Tüm alıştırma soruları

2195 soru

Soru 661Soru

What is the remainder when 5995^{99} is divided by 1313?

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Cevap: 8

Cevap

The remainder when 5995^{99} is divided by 1313 is 88.
To find the remainder of 5995^{99} divided by 1313, find the repeating pattern of remainders for powers of 55 modulo 1313. The first four powers yield remainders 5,12,8,5, 12, 8, and 11. Since 541(mod13)5^4 \equiv 1 \pmod{13}, the remainders repeat every 44 powers. Dividing the exponent 9999 by 44 gives 99=4×24+399 = 4 \times 24 + 3. The remainder of 33 indicates that 5995^{99} has the exact same remainder modulo 1313 as 535^3, which is 88.

Adım Adım Çözüm

1
Analyze the remainders of powers of 5 modulo 13 to identify the repeating period length.
515(mod13)5^1 \equiv 5 \pmod{13}, 5212(mod13)5^2 \equiv 12 \pmod{13}, 538(mod13)5^3 \equiv 8 \pmod{13}, and 541(mod13)5^4 \equiv 1 \pmod{13}. The sequence of remainders repeats every 4 powers.
Integral powers modulo a positive integer exhibit periodic behavior.
2
Compute the remainder of the exponent 99 divided by the cycle length 4.
99÷4=2499 \div 4 = 24 with a remainder of 33.
The exponent's remainder modulo the cycle length determines which element in the periodic cycle gives the equivalent value.
3
Evaluate the 3rd term in the remainder cycle.
The 3rd element in the sequence (5,12,8,1)(5, 12, 8, 1) is 88.
An exponent congruent to 3(mod4)3 \pmod 4 yields the same remainder as 53(mod13)5^3 \pmod{13}.

Anahtar Kavram

Modular Arithmetic and Cyclicity of Powers
Soru 662Soru

At a chemical refining plant, a raw liquid compound containing Substance X, Substance Y, and an inert solvent is processed in two sequential purification phases. Initially, Substance X accounts for 0.250.25 of the total weight of the compound, and Substance Y accounts for 25\frac{2}{5} of the remaining weight, with the inert solvent comprising the rest. In Phase 1, 20%20\% of Substance X and 30%30\% of Substance Y are removed, while all of the inert solvent is retained. In Phase 2, a certain percentage p%p\% of the inert solvent present after Phase 1 is removed, while no other substances are removed. If Substance X represents exactly 40%40\% of the total weight of the compound remaining after Phase 2, what is the value of pp?

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Cevap: 80

Cevap

The value of pp is 80.
Assuming a total initial weight of 100 units, Substance X is 25 units and the remaining weight is 75 units. Substance Y is 25\frac{2}{5} of 75, which equals 30 units, leaving 45 units of inert solvent. After Phase 1, 20 units of Substance X and 21 units of Substance Y remain, along with the full 45 units of solvent. In Phase 2, Substance X (20 units) becomes 40%40\% of the total mixture, making the final total weight 200.40=50\frac{20}{0.40} = 50 units. Since Substance X and Substance Y together account for 20+21=4120 + 21 = 41 units, the remaining solvent after Phase 2 must be 5041=950 - 41 = 9 units. Reducing solvent from 45 units down to 9 units requires removing 45945=3645=80%\frac{45 - 9}{45} = \frac{36}{45} = 80\% of the solvent. Thus, p=80p = 80.

Adım Adım Çözüm

1
Determine initial component amounts using decimal and fractional breakdown.
In a 100-unit mixture, Substance X = 25 units, Substance Y = 30 units, and Inert Solvent = 45 units.
Substance X is 0.250.25 of the total (2525 units). Of the remaining 7575 units, Substance Y is 25×75=30\frac{2}{5} \times 75 = 30 units. The rest (7530=4575 - 30 = 45 units) is inert solvent.
2
Calculate remaining component amounts after Phase 1 percentage reductions.
Substance X = 20 units, Substance Y = 21 units, Inert Solvent = 45 units.
Removing 20%20\% of Substance X leaves 25×0.80=2025 \times 0.80 = 20 units. Removing 30%30\% of Substance Y leaves 30×0.70=2130 \times 0.70 = 21 units. No solvent is removed in Phase 1.
3
Determine the final total mixture weight using the final percentage of Substance X.
Final total mixture weight = 50 units.
Substance X (20 units) represents 40%40\% (0.400.40) of the final mixture after Phase 2, so the total weight is 200.40=50\frac{20}{0.40} = 50 units.
4
Formulate and solve the linear equation for pp.
p=80p = 80.
The total weight is the sum of all remaining components: 20+21+45(1p100)=5020 + 21 + 45\left(1 - \frac{p}{100}\right) = 50. Solving 41+45(1p100)=5041 + 45\left(1 - \frac{p}{100}\right) = 50 yields 45(1p100)=945\left(1 - \frac{p}{100}\right) = 9, so 1p100=0.201 - \frac{p}{100} = 0.20, giving p=80p = 80.

Anahtar Kavram

Multi-step percentage change, fractional remaining parts, and algebraic mixture equations
Soru 663Soru

An inventory manager is organizing items into storage boxes. When the items are packed into boxes of 2424, 3636, or 4545 items each, there are always 1111 items remaining unpacked. If the total number of items is between 700700 and 1,0001,000, how many items are there in total?

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Cevap: 731

Cevap

731
The number of items NN satisfies N11(mod24)N \equiv 11 \pmod{24}, N11(mod36)N \equiv 11 \pmod{36}, and N11(mod45)N \equiv 11 \pmod{45}. Consequently, N11N - 11 is a multiple of LCM(24,36,45)\text{LCM}(24, 36, 45). The prime factorizations are 24=23×324 = 2^3 \times 3, 36=22×3236 = 2^2 \times 3^2, and 45=32×545 = 3^2 \times 5. The LCM is 23×32×5=3602^3 \times 3^2 \times 5 = 360. Thus, N=360k+11N = 360k + 11 for some positive integer kk. Testing values of kk within the range 700<N<1,000700 < N < 1,000 yields k=2k = 2, giving N=360(2)+11=731N = 360(2) + 11 = 731.

Adım Adım Çözüm

1
Find the Least Common Multiple (LCM) of the divisors 24, 36, and 45.
LCM(24, 36, 45) = 360
Any number that leaves a constant remainder when divided by a set of numbers must be a constant remainder greater than a common multiple of those numbers.
2
Formulate the general equation for the total number of items N.
N = 360k + 11
Adding the common remainder 11 to any multiple of the LCM yields a valid total number of items.
3
Evaluate integer values of k to satisfy the constraint 700 < N < 1,000.
k = 2 gives N = 731
k = 1 gives 371, which is too small, and k = 3 gives 1,091, which exceeds 1,000.

