Let K = 7^{83} - 4^{42}. What is the remainder when -K is divided by 5? — GMAT

Let $K = 7^{83} - 4^{42}$ . What is the remainder when $-K$ is divided by $5$ ?

To find the remainder of

-K

divided by 5, first calculate

K \pmod 5

. By analyzing powers of 7 modulo 5,

7^1 \equiv 2, 7^2 \equiv 4, 7^3 \equiv 3, 7^4 \equiv 1

, so

7^{83} = 7^{4(20)+3} \equiv 3 \pmod 5

. For powers of 4 modulo 5,

4^1 \equiv 4, 4^2 \equiv 1

, so

4^{42} \equiv 1 \pmod 5

. Therefore,

K \equiv 3 - 1 = 2 \pmod 5

. Consequently,

-K \equiv -2 \pmod 5

. Adding the divisor 5 gives a non-negative remainder of 3.

Find the remainder of

7^{83}

when divided by

5

7^1 \equiv 2

7^2 \equiv 4

7^3 \equiv 3

7^4 \equiv 1 \pmod 5

. Since

83 = 4 \times 20 + 3

7^{83} \equiv 7^3 \equiv 3 \pmod 5

Powers of 7 follow a 4-term repeating remainder cycle modulo 5: (2, 4, 3, 1).

Find the remainder of

4^{42}

when divided by

5

4^1 \equiv 4

4^2 \equiv 1 \pmod 5

. Since 42 is even,

4^{42} \equiv 1 \pmod 5

Powers of 4 follow a 2-term repeating remainder cycle modulo 5: (4, 1).

Compute

K \pmod 5

K \equiv 3 - 1 = 2 \pmod 5

Modular subtraction allows evaluating terms independently.

Determine the positive remainder of

-K \pmod 5

-K \equiv -2 \equiv -2 + 5 = 3 \pmod 5

Standard division algorithm requires the remainder

r

when dividing by 5 to satisfy

0 \le r < 5

Modular cyclicity and negative remainder conversion

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