Problem Set 2

PROBLEMS

1.) Find a linear function f(x) such that f(1) = 42 and f(2) = 47.

2.) Solve for x: 4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 170.

3.) Prove that the product of 3 consecutive numbers is always divisible by 6.

4.) Prove that if p is prime, a and b are integers, and a \equiv b\mod p, then a^p \equiv b^p \mod p.

SOLUTIONS AND PROOFS

Post Date: October 20, 2009

1. Solution: This is just the same as saying, find the equation of the line passing through (1,42) and (2,1337). So, by point slope formula, we have, y - y_1 = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\Rightarrow y - 42 = \displaystyle\frac{1337 - 42}{2 - 1}(x - 1). \Rightarrow y = f(x) = -18x + 92.

2.) Solution:4^{x+1} + 4^{x+2} +4^{x+3} +4^{x+4} = 4^x(1 + 4 +4^2 +4^3) = 170 \Rightarrow 4^x(85) = 170\Rightarrow 4^x = 2 \Rightarrow 2^{2x}=2^1 \Rightarrow x = \displaystyle\frac{1}{2}.

3.) Proof: A number is divisible by 6 if it is divisible by 2 and 3. A product of 3 consecutive numbers is divisible by 2 because at least one of them is even, so it remains to show it is divisible by 3.

If a number is divided by 3, its possible remainders are 0, 1, and 2.  Assume n, n +1 and n+2 be the three consecutive numbers, and r be the remainder if n is divided by 3.

Case 1: If r=0, we are done.

Case 2: If r = 1, then n + 2 \Rightarrow r=0

Case 3: If r = 2, then n + 1 \Rightarrow r = 0.

Since the product of the three consecutive numbers is even, and for each case of r, one of the consecutive numbers is divisible by 3, the product of three consecutive numbers is divisible by 6. \blacksquare

4.) Proof: From definition, a^p \equiv b^p \mod p \Leftrightarrow b = a + kp for some k \in \mathbb{Z}.

Raising both sides of the equation to p, we have b^p = (a + kp)^p. By the binomial theorem,  b^p = (a + kp)^p = a^p + \displaystyle {p \choose 1}a^{p-1}kp + \displaystyle{p \choose 2}a^{p-2}k^2p^2 + \ldots + k^pp^p.

Notice that every term aside from a^p is divisible by p^2. (Why?). Therefore,  a^p \equiv 0 \mod p^2 .

Hence, then a^p \equiv b^p \mod p. \blacksquare

Related Posts Plugin for WordPress, Blogger...

Leave a Reply