduce the base 56 modulo 13.
564(mod13)56 \equiv 4 \pmod{13}
Modular arithmetic allows simplifying the base before exponentiation.
2
Find a pattern to easily evaluate 4103(mod13)4^{103} \pmod{13}.
43=641(mod13)4^3 = 64 \equiv -1 \pmod{13}
Finding a small power that is congruent to 1 or -1 simplifies large exponents significantly.
3
Evaluate 4103(mod13)4^{103} \pmod{13} using the established pattern.
4103=(43)34×41(1)34×4=4(mod13)4^{103} = (4^3)^{34} \times 4^1 \equiv (-1)^{34} \times 4 = 4 \pmod{13}
Exponent rules allow breaking down 103 into 3×34+13 \times 34 + 1.
4
Reduce the subtracted constant 45 modulo 13.
45=13×3+66(mod13)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.
46=24 - 6 = -2
Substitute the individual modulo results back into the original expression structure.
6
Convert the negative remainder into a positive remainder.
2+13=11-2 + 13 = 11
The question asks for a strictly positive remainder, which is standard in modular division contexts.

Anahtar Kavram

Modular Arithmetic and Negative Remainders
Soru 78Soru

A cybersecurity protocol requires transmitting a data file of exactly 54005400 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 1212. Based on these conditions, how many different packet sizes are possible?

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Cevap: 18

Cevap

18
To find the number of factors of 54005400 that are multiples of 1212, we first divide 54005400 by 1212, which gives 450450. The number of possible packet sizes is exactly equal to the number of divisors of 450450. The prime factorization of 450450 is 21×32×522^1 \times 3^2 \times 5^2. Using the divisor-counting formula (adding 11 to each exponent and multiplying), we find there are (1+1)×(2+1)×(2+1)=2×3×3=18(1+1) \times (2+1) \times (2+1) = 2 \times 3 \times 3 = 18 valid packet sizes.

Adım Adım Çözüm

1
Find the prime factorization of the total file size (54005400).
5400=54×100=23×33×525400 = 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 SS.
SS must be a divisor of 54005400 and a multiple of 1212 (22×312^2 \times 3^1). This means S=12×kS = 12 \times k.
Translates the real-world constraints into a mathematical relationship.
3
Find the maximum possible value and prime factorization of the multiplier kk.
kk must be a divisor of 540012=450\frac{5400}{12} = 450. The prime factorization of 450450 is 21×32×522^1 \times 3^2 \times 5^2.
The number of valid packet sizes SS is directly equal to the number of divisors of kk.
4
Calculate the total number of divisors of 450450.
(1+1)×(2+1)×(2+1)=2×3×3=18(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).

Anahtar Kavram

Finding the number of factors of a number that are multiples of another given number.
Soru 79Soru

During a materials testing process, the density of four composite alloys (A, B, C, and D) is recorded in grams per cubic centimeter (g/cm3\text{g/cm}^3):

- Alloy A: 2.152.\overline{15}
- Alloy B: 157\frac{15}{7}
- Alloy C: 2.152.1\overline{5}
- Alloy D: 2813\frac{28}{13}

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

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

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...2.142...) is the smallest. The remaining values all begin with 2.152.15, but expanding the subsequent decimal places reveals that Alloy A is 2.1515...2.1515..., Alloy D is 2.1538...2.1538..., and Alloy C is 2.1555...2.1555.... This yields the exact ascending sequence: Alloy B < Alloy A < Alloy D < Alloy C.

Adım Adım Çözüm

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 (0.150.\overline{15}), B (17\frac{1}{7}), C (0.150.1\overline{5}), and D (213\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...0.1515..., B 0.1428...\approx 0.1428..., C = 0.1555...0.1555..., D 0.1538...\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 (1599=533\frac{15}{99} = \frac{5}{33}), B (17\frac{1}{7}), C (1490=745\frac{14}{90} = \frac{7}{45}), D (213\frac{2}{13}). Comparing B and A: 1×33=331 \times 33 = 33, 7×5=357 \times 5 = 35, so B < A. Comparing A and D: 5×13=655 \times 13 = 65, 33×2=6633 \times 2 = 66, so A < D. Comparing D and C: 2×45=902 \times 45 = 90, 13×7=9113 \times 7 = 91, so D < C.
Cross-multiplication of fractions provides absolute mathematical proof of the inequalities: 17<533<213<745\frac{1}{7} < \frac{5}{33} < \frac{2}{13} < \frac{7}{45}.

Anahtar Kavram

Comparing mixed fractions and recurring decimals via decimal expansion and cross-multiplication
Soru 80Soru

A wholesale merchant has three batches of specialty syrups measuring 635\frac{63}{5} liters, 7225\frac{72}{25} liters, and 8110\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?

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Cevap: 0.18

Cevap

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 HCF=HCF of numeratorsLCM of denominators\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}. The HCF of the numerators (63,72,8163, 72, 81) is 99, and the LCM of the denominators (5,25,105, 25, 10) is 5050. The resulting maximum capacity is 950\frac{9}{50}, which equals 0.180.18 liters.

Adım Adım Çözüm

1
Determine the mathematical operation required.
Calculate the HCF of 635\frac{63}{5}, 7225\frac{72}{25}, and 8110\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.
HCF of fractions=HCF of numeratorsLCM of denominators\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.
HCF(63,72,81)=9\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.
LCM(5,25,10)=50\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.
950=0.18\frac{9}{50} = 0.18
Dividing the HCF of the numerators by the LCM of the denominators yields the exact capacity in liters.

Anahtar Kavram

HCF and LCM of fractions
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Basic Numeracy Alıştırma Soruları — State PSC Exam — Sayfa 4 | Examkin