Number Properties and Arithmetic

232 soru

Soru 181Soru

A cloud data center processes an incoming raw data stream through sequential filtering stages. In Stage 1, 38\frac{3}{8} of the incoming raw data volume is discarded as noise, and 0.200.20 of the remaining data is flagged for long-term archiving. In Stage 2, 45\frac{4}{5} of the data not flagged for long-term archiving is processed into active storage, while the rest is discarded. If the volume of data processed into active storage in Stage 2 is 5050 terabytes greater than the volume of data discarded in Stage 1, what was the initial volume of the raw data stream, in terabytes?

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Cevap: 2,000

Cevap

2,000 terabytes
Let the initial data volume be XX. Discarded in Stage 1 is 38X\frac{3}{8}X, leaving 58X\frac{5}{8}X. Archiving takes 20%20\% (0.20=150.20 = \frac{1}{5}) of this remainder, which is 15×58X=18X\frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X. The data entering Stage 2 is 58X18X=12X\frac{5}{8}X - \frac{1}{8}X = \frac{1}{2}X. Active storage in Stage 2 is 45×12X=25X\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X. Subtracting Stage 1 discarded volume from Stage 2 active storage volume gives 25X38X=1640X1540X=140X\frac{2}{5}X - \frac{3}{8}X = \frac{16}{40}X - \frac{15}{40}X = \frac{1}{40}X. Setting 140X=50\frac{1}{40}X = 50 yields X=2,000X = 2,000 terabytes.

Adım Adım Çözüm

1
Define the variable and compute Stage 1 discarded volume and remaining volume.
Discarded in Stage 1 = 38X\frac{3}{8}X; Remaining after Stage 1 discard = X38X=58XX - \frac{3}{8}X = \frac{5}{8}X.
Establishing quantities in terms of the total initial volume XX allows setting up a single-variable linear equation.
2
Calculate the volume flagged for archiving and the volume available for Stage 2.
Archived volume = 0.20×58X=15×58X=18X0.20 \times \frac{5}{8}X = \frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X. Available for Stage 2 = \frac{5}{8}X - \frac{1}{8}X = \frac{4}{8}X = \frac{1}{2}X$.
The 0.200.20 decimal must be converted to a fraction (15\frac{1}{5}) and applied to the remaining 58X\frac{5}{8}X base.
3
Calculate the volume processed into active storage in Stage 2.
Active storage volume = 45×12X=25X\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X.
Stage 2 processes 45\frac{4}{5} of the data that entered Stage 2.
4
Set up the algebraic equation comparing active storage volume in Stage 2 to discarded volume in Stage 1.
25X38X=50\frac{2}{5}X - \frac{3}{8}X = 50.
The problem states that active storage volume in Stage 2 is 5050 terabytes greater than the Stage 1 discarded volume.
5
Find a common denominator and solve for XX.
(16401540)X=50    140X=50    X=2,000\left(\frac{16}{40} - \frac{15}{40}\right)X = 50 \implies \frac{1}{40}X = 50 \implies X = 2,000.
Converting fractions to a denominator of 4040 yields 140X=50\frac{1}{40}X = 50, so multiplying by 4040 gives X=2,000X = 2,000 terabytes.

Anahtar Kavram

Multi-step arithmetic operations involving sequential fractions, decimals, and percent bases
Tahmini Süre:2m 0s
Soru 182Soru

Let K=783442K = 7^{83} - 4^{42}. What is the remainder when K-K is divided by 55?

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Cevap: 3

Cevap

3
To find the remainder of K-K divided by 5, first calculate K(mod5)K \pmod 5. By analyzing powers of 7 modulo 5, 712,724,733,7417^1 \equiv 2, 7^2 \equiv 4, 7^3 \equiv 3, 7^4 \equiv 1, so 783=74(20)+33(mod5)7^{83} = 7^{4(20)+3} \equiv 3 \pmod 5. For powers of 4 modulo 5, 414,4214^1 \equiv 4, 4^2 \equiv 1, so 4421(mod5)4^{42} \equiv 1 \pmod 5. Therefore, K31=2(mod5)K \equiv 3 - 1 = 2 \pmod 5. Consequently, K2(mod5)-K \equiv -2 \pmod 5. Adding the divisor 5 gives a non-negative remainder of 3.