Anahtar Kavram

Least Common Multiple (LCM) with Constant Remainder
Soru 664Soru

At the beginning of a fiscal year, a technology company allocated its total operating budget among three divisions: Research & Development, Marketing, and Operations. Research & Development received 0.450.45 of the total budget, Marketing received 13\frac{1}{3} of the total budget, and Operations received the remaining portion. Mid-year, the company adjusted the budget by increasing the Research & Development allocation by 20%20\% of its initial amount and reducing the Marketing allocation by 14\frac{1}{4} of its initial amount, while the Operations allocation remained unchanged. If these mid-year adjustments resulted in a net increase of $30,000\$30,000 in the company's total operating budget, what was the company's total operating budget at the beginning of the fiscal year?

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Cevap: $4,500,000

Cevap

$4,500,000
The option specifying 4,500,000 is correct. The increase in the Research & Development division is 20\%of of 0.45 B ,whichequals, which equals 0.09 B (or (or \frac{9}{100} B ).ThereductioninMarketingis). The reduction in Marketing is \frac{1}{4}of of \frac{1}{3} B ,whichequals, which equals \frac{1}{12} B .Combiningtheseusingacommondenominatorgivesanetincreaseof. Combining these using a common denominator gives a net increase of \frac{27}{300} B - \frac{25}{300} B = \frac{2}{300} B = \frac{1}{150} B .Setting. Setting \frac{1}{150} B = \30,00030,000 yields B=$4,500,000B = \$4,500,000.

Adım Adım Çözüm

1
Express initial allocations as fractions of total budget B
Research & Development initial share = 0.45B=920B0.45 B = \frac{9}{20} B; Marketing initial share = 13B\frac{1}{3} B
Converting decimals to fractions simplifies exact combined calculations.
2
Calculate the monetary change in each division relative to total budget B
R&D increase = 20%×0.45B=0.20×0.45B=0.09B=9100B20\% \times 0.45 B = 0.20 \times 0.45 B = 0.09 B = \frac{9}{100} B. Marketing decrease = 14×13B=112B\frac{1}{4} \times \frac{1}{3} B = \frac{1}{12} B. Operations change = 00.
Each change applies specifically to the division's initial allocation, not the total budget.
3
Find the net change in the total budget by taking the difference between the increase and the decrease
Net change = (9100112)B=(2730025300)B=2300B=1150B\left(\frac{9}{100} - \frac{1}{12}\right) B = \left(\frac{27}{300} - \frac{25}{300}\right) B = \frac{2}{300} B = \frac{1}{150} B
Determining the net fraction allows direct equation setup with the given dollar increase.
4
Solve for the initial total operating budget B using the given net increase of $30,000
\frac{1}{150} B = 30,000 \implies B = 30,000 \times 150 = 4,500,000
Multiplying both sides by 150 yields the original total budget.

Anahtar Kavram

Combining decimal and fraction arithmetic to determine net percentage changes across component allocations.

Alternatif Yöntem

Convert all components to a common denominator out of 300 parts initially. Total budget = 300 parts. R&D = 135 parts (0.45). Marketing = 100 parts (1/3). Operations = 65 parts. R&D increase = 20% of 135 = +27 parts. Marketing decrease = 1/4 of 100 = -25 parts. Net change = +2 parts = 30,000.Therefore1part=30,000. Therefore 1 part = 15,000, and 300 parts = $4,500,000.
Tahmini Süre:1m 45s
Soru 665Soru

Let N=1943+1143643N = 19^{43} + 11^{43} - 6^{43}. What is the remainder when NN is divided by 77?

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Cevap: 3

Cevap

3
Reducing each base modulo 7 simplifies the expression significantly: 192(mod7)19 \equiv -2 \pmod{7}, 114(mod7)11 \equiv 4 \pmod{7}, and 61(mod7)6 \equiv -1 \pmod{7}. Utilizing the cyclicity period of 3 for powers of 2 modulo 7 (231(mod7)2^3 \equiv 1 \pmod{7}), 1943(243)25(mod7)19^{43} \equiv -(2^{43}) \equiv -2 \equiv 5 \pmod{7}. Similarly, 1143443=286=(23)28×44(mod7)11^{43} \equiv 4^{43} = 2^{86} = (2^3)^{28} \times 4 \equiv 4 \pmod{7}. Lastly, 643(1)43=1(mod7)6^{43} \equiv (-1)^{43} = -1 \pmod{7}. Combining these yields N5+4(1)=103(mod7)N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}.

Adım Adım Çözüm

1
Reduce each base modulo 7
192(mod7)19 \equiv -2 \pmod{7}, 114(mod7)11 \equiv 4 \pmod{7}, and 61(mod7)6 \equiv -1 \pmod{7}
Simplifying bases using modular arithmetic reduces large power evaluations to smaller equivalents.
2
Evaluate 1943(mod7)19^{43} \pmod{7} using cyclicity
194325(mod7)19^{43} \equiv -2 \equiv 5 \pmod{7}
Since 231(mod7)2^3 \equiv 1 \pmod{7} and 43=3(14)+143 = 3(14) + 1, 2432(mod7)2^{43} \equiv 2 \pmod{7}, so (2)43=(243)2(mod7)(-2)^{43} = -(2^{43}) \equiv -2 \pmod{7}.
3
Evaluate 1143(mod7)11^{43} \pmod{7} using powers of 2
11434(mod7)11^{43} \equiv 4 \pmod{7}
1143443=286=(23)28×22128×4=4(mod7)11^{43} \equiv 4^{43} = 2^{86} = (2^3)^{28} \times 2^2 \equiv 1^{28} \times 4 = 4 \pmod{7}.
4
Evaluate 643(mod7)6^{43} \pmod{7}
6431(mod7)6^{43} \equiv -1 \pmod{7}
Any odd power of 1-1 equals 1-1.
5
Combine all components to determine the final remainder
N5+4(1)=103(mod7)N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}
Summing the individual remainders modulo 7 gives a final positive remainder of 3.

Anahtar Kavram

Modular Arithmetic and Cyclicity of Powers
Soru 666Soru

A positive integer nn is divisible by 2020 but is not divisible by 88. If nn has exactly 1515 positive divisors and 10n10n has exactly 2424 positive divisors, what is the value of nn?

