Practice Exercise Solutions

What are 20%3, -20%3, 20%-3, and -20%-3? Verify your work by demonstrating x = (x/y)y + r where r is your answer.

20/3 = -20/-3 = 6 -20/3 = 20/-3 = -6

20 = (20/3)(3) + 2, so 20%3 = r = 2. -20 = (-20/3)(3) + -2, so -20%3 = r = -2. 20 = (20/-3)(-3) + 2, so 20%-3 = r = 2. -20 = (-20/-3)(-3) + -2, so -20%-3 = r = -2.

Use the division algorithm to find 20 mod 3, and -20 mod 3. Verify your work by demonstrating x = yq + r.

20 = (3)(6) + 2 and 0 <= 2 < 3, so 20 mod 3 = 2. -20 = (3)(-7) + 1 and 0 <= 1 < 3, so -20 mod 3 = 1.

Use the division algorithm to find 20 mod 4, and -20 mod 4. Verify your work by demonstrating x = yq + r.

20 = (4)(5) + 0 and 0 <= 0 < 4, so 20 mod 4 = 0. -20 = (4)(-5) + 0 and 0 <= 0 < 4, so -20 mod 4 = 0.

Calculate $3^{10} \bmod 10$ by using laws of exponents and the above theorem to mod intermediate results and simplify the computation. Verify your answer using WolframAlpha.

There is no right answer to this problem. Simply demonstrate that you understand that you can break the problem into smaller ones and then mod the smaller ones. So, 3¹⁰ mod 10 = (3)(3³)(3³)(3³) mod 10 = (3 mod 10)(3³ mod 10)(3³ mod 10)(3³ mod 10) mod 10 = (3)(7)(7)(7) mod 10 = = (3)(7² mod 10)(7) mod 10 = (3)(9)(7) mod 10 = 189 mod 10 = 9.

Typing "3^10 mod 10" at wolframalpha.com gives the same result.

Calculate $7^8 \bmod 10$ using the same method.

7⁸ mod 10 = (7² mod 10)⁴ mod 10 = (9⁴) mod 10 = (9² mod 10)² mod 10 = 1.

Typing "7^8 mod 10" at wolframalpha.com gives the same result.

Calculate $(3^{10} + 7^8) \bmod 10$.

(3¹⁰ + 7⁸) mod 10 = [(3¹⁰ mod 10) + (7⁸ mod 10)] mod 10 = (9 + 1) mod 10 = 0.

Typing "(3^10 + 7^8) mod 10" at wolframalpha.com gives the same result.

What are $\lfloor \pi \rfloor$, $\lfloor -\pi \rfloor$, $\lceil \pi \rceil$, $\lceil -\pi \rceil$?

3, -4, 4, -3.

Typing "ceiling(pi)", etc, at wolframalpha.com gives the same result.