Adım Adım Çözüm

1
Find the remainder of 7837^{83} when divided by 55.
7127^1 \equiv 2, 7247^2 \equiv 4, 7337^3 \equiv 3, 741(mod5)7^4 \equiv 1 \pmod 5. Since 83=4×20+383 = 4 \times 20 + 3, 783733(mod5)7^{83} \equiv 7^3 \equiv 3 \pmod 5.
Powers of 7 follow a 4-term repeating remainder cycle modulo 5: (2, 4, 3, 1).
2
Find the remainder of 4424^{42} when divided by 55.
4144^1 \equiv 4, 421(mod5)4^2 \equiv 1 \pmod 5. Since 42 is even, 4421(mod5)4^{42} \equiv 1 \pmod 5.
Powers of 4 follow a 2-term repeating remainder cycle modulo 5: (4, 1).
3
Compute K(mod5)K \pmod 5.
K31=2(mod5)K \equiv 3 - 1 = 2 \pmod 5.
Modular subtraction allows evaluating terms independently.
4
Determine the positive remainder of K(mod5)-K \pmod 5.
K22+5=3(mod5)-K \equiv -2 \equiv -2 + 5 = 3 \pmod 5.
Standard division algorithm requires the remainder rr when dividing by 5 to satisfy 0r<50 \le r < 5.

Anahtar Kavram

Modular cyclicity and negative remainder conversion
Tahmini Süre:2m 0s
Soru 183Soru

If the sum of three consecutive even integers is 4242, what is the smallest of these integers?

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Cevap: 1212

Cevap

1212
The arithmetic mean (average) of three consecutive even integers is equal to the middle integer. Dividing the total sum 4242 by 33 yields 1414, which is the middle integer. Subtracting 22 gives the smallest integer, 1212.

Adım Adım Çözüm

1
Define the three consecutive even integers algebraically
Let the three consecutive even integers be nn, n+2n + 2, and n+4n + 4, where nn represents the smallest integer.
Consecutive even integers differ by 22.
2
Set up the equation for their sum
n+(n+2)+(n+4)=42    3n+6=42n + (n + 2) + (n + 4) = 42 \implies 3n + 6 = 42
The problem states that the sum of the three integers equals 4242.
3
Solve for nn
3n=36    n=123n = 36 \implies n = 12
Subtract 66 from both sides and divide by 33 to find the smallest integer.

Anahtar Kavram

Properties and Sums of Consecutive Even Integers
Tahmini Süre:45s
Soru 184Soru

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 185Soru

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 186Soru

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 187Soru

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 188Soru

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 189Soru

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 190Soru

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 191Soru

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 192Soru

If xx is a real number such that 92x+127x13x+23=243\sqrt[3]{\frac{9^{2x+1} \cdot 27^{x-1}}{3^{x+2}}} = 243, what is the value of xx?

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Cevap: 3

Cevap

The value of xx is 3.
Converting all terms inside the radical to powers of 3 yields 34x+233x33x+23=36x33=32x1\sqrt[3]{\frac{3^{4x+2} \cdot 3^{3x-3}}{3^{x+2}}} = \sqrt[3]{3^{6x-3}} = 3^{2x-1}. Setting 32x1=243=353^{2x-1} = 243 = 3^5 gives 2x1=52x - 1 = 5, which solves to x=3x = 3.

