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Zorluk: ZorRemainders and Units Digit Cyclicity

Let N=1943+1143−643N = 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: 19≡−2(mod7)19 \equiv -2 \pmod{7}, 11≡4(mod7)11 \equiv 4 \pmod{7}, and 6≡−1(mod7)6 \equiv -1 \pmod{7}. Utilizing the cyclicity period of 3 for powers of 2 modulo 7 (23≡1(mod7)2^3 \equiv 1 \pmod{7}), 1943≡−(243)≡−2≡5(mod7)19^{43} \equiv -(2^{43}) \equiv -2 \equiv 5 \pmod{7}. Similarly, 1143≡443=286=(23)28×4≡4(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 N≡5+4−(−1)=10≡3(mod7)N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}.

Adım Adım Çözüm

1
Reduce each base modulo 7
19≡−2(mod7)19 \equiv -2 \pmod{7}, 11≡4(mod7)11 \equiv 4 \pmod{7}, and 6≡−1(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
1943≡−2≡5(mod7)19^{43} \equiv -2 \equiv 5 \pmod{7}
Since 23≡1(mod7)2^3 \equiv 1 \pmod{7} and 43=3(14)+143 = 3(14) + 1, 243≡2(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
1143≡4(mod7)11^{43} \equiv 4 \pmod{7}
1143≡443=286=(23)28×22≡128×4=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}
643≡−1(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
N≡5+4−(−1)=10≡3(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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