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Cevap: 2500

Cevap

2500
To determine nn, analyze its prime factorization. Divisibility by 20=225120 = 2^2 \cdot 5^1 requires that nn contains 22 raised to at least the power of 22 and 55 raised to at least the power of 11. The condition that nn is not divisible by 8=238 = 2^3 restricts the exponent of 22 to exactly 22. Thus, n=225bKn = 2^2 \cdot 5^b \cdot K', where b1b \ge 1 and KK' is a product of powers of distinct prime factors other than 22 and 55. The number of positive divisors of nn is d(n)=(2+1)(b+1)K=3(b+1)K=15d(n) = (2+1)(b+1)K = 3(b+1)K = 15, which simplifies to (b+1)K=5(b+1)K = 5. Because b1b \ge 1, we have b+12b+1 \ge 2. Since 55 is a prime number, its only divisor greater than or equal to 22 is 55. Thus, b+1=5b+1 = 5, giving b=4b = 4, and K=1K = 1, which means nn has no prime factors other than 22 and 55. Therefore, n=2254=4625=2500n = 2^2 \cdot 5^4 = 4 \cdot 625 = 2500. We verify that 10n=235510n = 2^3 \cdot 5^5 has (3+1)(5+1)=24(3+1)(5+1) = 24 positive divisors, confirming the solution.

Adım Adım Çözüm

1
Determine the power of 2 in the prime factorization of nn
The exponent of 22 in nn is exactly 22
Because nn is a multiple of 20=22520 = 2^2 \cdot 5 but not a multiple of 8=238 = 2^3, 222^2 divides nn but 232^3 does not.
2
Set up the divisor counting formula for nn
d(n)=(2+1)(b+1)K=15    (b+1)K=5d(n) = (2+1)(b+1)K = 15 \implies (b+1)K = 5, where b1b \ge 1 is the exponent of 55 and KK represents the product of terms from any additional prime factors
The total number of positive divisors of an integer N=p1a1p2a2N = p_1^{a_1} p_2^{a_2} \dots is given by (a1+1)(a2+1)(a_1+1)(a_2+1)\dots
3
Solve for the exponents and prime factors of nn
b=4b = 4 and K=1K = 1, giving n=2254n = 2^2 \cdot 5^4
Since nn is divisible by 2020, 55 is a prime factor of nn, so b1b \ge 1, which implies b+12b+1 \ge 2. Since 55 is prime, its only factor greater than 11 is 55, forcing b+1=5b+1 = 5 and K=1K = 1.
4
Verify d(10n)d(10n) and evaluate nn
10n=235510n = 2^3 \cdot 5^5 has (3+1)(5+1)=24(3+1)(5+1) = 24 divisors, and n=4625=2500n = 4 \cdot 625 = 2500
Multiplying nn by 10=2510 = 2 \cdot 5 increases the exponent of 22 from 22 to 33 and the exponent of 55 from 44 to 55.

Anahtar Kavram

Divisor Count Formula and Prime Factorization Constraints
Soru 667Soru

Let S=k=182(1)kk7kS = \sum_{k=1}^{82} (-1)^k \cdot k \cdot 7^k. What is the remainder when S-S is divided by 1010?

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Cevap: 9

Cevap

The remainder when S-S is divided by 1010 is 99.
Grouping the 82 terms into 20 blocks of 4 shows that each block contributes 6(mod10)6 \pmod{10}, summing to 1200(mod10)120 \equiv 0 \pmod{10}. The 81st and 82nd terms contribute 33 and 88 respectively, giving S111(mod10)S \equiv 11 \equiv 1 \pmod{10}. Therefore, S19(mod10)-S \equiv -1 \equiv 9 \pmod{10}.

Adım Adım Çözüm

1
Analyze the units digit cyclicity of 7k(mod10)7^k \pmod{10}.
The sequence of units digits for 7k7^k repeats every 4 powers: 7177^1 \equiv 7, 7297^2 \equiv 9, 7337^3 \equiv 3, 741(mod10)7^4 \equiv 1 \pmod{10}.
Units digits of powers of 7 follow a repeating cycle of length 4.
2
Evaluate the terms modulo 10 for a general block of 4 consecutive integers k=4m+1,4m+2,4m+3,4m+4k = 4m+1, 4m+2, 4m+3, 4m+4.
The four terms evaluate to: (1)4m+1(4m+1)74m+1(4m+1)(7)2m+3(mod10)(-1)^{4m+1}(4m+1)7^{4m+1} \equiv -(4m+1)(7) \equiv 2m+3 \pmod{10}; (1)4m+2(4m+2)74m+2(4m+2)(9)6m+8(mod10)(-1)^{4m+2}(4m+2)7^{4m+2} \equiv (4m+2)(9) \equiv 6m+8 \pmod{10}; (1)4m+3(4m+3)74m+3(4m+3)(3)8m+1(mod10)(-1)^{4m+3}(4m+3)7^{4m+3} \equiv -(4m+3)(3) \equiv 8m+1 \pmod{10}; and (1)4m+4(4m+4)74m+4(4m+4)(1)4m+4(mod10)(-1)^{4m+4}(4m+4)7^{4m+4} \equiv (4m+4)(1) \equiv 4m+4 \pmod{10}.
Modular arithmetic simplifies each term according to its position in the 4-term cycle.
3
Sum the four terms in a single block of 4.
(2m+3) + (6m+8) + (8m+1) + (4m+4) = 20m + 16 \equiv 6 \pmod{10}.
Every complete block of 4 consecutive terms contributes exactly 6 modulo 10, independent of mm.
4
Calculate the total sum S(mod10)S \pmod{10} for k=1k=1 to 8282.
82 terms contain 20 full blocks of 4 terms (k=1k=1 to 8080) plus 2 remaining terms (k=81k=81 and k=82k=82). The 20 blocks contribute 20×6=1200(mod10)20 \times 6 = 120 \equiv 0 \pmod{10}. For m=20m=20, k=81k=81 gives 2(20)+3=433(mod10)2(20)+3 = 43 \equiv 3 \pmod{10}, and k=82k=82 gives 6(20)+8=1288(mod10)6(20)+8 = 128 \equiv 8 \pmod{10}. Thus S0+3+8=111(mod10)S \equiv 0 + 3 + 8 = 11 \equiv 1 \pmod{10}.
Summing the contributions of all complete blocks and the remaining boundary terms gives S(mod10)S \pmod{10}.
5
Determine the positive remainder of S-S when divided by 10.
Since S1(mod10)S \equiv 1 \pmod{10}, S1101=9(mod10)-S \equiv -1 \equiv 10 - 1 = 9 \pmod{10}.
By definition of integer division, the remainder rr must satisfy 0r<100 \le r < 10.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic with Negative Integer Remainders
Soru 668Soru

If 2x3=7|2x - 3| = 7, which of the following is a possible value of xx?

