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Zorluk: ZorModular Arithmetic

If xx is the smallest positive integer satisfying the modular congruence 2x7(mod11)2^x \equiv 7 \pmod{11}, what is the value of (3x24x+5)(mod11)(3x^2 - 4x + 5) \pmod{11} expressed in standard non-negative remainder form?

Cevap: 3

Cevap

The smallest positive integer exponent satisfying 2x7(mod11)2^x \equiv 7 \pmod{11} is x=7x = 7. Substituting x=7x = 7 into 3x24x+53x^2 - 4x + 5 yields 124124, which simplifies to 3(mod11)3 \pmod{11}.
Evaluating powers of 2 modulo 11 shows that 27=1287(mod11)2^7 = 128 \equiv 7 \pmod{11}, giving x=7x = 7. Substituting x=7x = 7 into 3x24x+53x^2 - 4x + 5 gives 124124, which leaves a remainder of 33 when divided by 1111.

Adım Adım Çözüm

1
Find the smallest positive integer exponent xx satisfying 2x7(mod11)2^x \equiv 7 \pmod{11}
x=7x = 7
Evaluating consecutive powers of 2 modulo 11 shows 2122^1 \equiv 2, 2242^2 \equiv 4, 2382^3 \equiv 8, 2452^4 \equiv 5, 25102^5 \equiv 10, 2692^6 \equiv 9, and 2772^7 \equiv 7, making x=7x = 7 the smallest positive integer power.
2
Substitute x=7x = 7 into the expression 3x24x+53x^2 - 4x + 5
124
Direct substitution gives 3(7)24(7)+5=3(49)28+5=14728+5=1243(7)^2 - 4(7) + 5 = 3(49) - 28 + 5 = 147 - 28 + 5 = 124.
3
Reduce 124 modulo 11 to standard non-negative remainder form
3
Dividing 124 by 11 yields a quotient of 11 with a remainder of 3 (124=11×11+3124 = 11 \times 11 + 3).

Anahtar Kavram

Modular Exponentiation and Algebraic Evaluation in Modular Arithmetic
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