Adım Adım Çözüm

1
Express all exponential terms inside the radical using a common base of 3.
92x+1=34x+29^{2x+1} = 3^{4x+2} and 27x1=33x327^{x-1} = 3^{3x-3}.
Unifying bases allows the application of exponent multiplication and division rules.
2
Simplify the algebraic expression inside the cube root.
34x+233x33x+2=3(4x+2)+(3x3)(x+2)=36x3\frac{3^{4x+2} \cdot 3^{3x-3}}{3^{x+2}} = 3^{(4x+2) + (3x-3) - (x+2)} = 3^{6x-3}.
Adding exponents for multiplication of like bases and subtracting for division.
3
Simplify the radical expression using fractional exponent rules.
36x33=(36x3)1/3=32x1\sqrt[3]{3^{6x-3}} = (3^{6x-3})^{1/3} = 3^{2x-1}.
Taking the nn-th root of a power corresponds to dividing the exponent by nn.
4
Solve for xx by equating the exponent to the prime factorized form of 243.
32x1=35    2x1=5    x=33^{2x-1} = 3^5 \implies 2x - 1 = 5 \implies x = 3.
Since the bases are identical and non-zero, the exponents must be equal.

Anahtar Kavram

Exponent and radical rules with common base conversion
Soru 193Soru

An investment fund allocated its total initial capital among three venture portfolios: Healthcare, Clean Tech, and Artificial Intelligence. Initially, 25\frac{2}{5} of the capital was allocated to Healthcare and 0.350.35 was allocated to Clean Tech, with the remaining fraction allocated to Artificial Intelligence. During a mid-year portfolio rebalancing, the amount in Healthcare was increased by 25%25\%, the amount in Clean Tech was decreased by 20%20\%, and the amount in Artificial Intelligence remained unchanged. If the total capital across all three portfolios after rebalancing was $5,150,000\$5,150,000, what was the initial amount, in dollars, allocated to Artificial Intelligence?

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Cevap: $1,250,000

Cevap

The initial amount allocated to Artificial Intelligence was $1,250,000 dollars.
The correct option of 1,250,000isobtainedbyfirstcalculatingtheinitialproportionallocatedtoArtificialIntelligenceas1,250,000 is obtained by first calculating the initial proportion allocated to Artificial Intelligence as 1 - 0.40 - 0.35 = 0.25 .ApplyingthepercentageadjustmentsyieldsHealthcareat. Applying the percentage adjustments yields Healthcare at 0.50C ,CleanTechat, Clean Tech at 0.28C ,andArtificialIntelligenceat, and Artificial Intelligence at 0.25C .Thesumoftheseupdatedallocationsgives. The sum of these updated allocations gives 1.03C = \5,150,0005,150,000, which implies C=$5,000,000C = \$5,000,000. Taking 25%25\% of $5,000,000\$5,000,000 yields $1,250,000\$1,250,000.

Adım Adım Çözüm

1
Determine the initial decimal fraction allocated to each portfolio.
Healthcare = 25=0.40\frac{2}{5} = 0.40; Clean Tech = 0.350.35; Artificial Intelligence = 1(0.40+0.35)=0.251 - (0.40 + 0.35) = 0.25.
The sum of all initial portfolio fractions must equal 11.
2
Express the post-rebalancing portfolio values in terms of the initial total capital CC.
Healthcare = 0.40C×(1+0.25)=0.50C0.40C \times (1 + 0.25) = 0.50C; Clean Tech = 0.35C×(10.20)=0.28C0.35C \times (1 - 0.20) = 0.28C; Artificial Intelligence = 0.25C0.25C.
A 25%25\% increase scales a value by 1.251.25, and a 20%20\% decrease scales a value by 0.800.80.
3
Sum the new portfolio values to set up an equation for total capital CC.
0.50C+0.28C+0.25C=1.03C=5,150,0000.50C + 0.28C + 0.25C = 1.03C = 5,150,000.
The sum of all rebalanced portfolios equals the given final capital of $5,150,000\$5,150,000.
4
Solve for initial total capital CC and calculate the Artificial Intelligence allocation.
C=5,150,0001.03=5,000,000C = \frac{5,150,000}{1.03} = 5,000,000; Artificial Intelligence = 0.25×5,000,000=1,250,0000.25 \times 5,000,000 = 1,250,000.
Multiplying the initial total capital by 0.250.25 yields the original Artificial Intelligence allocation.