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Cevap: 55

Cevap

The value 55 is a valid solution to the given absolute value equation.
The absolute value equation 2x3=7|2x - 3| = 7 splits into two linear equations: 2x3=72x - 3 = 7 and 2x3=72x - 3 = -7. Solving the first equation gives 2x=102x = 10, so x=5x = 5. Solving the second gives 2x=42x = -4, so x=2x = -2. Among the listed options, 55 is the correct solution.

Adım Adım Çözüm

1
Set up the two cases for the absolute value equation 2x3=7|2x - 3| = 7.
Case 1: 2x3=72x - 3 = 7; Case 2: 2x3=72x - 3 = -7.
An absolute value expression u=c|u| = c (where c0c \geq 0) resolves to u=cu = c or u=cu = -c.
2
Solve Case 1: 2x3=72x - 3 = 7.
2x=10    x=52x = 10 \implies x = 5.
Add 3 to both sides and divide by 2.
3
Solve Case 2: 2x3=72x - 3 = -7.
2x=4    x=22x = -4 \implies x = -2.
Add 3 to both sides and divide by 2.
4
Compare the solutions (x=5x = 5 and x=2x = -2) with the given choices.
The value 55 is present among the options.
Identifies the correct choice matching one of the calculated solutions.

Anahtar Kavram

Solving Linear Absolute Value Equations
Tahmini Süre:45s
Soru 669Soru

To reduce municipal carbon emissions and alleviate gridlock, City X introduced a daily congestion toll on private passenger automobiles entering its commercial core during morning peak hours. Analysts argue that this policy will successfully decrease overall motor vehicle traffic along the primary access corridors leading into the commercial core by at least 25 percent. Their reasoning is that the added financial cost will compel a significant portion of daily automobile commuters to switch to the city's underutilized suburban rail network. Which of the following, if true, most seriously weakens the analysts' argument?

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Cevap: Because the toll applies exclusively to private passenger automobiles, commercial delivery fleets and ride-hailing services plan to sharply increase peak-hour trips along those same access corridors to meet shifted consumer demand.

Cevap

The argument is most seriously weakened by the statement that commercial delivery fleets and ride-hailing services will increase peak-hour trips along the access corridors, replacing reduced private cars with other motor vehicles.
The correct answer demonstrates that while private passenger cars may decrease due to the toll, another category of motor vehicles—commercial fleets and ride-hailing cars—will significantly increase operations on the exact same access corridors. This directly invalidates the analysts' assumption that reduced private automobile commuter traffic guarantees a 25 percent drop in total motor vehicle volume.

Adım Adım Çözüm

1
Identify the main conclusion and premises.
Conclusion: Overall motor vehicle traffic along primary access corridors leading into the commercial core will decrease by at least 25%. Premise: A daily toll on private passenger cars will cause commuters to switch to suburban rail.
Deconstructing the argument isolates the leap between private car commuter behavior and total vehicle traffic volume.
2
Identify unstated assumptions in the argument's logic.
The argument assumes that private passenger vehicles are the sole major contributor to traffic volume on those corridors, or that decrease in private vehicles will not be replaced by other types of vehicular traffic.
Weakening questions require finding new information that shatters an unstated assumption.
3
Evaluate option impacts against the core assumption.
Showing that exempt vehicle categories (delivery fleets and ride-hailing cars) will increase trip frequency along the exact same corridors directly counteracts the projected net traffic reduction.
Demonstrating an offsetting surge in non-tolled motor vehicles directly breaks the link between fewer private cars and lower overall corridor traffic volume.

Anahtar Kavram

Evaluating Causal and Extrapolation Arguments by Identifying Offsetting Alternative Factors
Soru 670Soru

Let N=233a5bN = 2^3 \cdot 3^a \cdot 5^b, where aa and bb are positive integers. If NN has exactly 60 positive integer divisors that are multiples of 10 and exactly 40 positive integer divisors that are multiples of 15, how many positive integer divisors of NN are multiples of 12?

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Cevap: 22

Cevap

22
The correct answer is 22. From the problem statement, divisors divisible by 10 require x{1,2,3}x \in \{1, 2, 3\}, y{0,,a}y \in \{0, \dots, a\}, and z{1,,b}z \in \{1, \dots, b\}, leading to 3(a+1)b=603(a+1)b = 60, or (a+1)b=20(a+1)b = 20. Divisors divisible by 15 require x{0,1,2,3}x \in \{0, 1, 2, 3\}, y{1,,a}y \in \{1, \dots, a\}, and z{1,,b}z \in \{1, \dots, b\}, leading to 4ab=404ab = 40, or ab=10ab = 10. Substituting ab=10ab = 10 into ab+b=20ab + b = 20 yields b=10b = 10 and a=1a = 1. To be a multiple of 12=223112 = 2^2 \cdot 3^1, a divisor must have x{2,3}x \in \{2, 3\} (2 choices), y{1}y \in \{1\} (1 choice), and z{0,1,,10}z \in \{0, 1, \dots, 10\} (11 choices). The total number of such divisors is 2×1×11=222 \times 1 \times 11 = 22.

Adım Adım Çözüm

1
Set up the equation for positive divisors of NN that are multiples of 10.
Multiples of 10 must be divisible by 21512^1 \cdot 5^1. Thus, for a divisor 2x3y5z2^x \cdot 3^y \cdot 5^z, we need 1x31 \le x \le 3 (3 choices), 0ya0 \le y \le a (a+1a+1 choices), and 1zb1 \le z \le b (bb choices). The total number of such divisors is 3(a+1)b=603(a + 1)b = 60, which simplifies to (a+1)b=20(a + 1)b = 20.
Divisibility by 10 requires at least one factor of 2 and at least one factor of 5.
2
Set up the equation for positive divisors of NN that are multiples of 15.
Multiples of 15 must be divisible by 31513^1 \cdot 5^1. Thus, we need 0x30 \le x \le 3 (4 choices), 1ya1 \le y \le a (aa choices), and 1zb1 \le z \le b (bb choices). The total number of such divisors is 4ab=404ab = 40, which simplifies to ab=10ab = 10.
Divisibility by 15 requires at least one factor of 3 and at least one factor of 5.
3
Solve the system of equations for aa and bb.
Expanding (a+1)b=20(a + 1)b = 20 yields ab+b=20ab + b = 20. Substituting ab=10ab = 10 into this equation gives 10+b=20    b=1010 + b = 20 \implies b = 10. Since ab=10ab = 10, we find a(10)=10    a=1a(10) = 10 \implies a = 1.
Determining the exact exponents aa and bb allows us to fully specify N=2331510N = 2^3 \cdot 3^1 \cdot 5^{10}.
4
Calculate the number of divisors of NN that are multiples of 12.
Multiples of 12 must be divisible by 12=223112 = 2^2 \cdot 3^1. For a divisor 2x3y5z2^x \cdot 3^y \cdot 5^z, we require 2x32 \le x \le 3 (2 choices: 22,232^2, 2^3), 1ya=11 \le y \le a = 1 (1 choice: 313^1), and 0zb=100 \le z \le b = 10 (b+1=11b + 1 = 11 choices: 50,51,,5105^0, 5^1, \dots, 5^{10}). Total multiples of 12 = 2×1×11=222 \times 1 \times 11 = 22.
Applying the fundamental counting principle to the constrained prime factor exponent ranges.

