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Zorluk: ZorRemainders and Units Digit Cyclicity

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

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
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