Anahtar Kavram

Combining sequential percentage changes and decimal/fraction conversions to solve multi-step financial allocation word problems.
Soru 194Soru

For how many positive integers nn is the expression n2+7n+120n+3\frac{n^2 + 7n + 120}{n + 3} equal to an integer?

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Cevap: 9

Cevap

9 positive integers
By dividing the numerator by the denominator, the expression simplifies to n+4+108n+3n + 4 + \frac{108}{n + 3}. For this to yield an integer, (n+3)(n + 3) must be a positive divisor of 108. The number 108 has prime factorization 22332^2 \cdot 3^3, giving it (2+1)(3+1)=12(2+1)(3+1) = 12 total positive divisors. Since nn must be a positive integer (n1n \ge 1), n+3n + 3 must be at least 4. Eliminating the divisors 1, 2, and 3 leaves 9 valid values for n+3n + 3, which correspond to 9 unique positive integers nn.

Adım Adım Çözüm

1
Perform polynomial long division or algebraic manipulation on the numerator
\frac{n^2 + 7n + 120}{n + 3} = \frac{(n^2 + 3n) + (4n + 12) + 108}{n + 3} = \frac{n(n+3) + 4(n+3) + 108}{n + 3} = n + 4 + \frac{108}{n + 3}
Separating the expression into an integer term (n+4)(n + 4) and a proper fractional term 108n+3\frac{108}{n + 3} isolates the divisibility condition.
2
Determine the condition for the fractional term to be an integer
The expression is an integer if and only if (n+3)(n + 3) is a positive integer divisor of 108.
Since nn is a positive integer, (n+4)(n + 4) is always an integer, so 108n+3\frac{108}{n + 3} must also be an integer.
3
Calculate the total number of positive divisors of 108
Prime factorization: 108=2233108 = 2^2 \cdot 3^3. Total divisors = (2+1)(3+1)=34=12(2+1)(3+1) = 3 \cdot 4 = 12.
The formula for the number of positive divisors of p1ap2bp_1^{a} p_2^{b} is (a+1)(b+1)(a+1)(b+1).
4
Apply the constraint that nn is a positive integer (n1n \ge 1)
Since n1n \ge 1, we have n+34n + 3 \ge 4. The divisors of 108 are {1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108}. The divisors less than 4 are {1, 2, 3} (3 divisors).
If n+3n + 3 were 1, 2, or 3, nn would be 2-2, 1-1, or 00, none of which are positive integers.
5
Subtract invalid divisors from total divisors
Valid divisors = 123=912 - 3 = 9.
Each divisor d4d \ge 4 yields exactly one unique positive integer n=d3n = d - 3.

Anahtar Kavram

Algebraic Divisibility and Prime Factorization Divisor Counting
Soru 195Soru

A cargo logistics warehouse handles three types of freight: Standard, Express, and Fragile. At the start of the week, 25\frac{2}{5} of the total shipments in the warehouse were Standard freight, 0.350.35 of the total shipments were Express freight, and the remaining shipments were Fragile freight. During a midweek inventory adjustment, the number of Standard freight shipments increased by 25%25\%, the number of Express freight shipments decreased by 20%20\%, and the number of Fragile freight shipments remained unchanged at 125125 units. What was the total number of shipments in the warehouse after the midweek inventory adjustment?

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Cevap: 515

Cevap

515 shipments
Converting 25\frac{2}{5} to a decimal gives 0.400.40. The initial proportion of Fragile freight is 10.400.35=0.251 - 0.40 - 0.35 = 0.25. Since Fragile freight equals 125125 shipments, the initial total number of shipments is 1250.25=500\frac{125}{0.25} = 500. Prior to adjustment, the warehouse has 0.40×500=2000.40 \times 500 = 200 Standard shipments and 0.35×500=1750.35 \times 500 = 175 Express shipments. After the adjustment, Standard shipments increase to 200×1.25=250200 \times 1.25 = 250, Express shipments decrease to 175×0.80=140175 \times 0.80 = 140, and Fragile shipments remain 125125. Summing these post-adjustment counts yields 250+140+125=515250 + 140 + 125 = 515 shipments.