Anahtar Kavram

Divisor counting formula under prime factor constraint conditions
Soru 671Soru

A 6-letter security code is created by arranging all the letters in the word BANANABANANA. How many distinct security codes can be formed such that the two letters NN are not adjacent to each other?

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Cevap: 40

Cevap

40 distinct security codes can be formed such that the two NN's are not adjacent.
To find the number of arrangements where the two NN's are not adjacent, use complementary counting. First, compute total unrestricted permutations of BANANABANANA (1 BB, 3 AA's, 2 NN's), which gives 6!3!×2!=60\frac{6!}{3! \times 2!} = 60. Next, count the forbidden arrangements where the two NN's are together by grouping them into a single block (NN)(NN). Arranging BB, AA, AA, AA, and (NN)(NN) yields 5!3!=20\frac{5!}{3!} = 20. Subtracting forbidden arrangements from total arrangements gives 6020=4060 - 20 = 40.

Adım Adım Çözüm

1
Calculate the total number of distinct arrangements of the letters in BANANABANANA without restrictions.
The word BANANABANANA contains 6 letters in total: 1 BB, 3 AA's, and 2 NN's. The number of distinct permutations is 6!1!×3!×2!=7201×6×2=60\frac{6!}{1! \times 3! \times 2!} = \frac{720}{1 \times 6 \times 2} = 60.
Indistinguishable duplicate letters must be accounted for by dividing the total permutations by the factorial of each letter's count.
2
Calculate the number of arrangements where the two NN's ARE adjacent.
Treat the pair NNNN as a single block. The items to arrange are BB, AA, AA, AA, and (NN)(NN), which gives 5 items with 3 identical AA's. The number of arrangements is 5!1!×3!×1!=1206=20\frac{5!}{1! \times 3! \times 1!} = \frac{120}{6} = 20.
Grouping restricted elements into a single composite element simplifies calculating forbidden configurations.
3
Apply complementary counting to find the number of valid arrangements where the two NN's are not adjacent.
Total valid arrangements = Total arrangements - Adjacent arrangements = 6020=4060 - 20 = 40.
Complementary counting is the most efficient method when calculating non-adjacent placements.

Anahtar Kavram

Counting with Restrictions and Repetitions (Complementary Counting Principle)
Tahmini Süre:1m 30s
Soru 672Soru

A boutique investment fund allocates its capital among Technology, Healthcare, and Renewable Energy in the ratio 3:4:53 : 4 : 5, respectively. After receiving an additional $12\$12 million in new capital, the entire new amount is invested in Healthcare, changing the ratio of capital among Technology, Healthcare, and Renewable Energy to 3:6:53 : 6 : 5. What was the total initial capital, in millions of dollars, invested in the fund?

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Cevap: 7272

Cevap

7272 million dollars
Let the initial capital allocations be 3x3x, 4x4x, and 5x5x million dollars for Technology, Healthcare, and Renewable Energy, respectively. The initial total capital is 3x+4x+5x=12x3x + 4x + 5x = 12x. When $12\$12 million is added to Healthcare, its new amount becomes 4x+124x + 12. Comparing the unchanged Technology amount to the updated Healthcare amount gives the proportion 3x4x+12=36=12\frac{3x}{4x + 12} = \frac{3}{6} = \frac{1}{2}. Cross-multiplying yields 6x=4x+126x = 4x + 12, so 2x=122x = 12 and x=6x = 6. The total initial capital is 12×6=7212 \times 6 = 72 million dollars.

Adım Adım Çözüm

1
Represent initial allocations using a multiplier xx
Technology =3x= 3x, Healthcare =4x= 4x, Renewable Energy =5x= 5x
Ratios represent proportional parts, so multiplying each term by a common constant xx expresses the actual amounts.
2
Set up the equation following the addition of $12\$12 million to Healthcare
New Healthcare allocation =4x+12= 4x + 12. The new ratio is 3x:(4x+12):5x=3:6:53x : (4x + 12) : 5x = 3 : 6 : 5.
Only the Healthcare amount changes, while Technology and Renewable Energy allocations remain 3x3x and 5x5x respectively.
3
Solve for xx using the ratio between Technology and Healthcare
\frac{3x}{4x + 12} = \frac{3}{6} \implies \frac{3x}{4x + 12} = \frac{1}{2} \implies 6x = 4x + 12 \implies 2x = 12 \implies x = 6
Equating the algebraic ratio of Technology to Healthcare to the new simplified ratio allows solving for xx.
4
Calculate the total initial capital
\text{Total initial capital} = 3x + 4x + 5x = 12x = 12(6) = 72
Summing the initial parts (12x12x) and substituting x=6x = 6 yields the initial total.

Anahtar Kavram

Ratio scaling and constant-part proportion adjustments
Tahmini Süre:1m 30s
Soru 673Soru

How many distinct 6-digit positive integers can be formed using all of the digits 1,1,2,2,3,1, 1, 2, 2, 3, and 44 such that no two identical digits are adjacent to each other?

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Cevap: 84

Cevap

84
To find the number of 6-digit integers where no two identical digits are adjacent, apply complementary counting with the Principle of Inclusion-Exclusion. First, calculate the total distinct arrangements of the digits 1,1,2,2,3,41, 1, 2, 2, 3, 4, which is 6!2!2!=180\frac{6!}{2!2!} = 180. Next, count the forbidden outcomes. Treating the two 1s as a single block gives 5!2!=60\frac{5!}{2!} = 60 arrangements with adjacent 1s. Similarly, treating the two 2s as a single block gives 5!2!=60\frac{5!}{2!} = 60 arrangements with adjacent 2s. Arrangements where both the 1s are adjacent and the 2s are adjacent use two single blocks along with digits 3 and 4, giving 4!=244! = 24 arrangements. By Inclusion-Exclusion, the total number of forbidden arrangements is 60+6024=9660 + 60 - 24 = 96. Subtracting these from the total unrestricted arrangements gives 18096=84180 - 96 = 84.

