Let N = 19^{43} + 11^{43} - 6^{43}. What is the remainder when N is divided by 7? — GMAT

Let $N = 19^{43} + 11^{43} - 6^{43}$ . What is the remainder when $N$ is divided by $7$ ?

Answer: 3

Reducing each base modulo 7 simplifies the expression significantly:

19 \equiv -2 \pmod{7}

11 \equiv 4 \pmod{7}

, and

6 \equiv -1 \pmod{7}

. Utilizing the cyclicity period of 3 for powers of 2 modulo 7 (

2^3 \equiv 1 \pmod{7}

19^{43} \equiv -(2^{43}) \equiv -2 \equiv 5 \pmod{7}

. Similarly,

11^{43} \equiv 4^{43} = 2^{86} = (2^3)^{28} \times 4 \equiv 4 \pmod{7}

. Lastly,

6^{43} \equiv (-1)^{43} = -1 \pmod{7}

. Combining these yields

N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}

Reduce each base modulo 7

19 \equiv -2 \pmod{7}

11 \equiv 4 \pmod{7}

, and

6 \equiv -1 \pmod{7}

Simplifying bases using modular arithmetic reduces large power evaluations to smaller equivalents.

Evaluate

19^{43} \pmod{7}

using cyclicity

19^{43} \equiv -2 \equiv 5 \pmod{7}

Since

2^3 \equiv 1 \pmod{7}

and

43 = 3(14) + 1

2^{43} \equiv 2 \pmod{7}

, so

(-2)^{43} = -(2^{43}) \equiv -2 \pmod{7}

Evaluate

11^{43} \pmod{7}

using powers of 2

11^{43} \equiv 4 \pmod{7}

11^{43} \equiv 4^{43} = 2^{86} = (2^3)^{28} \times 2^2 \equiv 1^{28} \times 4 = 4 \pmod{7}

Evaluate

6^{43} \pmod{7}

6^{43} \equiv -1 \pmod{7}

Any odd power of

-1

equals

-1

Combine all components to determine the final remainder

N \equiv 5 + 4 - (-1) = 10 \equiv 3 \pmod{7}

Summing the individual remainders modulo 7 gives a final positive remainder of 3.

Modular Arithmetic and Cyclicity of Powers

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