Adım Adım Çözüm

1
Determine the initial fraction representing Fragile freight.
Fragile freight constitutes 0.250.25 (or 25%25\%) of the total initial shipments.
Converting 25\frac{2}{5} to a decimal yields 0.400.40. Subtracting the Standard (0.400.40) and Express (0.350.35) proportions from the total (1.001.00) leaves 1.000.400.35=0.251.00 - 0.40 - 0.35 = 0.25.
2
Calculate the total initial number of shipments in the warehouse.
Initial total shipments = 500500.
Since 25%25\% of the total initial shipments equals 125125 Fragile units, the initial total is 1250.25=500\frac{125}{0.25} = 500.
3
Calculate the initial quantities of Standard and Express shipments.
Initial Standard shipments = 200200, Initial Express shipments = 175175.
Standard shipments = 0.40×500=2000.40 \times 500 = 200, and Express shipments = 0.35×500=1750.35 \times 500 = 175.
4
Apply the inventory percentage changes to each category.
New Standard shipments = 250250, New Express shipments = 140140, Fragile shipments = 125125.
A 25%25\% increase in Standard shipments gives 200×1.25=250200 \times 1.25 = 250. A 20%20\% decrease in Express shipments gives 175×0.80=140175 \times 0.80 = 140. Fragile shipments remain 125125.
5
Sum the adjusted shipment counts to determine the net total.
Final total shipments = 515515.
Adding the three updated categories gives 250+140+125=515250 + 140 + 125 = 515.

Anahtar Kavram

Fractions, Decimals, and Percents Arithmetic
Tahmini Süre:1m 30s
Soru 196Soru

If k=233a7bk = 2^3 \cdot 3^a \cdot 7^b, where aa and bb are positive integers, and kk has exactly 36 positive integer divisors, how many positive integer divisors does 3k3k have?

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Cevap: 48

Cevap

48
The number of positive integer divisors of a prime-factored number p1e1p2e2pmemp_1^{e_1} p_2^{e_2} \cdots p_m^{e_m} is (e1+1)(e2+1)(em+1)(e_1+1)(e_2+1)\cdots(e_m+1). For k=233a7bk = 2^3 \cdot 3^a \cdot 7^b, the divisor count is 4(a+1)(b+1)=364(a+1)(b+1) = 36, which simplifies to (a+1)(b+1)=9(a+1)(b+1) = 9. Given a,b1a, b \ge 1, the only integer solution is a=2a = 2 and b=2b = 2. Multiplying kk by 3 yields 3k=2333723k = 2^3 \cdot 3^3 \cdot 7^2, which has (3+1)(3+1)(2+1)=48(3+1)(3+1)(2+1) = 48 positive divisors.

