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Zorluk: OrtaDivisibility Rules and Remainder Theorem

In a basic encryption algorithm, a numerical value VV is generated using the formula V=4791193V = 47^{91} - 19^{3}. The final security key is determined by the positive remainder when VV is divided by 1212. What is the value of the final security key?

  1. A
    8
  2. B
    6
  3. C
    2
  4. 4Cevap

Cevap

4
The correct answer is found by applying modular arithmetic rules to reduce each part of the expression modulo 12. First, 471(mod12)47 \equiv -1 \pmod{12}, so 4791(1)91=147^{91} \equiv (-1)^{91} = -1. Second, 197(mod12)19 \equiv 7 \pmod{12}, and since 72=491(mod12)7^2 = 49 \equiv 1 \pmod{12}, we get 737×1=77^3 \equiv 7 \times 1 = 7. Subtracting these yields 17=8-1 - 7 = -8. To find the equivalent positive remainder, we add the divisor: 8+12=4-8 + 12 = 4.

Adım Adım Çözüm

1
Find the remainder of 479147^{91} modulo 12.
Since 47=12×4147 = 12 \times 4 - 1, it is easier to use 471(mod12)47 \equiv -1 \pmod{12}. Therefore, 4791(1)91=1(mod12)47^{91} \equiv (-1)^{91} = -1 \pmod{12}.
Using negative remainders for numbers close to a multiple of the divisor simplifies large exponent calculations.
2
Find the remainder of 19319^{3} modulo 12.
Since 19=12×1+719 = 12 \times 1 + 7, we have 197(mod12)19 \equiv 7 \pmod{12}. Thus, 19373(mod12)19^3 \equiv 7^3 \pmod{12}. We know 72=491(mod12)7^2 = 49 \equiv 1 \pmod{12}. Multiplying by 7 gives 731×7=7(mod12)7^3 \equiv 1 \times 7 = 7 \pmod{12}.
Reducing the base before exponentiation makes the calculation manageable without needing to calculate 19319^3 completely.
3
Subtract the individual remainders.
(1)7=8(mod12)(-1) - 7 = -8 \pmod{12}.
Applying modular arithmetic properties to combine the terms in the original expression.
4
Convert the negative remainder to a positive remainder.
8+12=4-8 + 12 = 4.
The standard remainder definition requires a non-negative integer strictly less than the divisor.

Anahtar Kavram

Modular arithmetic, negative remainders, and power reduction rules.
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