A synchronized scheduling system operates on a repeating 19-millisecond cycle. A specific … — State PSC Exam

A synchronized scheduling system operates on a repeating 19-millisecond cycle. A specific event is triggered at a timestamp $T$ in milliseconds, given by the formula $T = 37^{45} - 17 \times 40^{23}$ . To find the exact position within the current cycle when the event occurs, the system calculates the positive remainder when $T$ is divided by $19$ . At what millisecond mark within the cycle does the event trigger?

Answer: 6 milliseconds

Answer

By applying modular arithmetic rules, Fermat's Little Theorem reduces the large exponents. The first term evaluates to

-1 \pmod{19}

and the second term evaluates to

12 \pmod{19}

. Their difference is

-13 \pmod{19}

, which corresponds to a positive remainder of

6

Step-by-Step Solution

Simplify the first term

37^{45}

modulo 19.

37^{45} \equiv -1 \pmod{19}

Since

37 = 19 \times 2 - 1

, it follows that

37 \equiv -1 \pmod{19}

. An odd power of

-1

-1

Simplify the base of the second term,

40^{23}

, modulo 19.

40^{23} \equiv 2^{23} \pmod{19}

Because

40 = 19 \times 2 + 2

, we can replace the base

40

with its remainder

2

Use Fermat's Little Theorem to reduce the exponent in

2^{23} \pmod{19}

2^{23} \equiv 13 \pmod{19}

Fermat's Little Theorem states

a^{p-1} \equiv 1 \pmod{p}

for a prime

p

. Here,

2^{18} \equiv 1 \pmod{19}

. Therefore,

2^{23} = 2^{18} \times 2^5 \equiv 1 \times 32 \equiv 13 \pmod{19}

Multiply by 17 and find the remainder of the second term.

17 \times 40^{23} \equiv 12 \pmod{19}

We can write

17 \equiv -2 \pmod{19}

. Then,

(-2) \times 13 = -26

. Adding a multiple of 19 (which is 38) gives

-26 + 38 = 12

Subtract the second term from the first and convert to a positive remainder.

The final remainder is

6

-1 - 12 = -13

. To get the positive remainder, add the modulus 19:

-13 + 19 = 6

Key Concept

Applying modular arithmetic rules, properties of negative remainders, and Fermat's Little Theorem to simplify large exponential expressions.

Estimated Time:1m 30s

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