Adım Adım Çözüm

1
Calculate the total number of unrestricted distinct 6-digit arrangements of the multiset {1, 1, 2, 2, 3, 4}.
Total arrangements = 6!2!×2!=7204=180\frac{6!}{2! \times 2!} = \frac{720}{4} = 180.
Dividing by 2!×2!2! \times 2! accounts for the indistinguishable repeated digits 1 and 2.
2
Define the forbidden events and calculate their individual counts.
Let AA be the set of arrangements with adjacent 1s, and BB be the set of arrangements with adjacent 2s.
For A|A|, treat (11)(11) as a single block: elements are {(11),2,2,3,4}\{(11), 2, 2, 3, 4\}, so A=5!2!=60|A| = \frac{5!}{2!} = 60.
For B|B|, treat (22)(22) as a single block: elements are {1,1,(22),3,4}\{1, 1, (22), 3, 4\}, so B=5!2!=60|B| = \frac{5!}{2!} = 60.
Grouping adjacent identical numbers into a single block simplifies counting restriction violations.
3
Calculate the intersection count AB|A \cap B| where both pairs are adjacent.
Treating both (11)(11) and (22)(22) as single blocks yields elements {(11),(22),3,4}\{(11), (22), 3, 4\}, so AB=4!=24|A \cap B| = 4! = 24.
Arrangements where both pairs of identical digits are adjacent are counted in both AA and BB and must be identified.
4
Apply the Principle of Inclusion-Exclusion to find the total forbidden arrangements and subtract from the total unrestricted arrangements.
AB=A+BAB=60+6024=96|A \cup B| = |A| + |B| - |A \cap B| = 60 + 60 - 24 = 96.
Valid arrangements = 18096=84180 - 96 = 84.
Complementary counting yields the exact number of arrangements where no two identical digits are adjacent.

Anahtar Kavram

Permutations of Multisets with Restrictions via Inclusion-Exclusion
Soru 674Soru

A 7-digit security passcode is to be formed using the digits 1,1,2,2,3,4,1, 1, 2, 2, 3, 4, and 55. How many distinct 7-digit passcodes can be formed such that neither the two 11 s nor the two 22 s are adjacent to each other?

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Cevap: 660

Cevap

660 distinct passcodes can be formed.
To find the number of arrangements where neither identical pair is adjacent, we use complementary counting. First, compute total unrestricted arrangements of the digits 1,1,2,2,3,4,51,1,2,2,3,4,5, which is 7!2!2!=1,260\frac{7!}{2!2!} = 1,260. Next, by treating (11)(11) as a single block, there are 6!2!=360\frac{6!}{2!} = 360 arrangements with adjacent 11 s. Similarly, treating (22)(22) as a single block yields 6!2!=360\frac{6!}{2!} = 360 arrangements with adjacent 22 s. Arrangements where both pairs are adjacent treat (11)(11) and (22)(22) as two single blocks alongside 3,4,53,4,5, giving 5!=1205! = 120 arrangements. Applying Inclusion-Exclusion gives 360+360120=600360 + 360 - 120 = 600 forbidden arrangements. Subtracting from the total gives 1,260600=6601,260 - 600 = 660.

Adım Adım Çözüm

1
Calculate the total number of distinct arrangements without any restrictions.
Total arrangements = 7!2!×2!=5,0404=1,260\frac{7!}{2! \times 2!} = \frac{5,040}{4} = 1,260.
There are 7 digits total, with the digit 1 repeated twice and the digit 2 repeated twice.
2
Calculate the number of arrangements where the two 11 s are adjacent.
N(adjacent 1s)=6!2!=7202=360N(\text{adjacent } 1\text{s}) = \frac{6!}{2!} = \frac{720}{2} = 360.
Treat the pair (11)(11) as a single unit, leaving 6 items: (11),2,2,3,4,5(11), 2, 2, 3, 4, 5 with two identical 22 s.
3
Calculate the number of arrangements where the two 22 s are adjacent.
N(adjacent 2s)=6!2!=7202=360N(\text{adjacent } 2\text{s}) = \frac{6!}{2!} = \frac{720}{2} = 360.
Treat the pair (22)(22) as a single unit, leaving 6 items: 1,1,(22),3,4,51, 1, (22), 3, 4, 5 with two identical 11 s.
4
Calculate the number of arrangements where both the two 11 s and two 22 s are simultaneously adjacent.
N(both adjacent)=5!=120N(\text{both adjacent}) = 5! = 120.
Treat (11)(11) and (22)(22) each as single distinct units, giving 5 distinct items: (11),(22),3,4,5(11), (22), 3, 4, 5.
5
Apply the Principle of Inclusion-Exclusion to find the total restricted count.
Valid passcodes = 1,260(360+360120)=1,260600=6601,260 - (360 + 360 - 120) = 1,260 - 600 = 660.
Subtract the union of forbidden cases from the total arrangements: N(total)[N(1s adj)+N(2s adj)N(both adj)]N(\text{total}) - [N(\text{1s adj}) + N(\text{2s adj}) - N(\text{both adj})].

Anahtar Kavram

Complementary Counting with Inclusion-Exclusion Principle on Permutations of Repeated Elements
Tahmini Süre:2m 0s
Soru 675Soru

For all real numbers aa and bb, the custom operation \star is defined by ab=(a+b)23ba \star b = (a + b)^2 - 3b. What is the value of 424 \star 2?

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Cevap: 30

Cevap

30
Substituting a=4a = 4 and b=2b = 2 into the definition (a+b)23b(a + b)^2 - 3b gives (4+2)23(2)=626=366=30(4 + 2)^2 - 3(2) = 6^2 - 6 = 36 - 6 = 30.

Adım Adım Çözüm

1
Identify the values assigned to the custom operator variables
a=4a = 4 and b=2b = 2
The expression 424 \star 2 corresponds to evaluating aba \star b where aa is the first operand and bb is the second operand.
2
Substitute a=4a = 4 and b=2b = 2 into the operator definition (a+b)23b(a + b)^2 - 3b
(4+2)23(2)(4 + 2)^2 - 3(2)
Direct substitution into the given algebraic definition.
3
Perform the operations in parentheses and exponentiation
626=3666^2 - 6 = 36 - 6
First add inside the parentheses (4+2=64 + 2 = 6), then square the sum (62=366^2 = 36), and multiply 3×2=63 \times 2 = 6.
4
Subtract to find the final value
30
Subtracting 66 from 3636 yields 3030.

Anahtar Kavram

Custom Operator Evaluation
Tahmini Süre:45s
Soru 676Soru

Three industrial machines, AA, BB, and CC, operating independently at their respective constant rates, can complete a production order of 1,2001,200 units in 1212 hours, 1515 hours, and 2020 hours, respectively. All three machines start working simultaneously on the order at 8:00 AM. At 10:00 AM, Machine AA experiences a mechanical failure and stops working permanently. Machine BB and Machine CC continue working together until 11:00 AM, at which point Machine BB's operating efficiency drops by 50%50\% due to overheating, while Machine CC continues at its original rate. At what time will the production order of 1,2001,200 units be fully completed?