Adım Adım Çözüm

1
Use the formula for the number of positive divisors of an integer.
The number of positive divisors of k=233a7bk = 2^3 \cdot 3^a \cdot 7^b is given by (3+1)(a+1)(b+1)=4(a+1)(b+1)(3+1)(a+1)(b+1) = 4(a+1)(b+1).
If N=p1e1p2e2pmemN = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}, the total number of positive integer divisors is (e1+1)(e2+1)(em+1)(e_1 + 1)(e_2 + 1) \cdots (e_m + 1).
2
Set the divisor count equal to 36 and solve for aa and bb.
4(a+1)(b+1)=36    (a+1)(b+1)=94(a+1)(b+1) = 36 \implies (a+1)(b+1) = 9.
Dividing both sides by 4 isolates the term containing aa and bb.
3
Determine the values of aa and bb using the constraint that aa and bb are positive integers.
Since a1a \ge 1 and b1b \ge 1, we have a+12a+1 \ge 2 and b+12b+1 \ge 2. The only factor pair of 9 where both factors are at least 2 is 3×33 \times 3. Thus, a+1=3    a=2a+1 = 3 \implies a = 2 and b+1=3    b=2b+1 = 3 \implies b = 2.
Positive integers must be greater than or equal to 1.
4
Write the prime factorization of 3k3k and compute its number of positive divisors.
3k=3(233272)=2333723k = 3 \cdot (2^3 \cdot 3^2 \cdot 7^2) = 2^3 \cdot 3^3 \cdot 7^2. The number of divisors is (3+1)(3+1)(2+1)=443=48(3+1)(3+1)(2+1) = 4 \cdot 4 \cdot 3 = 48.
Multiplying kk by 3 increases the exponent of 3 from 2 to 3.

Anahtar Kavram

Prime Factorization and Total Divisors Rule
Tahmini Süre:1m 30s
Soru 197Soru

At an urban hydroponic farm, a monthly harvest consists of three types of crops: leafy greens, herbs, and microgreens. Leafy greens account for 58\frac{5}{8} of the total harvest by weight, and herbs account for 0.250.25 of the total harvest by weight. If the yield of microgreens increases by 40%40\% in the following month while the yields of leafy greens and herbs remain constant, what fraction of the new total harvest by weight will be microgreens?

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Cevap: 16\frac{1}{6}

Cevap

The fraction of the new total harvest that will be microgreens is 16\frac{1}{6}.
Initially, leafy greens and herbs account for 58+0.25=58+28=78\frac{5}{8} + 0.25 = \frac{5}{8} + \frac{2}{8} = \frac{7}{8} of the total harvest weight, leaving microgreens with 178=181 - \frac{7}{8} = \frac{1}{8} of the total weight. When microgreens increase by 40%40\%, their new yield becomes 18×1.40=740\frac{1}{8} \times 1.40 = \frac{7}{40} of the original total harvest. The total harvest weight increases to 78+740=3540+740=4240\frac{7}{8} + \frac{7}{40} = \frac{35}{40} + \frac{7}{40} = \frac{42}{40}. The fraction of microgreens in this new total harvest is 7/4042/40=742=16\frac{7/40}{42/40} = \frac{7}{42} = \frac{1}{6}.

Adım Adım Çözüm

1
Convert all initial crop portions into consistent fractions to find the initial microgreens fraction.
Leafy greens = 58\frac{5}{8}. Herbs = 0.25=14=280.25 = \frac{1}{4} = \frac{2}{8}. Microgreens fraction = 1(58+28)=178=181 - \left(\frac{5}{8} + \frac{2}{8}\right) = 1 - \frac{7}{8} = \frac{1}{8}.
The sum of all three crop fractions must equal 11 (the total harvest).
2
Calculate the updated relative weight of microgreens after a 40%40\% increase.
New microgreens weight relative to original total = 18×(1+0.40)=18×75=740\frac{1}{8} \times (1 + 0.40) = \frac{1}{8} \times \frac{7}{5} = \frac{7}{40}.
Increasing a quantity by 40%40\% is equivalent to multiplying by 1.401.40 or 75\frac{7}{5}.
3
Determine the new total harvest weight relative to the original harvest.
New total harvest = Leafy greens + Herbs + New microgreens = 58+28+740=3540+740=4240=2120\frac{5}{8} + \frac{2}{8} + \frac{7}{40} = \frac{35}{40} + \frac{7}{40} = \frac{42}{40} = \frac{21}{20}.
Since the yields of leafy greens and herbs did not change, their combined relative weight remains 3540\frac{35}{40}.
4
Compute the fraction of the new total harvest that consists of microgreens.
Fraction = 7404240=742=16\frac{\frac{7}{40}}{\frac{42}{40}} = \frac{7}{42} = \frac{1}{6}.
Dividing the new microgreens weight by the new total harvest weight gives the updated proportion.