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Cevap: 4:48 PM

Cevap

The production order will be completed at 4:48 PM.
The correct answer of 4:48 PM is derived by dividing the job into three distinct stages. From 8:00 AM to 10:00 AM (2 hours), Machines A, B, and C produce 240 units/hour, yielding 480 units. From 10:00 AM to 11:00 AM (1 hour), Machines B and C produce 140 units/hour, yielding 140 units, leaving 580 units remaining. After 11:00 AM, Machine B operates at 40 units/hour and Machine C at 60 units/hour, producing a combined rate of 100 units/hour. Dividing 580 remaining units by 100 units/hour gives 5.8 hours (5 hours 48 minutes), placing completion at exactly 4:48 PM.

Adım Adım Çözüm

1
Determine individual production rates for each machine.
Machine AA produces 1,20012=100\frac{1,200}{12} = 100 units/hour. Machine BB produces 1,20015=80\frac{1,200}{15} = 80 units/hour. Machine CC produces 1,20020=60\frac{1,200}{20} = 60 units/hour.
Establishing hourly rates is necessary to calculate work completed in each time interval.
2
Calculate work completed during Stage 1 (8:00 AM to 10:00 AM).
Combined rate of A+B+C=100+80+60=240A + B + C = 100 + 80 + 60 = 240 units/hour. In 22 hours, work completed =240×2=480= 240 \times 2 = 480 units. Remaining units =1,200480=720= 1,200 - 480 = 720 units.
All three machines work together for 2 full hours before Machine A breaks down.
3
Calculate work completed during Stage 2 (10:00 AM to 11:00 AM).
Combined rate of B+C=80+60=140B + C = 80 + 60 = 140 units/hour. In 11 hour, work completed =140×1=140= 140 \times 1 = 140 units. Remaining units =720140=580= 720 - 140 = 580 units.
Machine A is inactive, leaving only Machines B and C operating at full capacity for 1 hour.
4
Calculate the time required for Stage 3 (from 11:00 AM until completion).
Machine BB's reduced rate =80×0.50=40= 80 \times 0.50 = 40 units/hour. New combined rate of B+C=40+60=100B + C = 40 + 60 = 100 units/hour. Time needed =580100=5.8= \frac{580}{100} = 5.8 hours =5= 5 hours and 4848 minutes.
Machine B operates at half speed while Machine C continues at full speed to finish the remaining 580 units.
5
Determine the final completion clock time.
11:00 AM +5+ 5 hours and 4848 minutes =4:48= 4:48 PM.
Adding the duration of the final stage to the starting time of 11:00 AM gives the exact completion time.

Anahtar Kavram

Multi-stage combined work problems require calculating individual rates, tracking partial work completed in each phase, and adjusting combined rates whenever active entities or their individual rates change.
Tahmini Süre:3m 0s
Soru 677Soru

An investment fund initially divides its total capital among three portfolios—Real Estate, Stocks, and Bonds—in the ratio 3:4:53 : 4 : 5. During a financial restructuring, a portion of capital is transferred directly from the Bonds portfolio to the Real Estate portfolio while the capital in Stocks remains unchanged, resulting in a new ratio of Real Estate to Stocks of 5:45 : 4. Subsequently, an additional $60,000\$60,000 in external capital is deposited into the Bonds portfolio, making the ratio of Stocks to Bonds 2:32 : 3. What was the total initial capital in the fund across all three portfolios?

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Cevap: $240,000

Cevap

The total initial capital in the fund was $240,000.
The correct answer of 240,000isobtainedbytrackingeachportfoliosbalanceacrossthetwosequentialchanges.Initially,RealEstate,Stocks,andBondsare240,000 is obtained by tracking each portfolio's balance across the two sequential changes. Initially, Real Estate, Stocks, and Bonds are 3x ,, 4x ,and, and 5x respectively(total respectively (total 12x ).WhenRealEstateincreasesto). When Real Estate increases to 5x relativetoStocks( relative to Stocks ( 4x ),therequiredcapitalcomesentirelyfromBonds,reducingBondsfrom), the required capital comes entirely from Bonds, reducing Bonds from 5x to to 3x .Adding. Adding \60,00060,000 to Bonds gives a new balance of 3x+60,0003x + 60,000. Establishing the new Stocks to Bonds ratio 4x3x+60,000=23\frac{4x}{3x + 60,000} = \frac{2}{3} yields x=20,000x = 20,000. Multiplying by the initial total of 12 parts yields 12×20,000=12 \times 20,000 = 240,000$.

Adım Adım Çözüm

1
Express initial portfolio amounts in terms of a multiplier xx
Real Estate = 3x3x, Stocks = 4x4x, Bonds = 5x5x. Total initial capital = 3x+4x+5x=12x3x + 4x + 5x = 12x.
The given initial ratio of Real Estate : Stocks : Bonds is 3:4:53 : 4 : 5.
2
Account for the internal capital transfer between Bonds and Real Estate
Real Estate becomes 5x5x, Stocks remains 4x4x, and Bonds becomes 3x3x.
Stocks remains 4x4x. Since the ratio of Real Estate to Stocks becomes 5:45 : 4, Real Estate must equal 5x5x. Because capital was only moved internally from Bonds to Real Estate, the sum of Real Estate and Bonds remains 3x+5x=8x3x + 5x = 8x, leaving Bonds with 8x5x=3x8x - 5x = 3x.
3
Set up an algebraic proportion after adding $60,000\$60,000 to Bonds
4x3x+60,000=23\frac{4x}{3x + 60,000} = \frac{2}{3}
The new ratio of Stocks to Bonds is given as 2:32 : 3.
4
Solve for the multiplier xx and determine the total initial capital
3(4x)=2(3x+60,000)12x=6x+120,0006x=120,000x=20,0003(4x) = 2(3x + 60,000) \Rightarrow 12x = 6x + 120,000 \Rightarrow 6x = 120,000 \Rightarrow x = 20,000. Total initial capital =12x=12×20,000== 12x = 12 \times 20,000 = 240,000$.
Cross-multiplying yields the value of xx, which is then scaled by the total initial ratio sum of 12 parts.

Anahtar Kavram

Multi-Step Ratio Adjustments and Internal vs. External Capital Changes
Tahmini Süre:2m 0s
Soru 678Soru

Company A charges a one-time fixed fee of $20\$20 plus $5\$5 per hour to rent a piece of machinery. Company B charges a one-time fixed fee of $35\$35 plus $2\$2 per hour to rent the same machinery. For how many hours of rental will the total cost charged by both companies be equal?