Anahtar Kavram

Fractions, Decimals, and Percents Arithmetic
Tahmini Süre:1m 30s
Soru 198Soru

Set AA consists of kk consecutive integers, where k>1k > 1. Set BB consists of 2k2k consecutive integers. The median of Set AA is equal to the least integer in Set BB, and the sum of all integers in Set AA is equal to the sum of all integers in Set BB. If the sum of the integers in Set BB is 153-153, what is the greatest integer in Set AA?

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Cevap: 13-13

Cevap

The greatest integer in Set AA is 13-13.
The problem establishes a relationship between two sets of consecutive integers using their medians and sums. Equating the sum formulas SA=k(a+k12)S_A = k\left(a + \frac{k-1}{2}\right) and SB=2k(a+3k22)S_B = 2k\left(a + \frac{3k-2}{2}\right) leads to a=35k2a = \frac{3-5k}{2}. Substituting this into the sum expression gives k2k2=153k - 2k^2 = -153, yielding k=9k = 9. Thus, the smallest element is a=21a = -21, and the greatest element is 21+91=13-21 + 9 - 1 = -13.

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1
Express the sum of Set AA in terms of kk and its smallest element aa.
The median of Set AA is MA=a+k12M_A = a + \frac{k-1}{2}, so the sum of Set AA is SA=kMA=k(a+k12)S_A = k \cdot M_A = k\left(a + \frac{k-1}{2}\right).
In any set of consecutive integers, the sum equals the number of terms multiplied by the average (median).
2
Express the sum of Set BB using the given conditions.
The least integer in Set BB is MA=a+k12M_A = a + \frac{k-1}{2}. Since Set BB has 2k2k consecutive integers, its median is MB=(a+k12)+2k12=a+3k22M_B = \left(a + \frac{k-1}{2}\right) + \frac{2k-1}{2} = a + \frac{3k-2}{2}. The sum of Set BB is SB=2kMB=2k(a+3k22)=k(2a+3k2)S_B = 2k \cdot M_B = 2k\left(a + \frac{3k-2}{2}\right) = k(2a + 3k - 2).
Set BB starts at the median of Set AA and contains 2k2k terms.
3
Equate SAS_A and SBS_B to solve for aa in terms of kk.
Since SA=SBS_A = S_B and k>1k > 1, we divide by kk: a+k12=2a+3k2    a=35k2a + \frac{k-1}{2} = 2a + 3k - 2 \implies a = \frac{3 - 5k}{2}.
The problem states that the sums of the two sets are equal.
4
Substitute a=35k2a = \frac{3 - 5k}{2} into SBS_B and set it equal to 153-153.
SB=k(2(35k2)+3k2)=k(35k+3k2)=k(12k)=k2k2S_B = k\left(2\left(\frac{3-5k}{2}\right) + 3k - 2\right) = k(3 - 5k + 3k - 2) = k(1 - 2k) = k - 2k^2. Setting k2k2=153    2k2k153=0k - 2k^2 = -153 \implies 2k^2 - k - 153 = 0. Factoring gives (2k+17)(k9)=0(2k + 17)(k - 9) = 0. Since kk must be a positive integer greater than 1, k=9k = 9.
Finding kk allows us to calculate the exact elements of Set AA.
5
Calculate the greatest integer in Set AA.
With k=9k = 9, the smallest integer a=35(9)2=21a = \frac{3 - 5(9)}{2} = -21. The greatest integer in Set AA is a+k1=21+91=13a + k - 1 = -21 + 9 - 1 = -13.
The kk-th term of a sequence of consecutive integers starting at aa is a+k1a + k - 1.

Anahtar Kavram

Properties of consecutive integer sets: sum equals number of terms times median; indexing terms in evenly spaced sets.
Soru 199Soru

What is the remainder when the integer N=383+783N = 3^{83} + 7^{83} is divided by 100100?

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Cevap: 70

Cevap

The remainder when N=383+783N = 3^{83} + 7^{83} is divided by 100100 is 70.
Finding the remainder when an integer is divided by 100100 is equivalent to finding its last two digits. Powers of 33 repeat their last two digits in cycles of 2020, and 3201(mod100)3^{20} \equiv 1 \pmod{100}. Thus 38333=27(mod100)3^{83} \equiv 3^3 = 27 \pmod{100}. Powers of 77 repeat their last two digits in cycles of 44, with 74=24011(mod100)7^4 = 2401 \equiv 1 \pmod{100}. Thus 78373=34343(mod100)7^{83} \equiv 7^3 = 343 \equiv 43 \pmod{100}. Combining these gives 27+43=7027 + 43 = 70.

Adım Adım Çözüm

1
Find the remainder of 3833^{83} modulo 100100 by identifying the cyclicity pattern of powers of 33.
3201(mod100)3^{20} \equiv 1 \pmod{100}, which implies 383=(320)4331427=27(mod100)3^{83} = (3^{20})^4 \cdot 3^3 \equiv 1^4 \cdot 27 = 27 \pmod{100}.
Since 3201(mod100)3^{20} \equiv 1 \pmod{100}, exponent 8383 can be reduced to 83(mod20)=383 \pmod{20} = 3.
2
Find the remainder of 7837^{83} modulo 100100 by identifying the cyclicity pattern of powers of 77.
74=24011(mod100)7^4 = 2401 \equiv 1 \pmod{100}, which implies 783=(74)207312034343(mod100)7^{83} = (7^4)^{20} \cdot 7^3 \equiv 1^{20} \cdot 343 \equiv 43 \pmod{100}.
Since 741(mod100)7^4 \equiv 1 \pmod{100}, exponent 8383 can be reduced to 83(mod4)=383 \pmod{4} = 3.
3
Sum the modular remainders to obtain the final remainder of NN.
N=383+78327+43=70(mod100)N = 3^{83} + 7^{83} \equiv 27 + 43 = 70 \pmod{100}.
The remainder of a sum is equal to the sum of the individual remainders modulo 100100.

Anahtar Kavram

Modular Arithmetic and Cyclicity of Last Two Digits
Soru 200Soru

What is the smallest positive integer mm such that the remainder when 6m6^m is divided by 1111 is equal to 88?

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Cevap: 7

Cevap

The smallest positive integer mm is 7.
Evaluating powers of 6 modulo 11 sequentially gives: 6166^1 \equiv 6, 6236^2 \equiv 3, 6376^3 \equiv 7, 6496^4 \equiv 9, 65106^5 \equiv 10, 6656^6 \equiv 5, and 678(mod11)6^7 \equiv 8 \pmod{11}. Therefore, m=7m = 7 is the smallest positive integer satisfying the condition.

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1
Calculate the remainder of powers of 6 when divided by 11 starting from m=1m = 1.
616(mod11)6^1 \equiv 6 \pmod{11}, 623(mod11)6^2 \equiv 3 \pmod{11}, 637(mod11)6^3 \equiv 7 \pmod{11}, 649(mod11)6^4 \equiv 9 \pmod{11}, 6510(mod11)6^5 \equiv 10 \pmod{11}, 665(mod11)6^6 \equiv 5 \pmod{11}, 678(mod11)6^7 \equiv 8 \pmod{11}.
Modular multiplication allows computing powers step-by-step by multiplying the previous remainder by 6 and finding the remainder modulo 11.
2
Identify the smallest exponent mm that yields a remainder of 8.
m=7m = 7
At m=7m = 7, 67=279,936=11×25,448+86^7 = 279,936 = 11 \times 25,448 + 8, which leaves a remainder of 8.

Anahtar Kavram

Modular Arithmetic and Cyclicity of Powers
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Number Properties and Arithmetic Alıştırma Soruları — GMAT — Sayfa 10 | Examkin