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Cevap: 5

Cevap

The total cost at both companies will be equal for a rental duration of 5 hours.
The cost charged by Company A is modeled by 20+5h20 + 5h and Company B by 35+2h35 + 2h. Setting these equal yields 20+5h=35+2h20 + 5h = 35 + 2h, which simplifies to 3h=153h = 15, giving h=5h = 5. Thus, 5 hours is the correct rental duration.

Adım Adım Çözüm

1
Define the variable and write total cost equations for each company.
Let hh represent the number of hours. Company A's cost is CA=20+5hC_A = 20 + 5h and Company B's cost is CB=35+2hC_B = 35 + 2h.
Linear cost models combine a fixed base cost with a variable hourly component.
2
Set the two cost equations equal to solve for hh.
20+5h=35+2h20 + 5h = 35 + 2h
The total costs are equal when CA=CBC_A = C_B.
3
Isolate the variable term hh on one side of the equation.
5h2h=35203h=15h=55h - 2h = 35 - 20 \Rightarrow 3h = 15 \Rightarrow h = 5
Subtracting 2h2h and 2020 from both sides simplifies the equation to solve directly for hh.

Anahtar Kavram

Setting up and solving single-variable linear equations from real-world cost contexts.
Tahmini Süre:45s
Soru 679Soru

If the quadratic equation x2kx+36=0x^2 - kx + 36 = 0 has two distinct positive integer roots, r1r_1 and r2r_2, and r1+2r_1 + 2 and r2+2r_2 + 2 are the roots of the quadratic equation x219x+m=0x^2 - 19x + m = 0, what is the value of mm?

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Cevap: 70

Cevap

The value of mm is 70.
By applying Vieta's formulas to both quadratic equations, we find that the sum of the original roots is r1+r2=15r_1 + r_2 = 15 and their product is r1r2=36r_1 r_2 = 36. Factoring x215x+36=(x3)(x12)=0x^2 - 15x + 36 = (x-3)(x-12) = 0 confirms the original roots are 33 and 1212. The new roots are 3+2=53+2=5 and 12+2=1412+2=14. Their product m=5×14=70m = 5 \times 14 = 70.

Adım Adım Çözüm

1
Apply Vieta's relations to the first quadratic equation
r1+r2=kr_1 + r_2 = k and r1r2=36r_1 \cdot r_2 = 36
For any quadratic equation x2+bx+c=0x^2 + bx + c = 0, the sum of roots is b-b and the product of roots is cc.
2
Apply Vieta's relations to the second quadratic equation
(r1+2)+(r2+2)=19    r1+r2=15(r_1 + 2) + (r_2 + 2) = 19 \implies r_1 + r_2 = 15
The coefficient of xx in x219x+m=0x^2 - 19x + m = 0 dictates that the sum of its roots equals 1919.
3
Determine the roots r1r_1 and r2r_2 and check integer constraints
r1=3r_1 = 3 and r2=12r_2 = 12
The integer factors of 3636 that sum to 1515 are 33 and 1212, satisfying all conditions.
4
Compute the constant term mm for the transformed equation
m=(3+2)(12+2)=5×14=70m = (3 + 2)(12 + 2) = 5 \times 14 = 70
The constant term mm equals the product of the transformed roots (r1+2)(r_1 + 2) and (r2+2)(r_2 + 2).

Anahtar Kavram

Vieta's Formulas and Quadratic Root Transformations
Tahmini Süre:2m 0s
Soru 680Soru

For all non-zero real numbers aa and bb, the custom operator Δ\Delta is defined by aΔb=a2b2aba \Delta b = \frac{a^2 - b^2}{ab}. A function f(x)f(x) is defined for all x0x \neq 0 and x1x \neq -1 by f(x)=xΔ(x+1)f(x) = x \Delta (x + 1). If kk is a positive constant such that f(k)=712f(k) = -\frac{7}{12}, what is the value of (k+2)Δk(k + 2) \Delta k?

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Cevap: 1615\frac{16}{15}

Cevap

The value of (k+2)Δk(k + 2) \Delta k is 1615\frac{16}{15}.
Applying the custom operator to f(x)f(x) yields f(x)=x2(x+1)2x(x+1)=2x1x2+xf(x) = \frac{x^2 - (x+1)^2}{x(x+1)} = \frac{-2x - 1}{x^2 + x}. Setting f(k)=712f(k) = -\frac{7}{12} gives 2k+1k2+k=712\frac{2k + 1}{k^2 + k} = \frac{7}{12}, which simplifies to 7k217k12=07k^2 - 17k - 12 = 0. Factoring (7k+4)(k3)=0(7k + 4)(k - 3) = 0 and selecting the positive root gives k=3k = 3. Evaluating (3+2)Δ3=5Δ3=523253=1615(3+2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \cdot 3} = \frac{16}{15}.

Adım Adım Çözüm

1
Express f(x)f(x) using the definition of the custom operator Δ\Delta.
f(x)=xΔ(x+1)=x2(x+1)2x(x+1)=x2(x2+2x+1)x(x+1)=2x1x2+xf(x) = x \Delta (x + 1) = \frac{x^2 - (x + 1)^2}{x(x + 1)} = \frac{x^2 - (x^2 + 2x + 1)}{x(x + 1)} = \frac{-2x - 1}{x^2 + x}
Substitute a=xa = x and b=x+1b = x + 1 into aΔb=a2b2aba \Delta b = \frac{a^2 - b^2}{ab}.
2
Set f(k)=712f(k) = -\frac{7}{12} and solve for the positive constant kk.
2k1k2+k=712    2k+1k2+k=712    12(2k+1)=7(k2+k)    7k217k12=0\frac{-2k - 1}{k^2 + k} = -\frac{7}{12} \implies \frac{2k + 1}{k^2 + k} = \frac{7}{12} \implies 12(2k + 1) = 7(k^2 + k) \implies 7k^2 - 17k - 12 = 0
Cross-multiply and rearrange into a standard quadratic equation.
3
Factor the quadratic equation to find the positive root kk.
(7k+4)(k3)=0    k=47(7k + 4)(k - 3) = 0 \implies k = -\frac{4}{7} or k=3k = 3. Since k>0k > 0, k=3k = 3.
The question specifies that kk is a positive constant.
4
Evaluate (k+2)Δk(k + 2) \Delta k for k=3k = 3.
(3+2)Δ3=5Δ3=52325×3=25915=1615(3 + 2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \times 3} = \frac{25 - 9}{15} = \frac{16}{15}
Substitute a=5a = 5 and b=3b = 3 into the custom operator definition.

Anahtar Kavram

Function definition substitution, non-commutative custom operators, and quadratic solving.
Tahmini Süre:2m 30s
ÖncekiSayfa 34 / 110Sonraki
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